9 ?3a CORNELL UNIVERSITY LIBRARIES Mathematica Ubrary White Hall CORNELL UNIVERSITY LIBRARY 3 1924 050 927 189 Date Due -A^errft^o ■ ^J [• ^^^^ ft?IHH-^ 'Jtft-t-iHtt) 11 2004 Due Fi ack Upon Recall or Leaving The University Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/cletails/cu31924050927189 INTEODUCTION TO QUATEENIONS, WITH NUMEROUS EXAMPLES. CCamlinlige; PRINTED BY C. J. CLAT. M.A- AT THE UNITERSITY PBESS. INTKODUCTION %>♦,, TO QUATERNIONS, WITH NUMEE0U3 EXAMPLES'. BY P. KELLAND, MA., F.R.S., tOBMEBLT FELLOW OF QUEENS' COLLEGE, CAMBEIDGE ; AND P. G. TAIT, M.A., FORMERLY FELLOW OF ST PETER'S COLLEGE, CAMBRIDGE ; PBOPBSSOES IN THE DEPARTMENT OF MATHEMATICS IN THE UNIVERSITY OP EDINBURGH. Hon&on : MACMILLAN AND CO. • 1873- \A.ll Eights reserved,] ?. .\^6 In compliance with current copyright law. LBS Archival Products produced this replacement volume on paper that meets the ANSI Standard Z39.48-1984 to replace the irreparably deteriorated original. 1988 PEEFACE. The present Treatise is, as the title-page indicates, the joint production of Prof. Tait and myself. The preface I write in the first person, as this enables me to offer some personal explanations. For many years past I have been accustomed, no doubt very imperfectly, to introduce to my class the subject of Quaternions as part of elementary Algebra, more with the view of establishing principles than of applying processes. Experience has taught me that to induce a student to think for himself there is nothing so effectual as to lay before him the different stages of the development of a science in some- thing like the historical order. And justice alike to the stu- dent and the subject forbade that I should stop short at that jpoint where, more simply and more effectually than at any other, the intimate connexion between principles and pro- cesses is made manifest. Moreover in lecturing on the ground- work on which the mathematical sciences- are based, I could not but bring before my class the names of great men who spoke in other tongues and belonged to other nationalities than their own — Diophantus, Des Cartes, Lagrange, for in- stance — and it was not just to omit the name of one as VI PREFACE. great as any of them, Sir William Rowan Hamilton, who spoke their own tongue and claimed their own nationality. It is true the name of Hamilton has not had the impress of time to stamp it with the seal of immortality. And it must be admitted that a cautious policy which forbids to wander from the beaten paths, and encourages converse with the past rather than interference with the present, is the true policy of a teacher. But in the case before us, quite iiTespective of the nationality of the inventor, there is ample ground for introducing this subject of Quaternions into an elementary course of mathematics. It belongs to first principles and is their crowning and completion. It brings those principles face to face with operations, and thus not only satisfies the student of the mutual dependence of the two, but tends to carry him back to a clear apprehension of what he had probably failed to appreciate in the sub- ordinate sciences. Besides, there is no branch of mathematics in which results of such wide variety are deduced by one uniform process; there is no territory like this to be attacked and subjugated by a single weapon. And what is of the utmost importance in an educational point of view, the reader of this subject does not require to eocumber his memory with a host of conclusions already arrived at in order to advance. Every problem is more or less self- contained. This isi my apology for the present treatise. The work is, as I have sa.id, the joint production of Prof. Tait and myself. The preface I have written ,T7ithout consulting, my colleague^ as. I am. thus CQableti PREFACE. VU to say what could not otherwise have been said, that mathematicians owe a lasting debt of gratitude to Prof. Tait for the singleness of purpose and the self-denying zeal with which he has worked out the designs of his friend Sir Wm. Hamilton, preferring always the claims of the science and of its founder to the assertion of his own power and originality in its development. For my own part I must confess that my knowledge of Quaternions is due exclusively to him. The first work of Sir Wm. Hamilton, Lectures on Quaternions, was very dimly and im- perfectly understood by me and I dare say by others, until Prof. Tait published his papers on the subject in the Messenger of Mathematics. Then, and not till then, did the science in all its simplicity develope itself to me. Sub-- sequently Prof. Tait has published a work of great value and originality. An Elementary Treatise on Quaternions. The literature of the subject is completed in all but ■what relates to its physical applications, when I mention in addition Hamilton's second great work, Elements of Quater- nions, a posthumous work so far as publication is concerned, but one of which the sheets had been corrected by the author, and which bears all the impress of his genius. But it is far from elementary, whatever its title may seem tO' imply; nor is the work of Prof Tait altogether free from difficulties. Hamilton and Tait write for mathematicians, and they do well, but the time has come when it behoves some one to write for those who desire to become mathe- maticians. Friends and pupils have urged me to undertake this duty, and after consultation with Prof. Tait, who from VIU PREFACE, being my pupil in youth is my teacher in riper years, I have, in conjunction with him, and drawing unreservedly from his writings, endeavoured in the first nine chapters of this treatise to illustrate and enforce the principles of this beautiful science. The last chapter, which may be regarded as an introduction to the application of Quater- nions to the region beyond that of pure geometry, is due to Prof Tait alone. Sir W. Hamilton, on nearly the last completed page of his last work, indicated Prof Tait as eminently fitted to carry on happily and usefully the appli- cations, mathematical and physical, of Quaternions, and as likely to become in the science one of the chief successors of its inventor. With how great justice, the reader of this chapter and of Prof Tait's other writings on the subject will judge. PHILIP KELLAND. UNIVEIiSITY OF EDINBnKGH, October, 1873. CONTENTS. .CHAPTEB I., PAGES Intbodcoiobt 1 — 5 CHAPTEE II. Vectob Addition and Scbteactiok 6 — 31 Definition of a vector, with conclusions immediately resulting therefrom, Art. 1 — 6 ; examples, 7 ; definition of unit tector and TENSOR, ■with examples, 8 ; coplanarity of three doinitial vectors, with conditions requisite for their terminating in a straight line, and examples, 9—13 ; mean point, 14. Additional Examples to Chapter II. CHAPTEB III. Vector Multiplication and Division . . 32 — 57 Definition of multiplication, and first principles, Art. 15 — 18; fundamental theorems of multiplication, 19 — 22 ; examples, 23 ; definitions of division, versor and quaternion, 24 — 28 ; examples, 29 ; conjugate quaternions, 30 ; interpretation of formulae, 31. Additional Examples to Ceapteb III. X CONTENTS. CHAPTER IV. PAGES The Stbaight Line and Plane 68 — 71 Equations of a straight line and plane, 32, 33 ; modifications and results— length of perpendicular on a plane — condition that four points shall lie in the same plane, &e. 34 ; examples, 35. Additional Examples to Chapieb IV. CHAPTEB V. The Ciroi^ and Sphebb ........ 72—89 ' 'Equations of the circle, with examplefi, 36, 37 ; tangent to circle and chord of contact, 38, 39 ; examples, 40 ; equations of the sphere with examples, 41, 42. Additional Examples to Chapieb V. CHAPTEB VI. The Ellipse 90—104 Equations of the ellipse, 43 ; properties of p, 44 ; equation of tangent, 45 ; Cartesian equations, 46 ; 0~'p, ^/), &c. 47 ; properties of the ellipse with examples, 48 — 50. Additional Examples to Chapieb VI. CHAPTER VII. The Paeaeola and Htpebbola ,.....< 105—126 Equation of the parabola in terms of p with examples, 52 — 54 ; ; equations of the parabola, ellipse and hyperbola in a form corre- sponding to those with Cartesian co-ordinates, with examples, 55. Additional Examples to Chapieb VIL CONTENTS. XI CHAPTEB Vin. PAGES Central Sdkfaoes op the Second Okder 127 — 152 Equation of the ellipsoid, 56 ; tangent plane and perpendicular on it, 57, 58; polar plane, 59, 60; conjugate diameters and diame- tral planes, with examples, 60 — 64 ; the cone, 65, 66 ; examples on central surfaces, 67 ; Pascal's hexagram, 68. Additional Examples to Chapteb YIII. CHAPTEE IX. FOEMUUE AND THEIE APPLICATION 153 — 179 Formulse, 69, 70 ; examples, 71. Additional Examples to Chaptee IX. CHAPTEE X. Vector Equations op the First Deoreb 180 208 Appendix 209—227 INTRODUCTION TO QUATERNIONS. CHAPTER I. INTRODUCTORY. The science named Quaternions by its illustrious founder. Sir William Rowan Hamilton, is the last and the most beautiful ex- ample of extension by the removal of limitations. The Algebraic sciences are based on ordinary arithmetic, start- ing at first with all its restrictions, but gradually freeing themselves from one and another, until the parent science scarce recognises itself in its offspring. A student will best get an idea of the thing by considering one case of extension within the science of Arith- metic itself. There are two distinct bases of operation in that science — addition and multiplication. In the infancy of the science the latter was a mere repetition of the former. Multiplication was, ^in fact, an abbreviated form of equal additions. It is in this form that it occurs in the earliest writer on arithmetic whose works have come down to us — Euclid. Within the limits to which his prin- ciples extended, the reasonings and conclusions of Euclid in his seventh and following Books are absolutely perfect. The demon- stration of the rule for finding the greatest common measure of two numbers in Prop. 2, Book VII. is identically the same as that which is given in all modern treatises. But Euclid dares not venture on fractions. Their properties were probably all but un- known to him. Accordingly we look in vain for any demonstration of the properties of fractions in the writings of the Greek arith- meticians. For that we must come lower down. On the revival T. ^.'' 1 2 QUATERNIONS. [CHAP, of scieDce in the West, we are presented with categorical treatises on arithmetic. The first printed treatise is that of Lucas de Burgo in 1494. The author considers a fraction to be a quotient, and thus, as he expres.sly states, the order of operations becomes the reverse of that for whole numbers — multiplication precedes addi- tion, etc. In our own country we have a tolerably early writer ou arithmetic, Robert Record, who dedicated his work to King Edward the Sixth. The ingenious author exhibits his treatise in the form of a dialogue between master and scholar. The scholar battles long with this difficulty — rthat multiplying a thing should make it less. At first, the master attempts to explain the anomaly by reference to proportion, thus : that the product by a fraction bears the same proportion to the thing multiplied that the multiplying fraction does to unity. The scholar is not satisfied ; and accord- ingly the master goes on to say : " If I multiply by more than one, the thing is increased ; if I take it but once, it is not changed ; and if I take it less than once, it cannot be so much as it was before. Then, seeing that a fraction is less than one, if I multiply by a fraction, it follows that I do take it less than once," etc. The scholar thereupon replies, " Sir, I do thank you much for this reason ; and I trust that I do peiceive the thing." Need we add that the same difficulty which the scholar in the time of King Edward experienced, is experienced by every thinking boy of our own times; and the explanation afibrded him is precisely the same admixture of multiplication, proportion, and division which suggested itself to old Robert Record. Every schoolboy feels that to multiply by a fraction is not to multiply at all in the sense in which multiplication was originally presented to him, viz. as an abbreviation of equal additions, or of repetitions of the thing multi- plied. A totally new view of the process of multiplication has insensibly crept in by the advance from whole numbers to fractions. So new, so difierent is it, that we are satisfied Euclid in his logical and unbending march could never have attained to it. It is only by standing loose for a time to logical accuracy that extensions in the abstract sciences— extensions at any rate which stretch from one science to another— are eflTected. Thus Diophantus in his I-J INTRODUCTORY. 3 Treatise on Arithmetic (i. e. Arithmetic extended to Algebra) boldly lays it down as a definition or first principle of his science that "minus into minus makes plus." The science he is founding ia subject to this condition, and the results must be interpreted consistently with it. So far as this condition does not belong to ordinary arithmetic, so far the science extends beyond ordinary arithmetic : and this is the distance to which it extends — It makes subtraction to stand by itself, apart from addition ; or, at any rate, not dependent on it. We trust, then, it begins to be seen that sciences are extended by the removal of barriers, of limitations, of conditions, on which sometimes their very existence appears to depend. Fractional arithmetic was an impossibility so long as multiplication was re- garded as abbreviated addition ; the moment an extended idea was entertained, ever so illogically, that moment fractional arithmetic started into existence. Algebra, except as mere symbolized arith- metic, was an impossibility so long as the thought of subtraction was chained to the requirement of something adequate to subtract from. The moment Diophantus gave it a separate existence — boldly and logically as it happened — by exhibiting the law of minus in the forefront as the primary definition of his science, that moment algebra in its highest form became a possibility ; and indeed the foundation-stone was no sooner laid than a goodly building arose on it. The examples we have given, perhaps from their very simplicity, escape notice, but they are not less really examples of extension from science to science by the removal of a restriction. We have selected them in preference to the more familiar one of the extension of the meaning of an index, whereby it becomes a logarithm, because they prepare the way for a further extension in the same direction to which we are presently to advance. Observe, then, that in frac- tions and in the rule of signs, addition (or subtraction) is very slenderly connected with multiplication (or division). Arithmetic as Euclid left it stands on one support, addition only, inasmuch as with him multiplication is but abbreviated addition. Arithmetic in its extended form rests on two supports, addition and multipliear 1—2 4 . QUATERNIONS. [CHAP, tion, tlie one different froni the other. This is the first idea we want our reader to get a firm hold of; that multiplication is not necessarily addition, but an operation self-contained, self-interpret- able — springing originally out of addition ; but, when full-growu, existing apart from its parent. The second idea we want our reader to fix his mind on is this, .that when a science has been extended into a new form, certain limitations, which appeared to be of the nature of essential truths in the old science, are found to be utterly untenable ; that it is, in fact, by throwing these limitations aside that room is made for the growth of the new science. We have instanced AJgebra as a growth out of Arithmetic by the removal of the restriction that subtraction shall require something to subtract from. The word " subtraction" may indeed be inapproiwiate, as the word multiplication ap- peared to be to Record's scholar, who failed to see how the multi- plication of a thing could make it less. In the advance of the sciences the old terminology often becomes inappropriate ; but if the mind can extract the right idea from the sound or sight of 'a word, it is the part of wisdom to retain it. And so all the old words have been retained in the science of Quaternions to which we are now to advance. The fundamental idea on which the science is based is that of motion — of transference. Real motion is indeed not needed, any more than real superposition is needed in Euclid's Geometry. An appeal is made to mental transference in the one science, to mental superposition in the other. We are then to consider how it is possible to frame a new science which shall spring out of Arithmetic, Algebra, and Geometry, and shall add to them the idea of motion — of transference. It must be confessed the project we entertain is not a project due to the nineteenth century. The Geometry of Des Cartes was based on something very much resembling the idea of motion, and so far the mere introduction of the idea of transference was not of much value. The real advance was due to the thought of severing multiplication from addition, so that the one might be the representative of a kind of motion absolutely different from that which was represented by I.] INTRODUCTORY. 5 the other, yet capable of being combined with it. What the nine- teenth century has done, then, is to divorce addition from multipli- cation in the new form in which the two are presented, and to cause the one, in this new character, to signify motion forwards and backwards, the other motion round and round. "We do not purpose to give a history of the science, and shall accordingly content ourselves with saying, that the notion of sepa- rating addition from multiplication — attributing to the one, motion from a point, to the other motion about a point — had been floating in the minds of mathematicians for half a century, without producing many results worth recording, when the subject fell into the hands of a giant. Sir William Rowan Hamilton, who early found that his road was obstructed — he knew not by what obstacle — so that many points which seemed within his reach were really inaccessible. He had done a considerable amount of good work, obstructed as he was, when, about the year 1843, he perceived clearly the obstruction to his progress in the shape of an old law which, prior to that time, had appeared like a law of common sense. The law in question is known as the commutative law of multiplication. Presented in its simplest form it is nothing more than this, "five times three is the same as three times five;" more generally, it appears under the form of " ah = ha whatever a and h may represent." When it came distinctly into the mind of Hamilton that this law is not a necessity, with the extended signification of multiplication, he saw his way clear, and gave up the law. The barrier being removed, he entered on the new science as a warrior enters a besieged city through a practicable breach. The reader will find it easy to enter after him. CHAPTER II. VECTOE ADDITION AND SUBTRACTION. 1. Definition of a Vector. A vector is the representative of transference through, a given distance, in a given direction. Thus if AB be a straight line, the idea to be attached to " vector AB " is that of transference from A to B. . For the sake of definiteness we shall frequently abbreviate the phrase " vector AB " by a Greek letter, retaining in the meantitue (with one exception to be noted in the next chapter) the English letters to denote ordinary numerical quantities. If we now sitart from B and advance to C in the same direction, BC being equal to AB, we may, as in ordinary geometry, designate " vector BC " by the same symbol, which we adopted to designate " vector AB." Further, if we start from any other point in space, and advance from that point by the distance OX equal to and in the same direction as AB, we are at liberty to designate " vector OX " by the same symbol as that which represents AB. Other circumstances will determine the starting point, and iu- dividvialize the line to which a specific vector corresponds. Our definition is therefore subject to the following condition : — All lilies which are equal and draicn in the same direction are represented by the same vector symbol. We have purposely employed the phrase " drawn in the same direction" instead of "parallel," because we wish to guard the student against confounding " vector AB " with " vector BA." ART. 2.] VECTOR ADDITION AND SUBTRACTION, 7 2. In order to apply algebra to geometry, it is necessary to impose on geometry the condition that when a line measured in one direction is represented by a positive symbol, the same line measured in the opposite direction must be represented by the cor- responding negative s3'mbol. lu the science before us the same condition is equally requisite, and indeed the reason for it is even more manifest. Por if a transference from A to ^ be represented by + a, the transference which neutralizes this, and brings us back again to A, cannot be conceived to be represented by anything but — a, provided the symbols + and — are to retain any of their old algebraic meaning. The vector AB, then, being represented by + a, the vector BA will be represented by — a. 3. Further it is abundantly evident that so far as addition and subtraction of parallel vectors are concerned, all the laws of Algebra must be applicable. Thus (in Art. 1) AB + BC or a + a produces the same result at AG which is twice as great as AB, SnA. is there- fore properly represented by 2a ; and so on for all the rest. The distributive law of addition may then be assumed to hold in all its integrity so long at least as we deal with vectors which are parallel to one another. In fact there is no reason whatever, so far, why a should not be treated in every respect as if it were an ordinary algebraic quantity. It need scarcely be added that vectors in the same direction have the same proportion as the lines which corre- spond to them. "We have then advanced to the following — Lemma. All lines drawn in the same direction are, as vectors, to he represented hy numerical multiples of one and the same symbol, to which tlie ordinary laws of Algebra, so far as tJieir addi- tion, subtraction, and numerical multiplication are concerned, may he unreservedly applied. % 4. The converse is of course true, that if lines as vectors are represented by multiples of the same vector symbol, they ar^^ parallel. 8 QUATERNIONS. ' [CHAP. 11. It is only necessary to add to what Las preceded, tbat if BC be a line not in the same direction with p AB, then the vector BO cannot be represented by a or by any multiple of a. The vector symbol a must A B be limited to express transference in a certain direction, and can- not, at the same time, express transference in any other direction. To express "vector BC" then, another and quite independent symbol fi must be introduced. This symbol, being united to a by the signs + and — , the laws of algebra will, of course, apply to the combination. 5, If we now join AC, and thus form a triangle ABC, and if we denote vector AB by a, BC by y3, AC by 7, it is clear that we shall be presented with the equation a + ;8 = y. This equation appears at first sight to be a violation of Euclid I. 20 : " Any two sides of a triangle are together greater than the third side." But it is not really so. The anomalous ajjpearance arises from the fact that whilst we have extended the meaning of the symbol + beyond its arithmetical signification, we have said nothing about that of a symbol = . It is clearly necessary that the signification of this symbol shall be extended along with that of the other. It must now be held to designate, as it does perpetually in algebra, " equivalent to." This being premised, the equation above is freed from its anomalous appearance, and is perfectly con- sistent with everything in ordinary geometry. Expressed in words it reads thus : "A transference from A to B followed by a trans- ference from .B to is equivalent to a transference from A to C." 6. Axiom. If two vectors have not the same direction, it is impossible that the one can neutralize the other. This is quite obvious, for when a transference has been efiected from A to B, it is impossible to conceive that any amount of trans- ference whatever along BC can bring the moving point back to A. It follows as a consequence of this axiom, that if a, ^ be different actual vectors, i.e. finite vectors not in the same direction, and if ART. 7.] VECTOR ADDITION AND SUBTRACTION. 9 ma+np = 0, where m and n are numerical quantities ;" then must m = and n = 0. Another form of this consequence may be thus stated. If ma. + 71^= pa + qp, then must m=p, and n = q. 7, We now proceed to exemplify the principles so far as they have hitherto been laid down. It is scarcely necessary to remind the reader that we are assuming the applicability of all the rules of algebra and arithmetic, so far as we are yet in a position to draw on them ; and consequently that our demoustrations of certain of Euclid's elementary propositions must be accepted subject to this assumption. To avoid prolixity, we shall very frequently drop the word vector, at least in cases where, either from the introduction of a Greek letter as its representative, or from obvious considerations, it must be clear that the mere line is not meant. The reader -n-ill not fail to notice that the method of demonstration consists mainly in reach- ing the same point by two different routes. (See remark on Ex. 9.) Examples. Ex. 1. Tlie straight lines wldchjoin tlie extremities of equal and parallel straight lines towards the same parts are themselves equal and parallel. Let ABhe equal and parallel to CD ; to prove that .4(7 is equal and parallel to.gZ>. Let vector AB be represented by a, then (Art. 1) vector CD is also repre- ^ sented by a. If now vector CA be represented by yS, vector DB by y, we shall have (Art. 5) vector CB=GA + AB--= p + a, and vector CB = CD + DB = a + y ; . •. /3 + a = a + y, and P = y; so that /? and y are the same vector symbol j consequently (Art 1) 10 QUATERNIONS. [CHAP. II. the lines which they represent are equal and parallel; i.e. CA is equal and parallel to £D. Ex. 2. The opposite sides of a parallelogram are equal ; and the diagonals bisect each other. Since AB is parallel to CD, if vector AB be represented by a, vector CD will be represented by some numerical multiple of a (Art. 3), call it ma. And since GA is pai-allel to BB ; if vector CA be /8, then vector DB is wyS ; hence TBctor CB=CA+AB = fi + a, and =CD + DB = ma + n^; .: a + /3 = ma + w/3. Hence (Art. 6) m = 1, n = 1, i.e. the opposite sides of the paral- lelogram are equal. Again, as vectors, AO + OB = AB = CD = C0 + OD; And as AO is a vector along OD, and CO a vector along OB ; it follows (Art. 6) that vector AO is vector OD, and vector CO is OB; .: line AO=OD, CO = OB. Ex. 3. The sides about the equal angles of equiangular triangles are proportionals. Let the triangles ABC, ADE have a commoa angle A, then, because the angles D and B are equal, DE ia parallel to BC. Let vector AD be represented by a, DE by )3, then (Art. 3) AB is ma, BC n/S. .-. as vectors, AE = AD + DE = a + /3, AC = AB + BC = ma + np. ' Now AC is a multiple of AE, call it p (a + fi). .-. ma + nl3=p{a + P), atid m — p — n (Art. 6). EX. 4.] VECTOR ADDITION AND SUBTRACTION. 11 But line ^5 : AD = m, line 5(7 : DE = n, .\ A£ : AD :: BC : BE. Ex. 4. The hheeiors of the sides of a triangle meet in a point which trisects each of them. Let the sides of the triangle ABC be bisected in D, E, F; and let AD, BE meet in G. Let vector BD or DC be a, CE or EA {3, ^^ then, as vectors, BA=BG + GA = 2a+2p = 2{a. + p), B' DE = DC+CE = a + p, hence (Art. 4) BA is parallel to DE, and equal to 2DE. Again, BG + GA= BA = 2DE = 2{DG+ GE). Now vector BG is along GE, and vector GA along DG. .: (Art. 6) BG = 2GE, GA = 2GD, whence the same is true of the lines. Lastly, BG=\BE = UB0+GE) o = |(2a + /3); GG^BG-BG = |(2a + /3)-2a :(i3-a), 12 - QUATEBNIONS. [CHAP. II. GF=BF-BG J^BA-BG, = « + i8-|(2a + ^) lience CG is in the same straight line with GF, and equal to 2GF. Ex. 5. When, instead of D and E being the middle points of the sides, they are any points whatever in those sides, it is required to find G and the point in which GG produced meets AB. BG GA Let -=~^ = m, -=,=«; also let vector DC = a, vector CE= fi ; .: BC = ma, CA=nl3. Hence BF= BO + GF = ma+ p, I)A = a + np. Let BG = xBE, GA = yDA, then BA = BG + GA = x (ma + /?) + y (a + nfi). But BA = ma + nP, . : (Art. 6) xm + y = m, x + yn = n, . BG {m-\)n AG (w-l)m and X, 1. e. -jt-= = — p , y or -— = = -^ -rr- . ' BE mn -I ' " AD mn - 1 Again, let BF = pBA = p (ma + n/S). But BF=BO + GF 4 = ma + a multiple of CG = ma + z CG suppose r=ma + z{BG-BG} \{rn--\)n = wia + z 'At- (ma + /3) - ma v . The two values of BF being equated, and Art. 6 applied, there results _ n— 1 77» — 1 P=l-Z r-, p = Z , , EX. 6--] ■whence VECTOR ADDITION AND SUBTRACTION. 1 —p n-\ 13 p m — i ' AF _AE BD or AF.BD.CE= AE . CD . BF. Ex. 6. When, instead of as in Ex. 4, wJiere D, E, F are points taken within BC, CA, AB at distances . equal, to half those lines res2)ectively, they are points taken in BO, CA, AB produced, at t/ie same distances respectively from C, A, and B ; to find the inter- sections. Let tJie jjoiuts of iutersection be respectively &',, G , G . E Retaining the notation of Ex. 4, we have BI) = 3a, GE=ZP; and .-. BG^ = xBE = a;(2a + 3/3) and BG^^BD+DG^ = 3a + yDA ^3a+y{CA-CI)) = 3a + 2/(2/3-a); .". 2x=3-2/, 3x=2y, and x = .: ]hieEG=lEB. ' 7 •(1), T 14 QUATERNIONS. [CHAP. II. Similarly line i?'G', = yi?'(7, line DG^=^ DA, and from equation (1) £G^ = -„ (2a + 3/8). But BG^ = BA + AG^ = 2a + 2p + AG^; ■■■ ^ff, = |(2;8-a); 2 hence line AG^ = ^ line i)jl and similarly of the others. Ex. 7. The middle points of the lines which join the points of bisection of the opposite sides of a quadrilateral coincide, whether the four sides of the quadrilateral he in the same plane or not. Let ABGD be a quadrilateral ; E, H, G, F the middle points of AB, BC, CD, DA ; Xthe middle point oi £G. Let vector AB = a, AG = p, AD = y, then AE + EG = AD + DG gives \a^.EG = y + \{P.~y), z-ad. AX = AE + ^EG = ^{a + p + y), which being symmetrical is a, p, y in the same as the vector to the middle point of HF. X is called (Art. 14) the mean point of ABCD. ■i Ex. 8. The point of bisection of the line' which joins the middle points of the diagonals of a quadrilateral (plane or not) is the mean point. EX. 9.] VECTOR ADDITION AND SUBTRACTION. Let P, Q be the middle points of AC, £D, R tha.t of PQ. Retaining the notation of the last ex- ample we have ap=Ip, i.e. AQ = ^{AB + AI>). Similarly AE = ^{AP + AQ) = („ + /3 + y), i.e. H is the same point as X in the last example ; and is therefore the mean point oi ABCD. Ex. 9. AD is drawn bisecting BC in D and is produced to any point E ; AB, GE produced meet in P ; AC, BE in Q; PQ is parallel to BC. Jjet AB = a, AC = fi, AP = xa,AQ = yP, .: BC=P-a,AD = AB+]^BG, ^- and AE is a multiple of AB^z (a + /S") say. Then CP = pCE gives xa- p=p{z {a + (3)- (3}, .^ (Art. 6) x=pz, - 1 =px—p ; .'. p = x+l. Similarly BQ = qBE gives yj3-a = q{z{a + 13)- a], y=qz, -\ = qz-\. 16 ■. QUATERNIONS. .' [CHAP. IL and since z = — =- we have P 1 x = y, p = q; .: PQ = yl3 - xoi=x {(3- a) = zBC, hence the line PQ is parallel to BC. The method pursued in this example leads to the solution of all similar problems. It consists, as we have already stated, in reach- ing the points P and Q respectively by two different routes, — viz. through and through £ for P ; through B and through JE for Q — and comparing the results. Cor. 1. PB : SO :: p-l : 1 :: X : 1 :: AP : AB. Cor. 2. AB : AB ■.■..2z : 1 :: 2x : x +1 :: 2(^-1) -.p :: 2PB : PO, .: AD : BE :: PE + EG : PE-EG. Ex. 10. If DEFhe drawn cutting the sides of a triangle ; then will AD . BF. GE=AE. GF. BD. Let BD = a, DA =pa, AE = p, EC = qfi, then BG = BA + AG = {\ +p) ». + {\ + q) /3, and GF is a multiple of BC. Let GF=xBG = x{{l+p)a+{\+q)^\. But GF=GE + EF — EG+EF = -qP + y(j>a+P); .-. equating, we have x (1 +p) '=yp,x{\+q) = -q + y, whence x = {\+ x)pq, CFBF AD GE '■^" BG" BG- BD'-AE' .-. AD.BF\GE = AE.GF.BB. EX. 11.] VECTOR ADDITION AND SUBTRACTION. 17 Ex. 11. If from any point within a parallelogram, parallels be drawn to the sides, the corresponding diagonals of the two parallelograms thus formed, and of the original parallelogram, shall meet in the same point. Jjet PQ, BS meet in T • join TO, on. Let OA = a,OB = p,OQ = ma, OS = np, then QP=QC+GP = nP + {\-m)a, SR=-SG +GR = ma + {\-n)p, and TO=TQ-OQ=x{np+{\-m)a}-ma, also T0=TS-0S = y{ma. + {l-n)P}-nl3, equating, there results xn = y{\—n)-n; x{\-m) — m = ym ; m and T0 = -. mn , -, mn _ _ 1— m — M^"'*"' 1— TO — Ji" hence (Art. 4) TO, OD are in the same straight line. Cor. to : TD :: mn : (1 -m){\-n) :: OSGQ : GRDP. Ex. 12. The points of bisection of the three diagonals of a com- plete quadrilateral are in a straight line. 1. Q. 2 18 QUATEENIONS. [chap. II. P, Q, R the middle points of the diagonals of the complete quadrila- teral ABGD, are in a straight line. Let AB =^a,AI) = P, AE = ma,AF=nP; .: BF=nl3-a and BO = x {n^ - a), ED=^-ma and CD = y (^ -ma). How BG + C-D = BI)=AD-AB a gives ■whence and X (np — a) + y{P — ma.) = j3 — a, xn + y =\, X + my = 1, m— 1 ap=Iac mn — 1 ' m,— 1 2 I. mn— I ^ ' ) AQ-AP= AB-AP=- 1 m (n — 1) a + w (wi — 1) yS ' 2 mn — 1 ' 1 2{mn-\) {(m-l)a + («-l)/8}, {(m-l)a + («-l)^}, 2 {mn - 1) or vector PR is a multiple of vector PQ, and therefore they are in the same straight line. CoE. Line PQ : PR ■.: \ : mn ;: AB.AD : AE.AF , :: triangle ABD : triangle AEF. "We shaU presently exemplify a very elegant method due to Sir W. Hamilton of proving three points to be in the same straight line. ART. 8.] VECTOR ADDITION AND SUBTRACTION. 19 8. It is often, convenient to take a vector of the length of the nnit, and to express the vector under consideration as a numerical multiple of this unit. Of course it is not necessary that the unit should have any specified value ; all that is required is that when once assumed for any given problem, it must remain unchanged throughout the discussion of that problem. If the Kne AB be supposed to be a units in length, and the unit vector along AB hot designated by a, then will vector AB be aa (Art. 3). Sir WiUiam Hamilton has termed the length of the line in such cases, the Tensor of the vector ; so that the vector AB is the product of the tensor AB and the unit vector along AB. Thus if, as in the examples worked under the last article, we designate the vector AB by a, we may write a = Ta Ua, where Ta is an abbre- viation for ' Tensor of the vector a ; Ua for ' unit vector along a'. Examples. Ex. 1. If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the segments at the base shall lutve the same ratio that the other sides of the triangle have to one another. Take unit vectors along AB, A C, which call a, /3 respectively : construct a rhombus -5,^'7\ o APQR on theni and draw its diagonal AR. Then since the diagonals of a rhombus bi- ^ sect its angles, it is clear that the vector •" ^ AB which bisects the angle .4 is a multiple of AB the diagonal vector of the rhombus. Now AIi = AP + PR = AP + AQ = a+p, .: AD = x{a. + P). Now vector AB—ca, AG = b^; using c, b as in ordinary geometry for the lengths of AB, AG. Hence BD=AD-AB=x{a + P)-ca, and BD=^yBG = y{AC-AB) = y{bp-ca.), 2—2 20 QUATERNIONS. [CHAP. II. Equating, x-c = -yc, x = yh ; •■• y=m and BD : DC :: y : l-y :: c : b :: BA : AC. Cor. If a, /8 are unit vectors from A, and if S be another vector from A such that B = x{a + §) • then S bisects the angle between a and j3. Ex. 2. y/ie i/iree bisectors of the angles of a triangle meet in a point. Let AD, BE bisect A, B and meet in G, CG bisects C. Let units along AB, AC, BG be a, j8, y, then as in the last example, AG = x{a + P), BG = y{-a + y). But ay = hp — ca. -K-"^)- and CG = AG-AC = x{a-i-P)-blS, also CG = BG-BC, ( b^-c -1)^ 0(3 _ ni(p— \)a-p(rn, — V}y p — m .-. (m-n){p-l)OR + {n-p){m-\)OP + {p-m){n-\)OQ = 0. And also {m-n){p-\) + [n- p) {m-\) + (j>~ m) {n-\) = 0, whence (Art. 13) P, Q, R are in the same straight line. Ex. 2. If a quadrilateral he divided into two quadrilaterals hy any cutting line, the centres of the three shall lie in a straight line. Let P^Q^QgP^ be the quadrilateral divided into two by the line Pfi^. Let the diagonals of PjQj^^P^ meet in iZ, ; and so of the others : i?„ R^, R^ are the centres. 24 QUATERNIONS. [CHAP. II. Produce P^P^, Q3Q1 to meet in 0. Let unit vectors along OP, OQ be denoted by a, ^ ; and put OP, = m^a, OP^^m^a, OP^ = m^a; pQ^=n^P, OQ^ = nJ, OQ^=nJ; then OR^ = OP, + P^R^ = jw^a + a; {n^^ - m^a), and OR^=OQ^ + Q^R^ = n^p + y{mjx-n^^). Equating, "we have m, — m-^x = m^, and w^a; = «, — m,y ; _ (m, — mj) Wi and Oi? ^ ^1"^. (^1 - ^.) ° + ^,^2 ("^1 - "^,) /^ Similarly, Qj^ ^ ^i"»3 K - ^,) " + ^hn^ (>»3 - 'W.) ^ ■ .-. (?n,ra, - m^n^) m^n^ OR^ + (m^n^ - mjn^) m^n^ OR^ + (n\n^ - JWiWi) m^n^ OR^ = 0. And also + (m^n^ - rrijUj) m^n^ = 0, ■whence (Art. 13) R^, R^, R^ are in the same straight line. Cor. -ffij, i?j, R, will pass through provided the coefficients of a and j8 in the three vectors have the same proportion, i. e. ■provided A_-L.1_J_.. 1_1.1_1 Ex. 3. If AD, BE, CF he drawn cutting one another at any point G vnthin a triangle, then FD, BE, EF shall meet the third sides of the triangle produced in points which lie in a straight line. Also the 2}roduGed sides of the triangle shall he cut harmo- nically. EX. 8.] VECTOR ADDITION AND SUBTRACTION. If, as in Ex. 5, Art. 7, we put we get, as in that example, AF : BF :: n-l : m-\; ... BF= "'~\ (wa + w/3), and i?'i) = 5i>-5i?'=— ^i:ii-{(n-2)a-wi8}. DM= xFD, compared with. DM=DG-MC=a-yl3, gives Xm-\){n-2) _^ Jm-\)n mj.«_fj m + n — 2 " ' tn + n — 2 'n-2' and BM=BG-MC=ma —^B. n—2 Again, FE=FA + AE= n-l m + n — 2 {mo -(to- 2) /3}. 26 QTTATEBNIONS. [CHAP. H. And EL = xFE, compared with EL = GL-CE=ya-^, . m nr / \ m(m-l) BL={y + m)a= ^_^ ' a. Thirdly, DN= xDE = x{a+ (3), compared with 2)iV"= BN-BD^y {ma + njS) - (wi - 1) a, m — \ gives y = , and ^xV= — ^-(Ttia + nB). m — n^ NoV {m-\){n-1)BM+{m-n)BN -(m-2){n-l)BL = 0. Also (ot - 1) (ti - 2) + (m - w) - (m - 2) (m- 1) = ; therefore ^J/, 5i\^, BL are in a straight line (Art. 13). Further, CL = ^^CD, m— 2 ' 7» — 2 .-. CZ : CD :: ^i : BD, and ^X is cut harmonibally. Ex. 4. 57te poini of intersection of bisectors of the sides of a triangle from the opposite angles, the point of intersection of per- pendiculars on the sides from the opposite angles, and the point of intersection of perpendiculars on tlie sides from their middle points lie in a straight line which is trisected by the first of these points. 1°. Let unit vector CB = a, unit vector CA = (3, then Ex. 4, Art. 7, CG=\ {aa + 6^8). ART. 14.] VECTOR ADDITION AND SUBTRACTION, 2°. Let AH, BK perpendiculars on the sides intersect in 0, then EA=hP- 6a cos C, = b{l3- acosC), KB = a {a- p cos C). -8 Now CO = GA+AO, and also =CB + BO gives 6/3 + yb (P — aa cos C) = aa + xa{a — p cos C). b cos C — a ax = - sin'G and CO = ■ . "^ {(6 - a cos (7) a + (a - 6 cos C^ ;S}. 3°. Let perpendiculars from JD and E (Ex. 4, Art. 7) meet inX, then DX is a multiple of SA. .: GX=CD^DX=CE+EXgiye.s -^aa + vlfi-acoa C) = n b/B + z {a — (3 cos C), b — a cos C •'• ^" Ssin'C ' , _ -, (a - 6 cos (7) a + (6 — a cos C) 3 ^'^•^ ^^=^ filPC^ • .-. 2CX+CO-3CG = 0, and also 2 + 1-3 = 0, .•. X, 0, G are in a straight line. Also G0-GG=2 (GG - GX), or vector GO = 2 vector XG, .: G0 = 2GX, and G trisects XO. 14. The vector to the mean point of any polygon is the mean of the vectors to the angles of the polygon. 28 QUATERNIONS. [CHAP. II. 1°. Let be any point ; then in the figure of Ex. 4, Art. 7 ■we have^ calling OA, a, OB, fi and OC, y, OG = a + AG = p+BG = y+CG- = ^{a + p + y)+^{AG + BG+CG) because AG + BG+CG = \{AD + BE + GF) = I {{AB + AC) + (BA + BC) + {GA + CB)] = 0. 2°. If OA, OB, OG, OD be a, (3, y, 8, in the figure of Ex. 7, Art. 7, we have 0X= 0E+ //X= 011+ 1 {OF- OH) = i(Oi^+OZ/) = ^(a + y3 + 7 + S). 3°. In the more general case we may define the mean point in a manner analogous to that adopted in mechanics to define the centre of inertia of equal masses placed at the angular points of the figure. Thus, if we take any rectangular axes OX,OT, and designate by a, /3 unit vectors parallel to these axes ; and by p.,p^, &o. the vectors to the different points ; and if we write a;,, y^ ; x^, y^, (fee. for the Cartesian co-ordinates of the different points referred to those axes ; and define the mean point as the centime of inertia of equal masses placed at the angular points j the Cartesian co-ordinates of that point will be y ~ m ' and its vector p = xa + y/S. EX, 1.] VECTOR ADDITION AND SUBTRACTION. 29 Now p, = a;,a + y,/3, p. = xjx + y,P, (fee. m m m = xa + y/3, = p. Cor. 1. (p,-p) + (p,-p) + (p3-p) + &c = 0, i. e. the sum of the vectors of all the points, drawn from the mean point, = 0. The extension of the same theorem to three dimensions is obvious. CoK. 2. If we have another system of n points whose vectors are cr^ , tr,, &c. then the vector to the mean point is n If now T be the mean point of the whole system, we have m + n ' or (m + w) T — mp — «cr = 0, hence (13) t, p, o- terminate in a right line; or the general mean point is situated on the right line which connects the two partial mean points. Additional Examples to Chap. II. 1. If P, Q, a, S be points taken in the sides AB, BC, CD, DA of a parallelogram, so that AP : AB :: BQ : BC, &c., PQBS will form a parallelogram. 2. If the points be taken so that AP = CE, BQ=DS, the same is true. 3. The mean point of PQBS is in both cases the same as that oiABCD. 30 QUATERNIONS. [CHAP. II. 4. Ti PQ'R'S' l?e another parallelogram described as in Ex. 1, the intersections oiPQ, P'Q', &c. shall be in the angular points of a j)arallelogram EFGH constructed from PQRS as P'QR'S' is con- structed from ABCD. 5. The quadrilateral formed by bisecting the sides of a quadri- lateral and joining the successive i)oints of bisection is a parallelo- gram, with the same mean point. 6. If the same be true of any other equable division such as trisection, the original quadrilateral is a parallelogram. 7. If any line pass through the mean point of a number of points, the sum of the perpendiculars on this line from the differ- ent points, measured in the same direction, is zero. 8. From a point E in the common base AB of the two triangles ABC, ABD, straight lines are drawn parallel io AG, AD, meeting BG, BD a.t F, G ; shew that FG is parallel to GD. 9. From any point in the base of a triangle, straight lines are drawn imrallel to the sides : shew that the intersections of the diagonals of every parallelogram so formed lie in a straight line. 10. If the sides of a triangle be produced, the bisectors of the external angles meet the opposite sides in three points which lie in a straight line. 11. If straight lines bisect the interior and exterior angles at A of the triangle ABG in I> and E respectively j prove that BB, BG, BE form an harmonical progression. 12. The diagonals of a parallelepiped bisect one another, 13. The mean point o'f a tetrahedron is the mean point of the tetrahedron formed by joining the mean points of the triangular faces ; and also those of the edges. 14. If the figure of Ex. 11, Art. 7 be that of a gauche quad- rilateral (a term employed by Ohasles to signify that the triangles AOD, BOD are not in the same jilane), the lines QP, DO, BS will EX. 15.] VECTOR ADDITION AND SUBTRACTION. 31 meet in a point, provided AP OS ^AQ DR 15. If through any point within the triangle ABC, three straight lines MN, PQ, ES be drawn respectively parallel to the sides AB, AC, BC ; then will 3f.V PQ lis _^ AB '^ AC^ BC'"' 16. ABCD is a parallelogram; E, the point of bisection of AB ; prove that AC, BE being joined will trisect each other. 17. ABCD is a parallelogram; PQ any line parallel to DC ; PD, QG meet in S, PA, QB in R; prove that AD is parallel to RS CHAPTER III. VECTOR MULTIPLICATION AND DIVISION. 15. We trust we have made the reader understand by what we stated in our Introductory Chapter, that, whilst we retain for 'multiplication' all its old properties, so far as it relates to ordi- nary algebraical quantities, we are at liberty to attach to it any signification we please when we speak of the multiplication of ^ vector by or into another vector. Of course the interpretation of our results will depend on the definition, and may in some points differ from the interpretation of the results of multiplication of numerical quantities. It is necessary to start with one limitation. Whereas in Algebra we are accustomed to iise at random the phrases ' multiply by ' and ' multiply into ' as tantamount to the same thing, it is now impossible to do so. We must select one to the exclusion of the other. The phrase selected is 'multiply into'; thus we shall un- derstand that the first written symbol in a sequence is the operator on that which follows : in other words that a^S shall read ' a into /8', and denote a operating on fi. 16. As in the Cartesian Geometry, so ^ here we indicate the position of a point in space by its relation to three axes, mutually at right angles, which we designate the axes of X, y, and z respectively. For graphic representation the axes of x and y are drawn in the plane of the paper whilst that of z being perpendicular to that plane is drawn in perspective only. As in ordinary ART. 17.] VECTOR MTJLTIPLICATIdN AND DIVISION. S3 geometry we assume that when vectors measured forwards are represented by positive symbols, vectors measured backwards will be represented by the corresponding negative symbols. In the figure before us, the positive directions are fo^-wa/rds, upwards and outwards; the corresponding negative directions, backwards, downwards and inwards. With respect to vector rotation we assume that, looked at in. perspective in the figure before us, it is negative when in the direction of the motion of the hands of a watch, positive when in the contrary direction. In other words, we assume, as is done in modem works on Dynamics, that rotation is positive when it takes place from y to z, z to x, x to y. negative when it takes place in the contrary directions (see Tait, Art. 65). Unit vectors at right angles to each other. 17. Definition. If i, j, k be unit vectors along Ox, Oy, Oz respectively, the result of the multiplication of i into j or ij is defined to be the turning of j through a right angle in the plane perpendicular to i and in the positive direction ; in other words, the operation of i on j turns it round so as to make it coincide with k ; and therefore briefly ij = k. To be consistent it is requisite to admit that if i instead of operating on j had operated on any other unit vector perpendicular to i in the plane of yz, it would have turned it through a right angle in the same direction, so that ik can be nothing else than —j. Extending to other unit vectors the definition which we have illustrated by referring to i, it is evident that j operating on k must bring it round to i, or jk = i. Again, always remembering that the positive directions of rotation are y to z, z to x, x to y, we must have ki =j. 18. As we have stated, we retain in connection with this definition the old laws of numerical multiplication, whenever numerical quantities are mixed up with vector operations ; thus 2i.3j=Qij. Further, there can be no reason whatever, but the contrary, why the laws of addition and subtraction should undergo T. Q. 3 84 QUATERNIONS. [CHAP. III. any modification when the operations are subject to this new definition • we must clearly have i (j + k) = ij + ik. Finally, as we are to regard the operations of this new de- finition as operations of multiplication — magnitude and motion of rotation being united in one vector symbol as multiplier, just as magnitude and motion of translation were united in 'one vector symbol in the last chapter — we are bound to retain all the laws of algebraic multiplication so far as they do not give results inconsistent with each other. In no other way can the conclusions be made to compare with those deduced from the corresponding operations in the previous science. Thus we retain what Sir William Hamilton terms the associative law of multiplication : the law which assumes that it is indifferent in what way operations are grouped, provided the order be not changed ; the law which makes it indifferent whether we consider ahc to be a X he or ah x c. This law is assumed to be applicable to multiplication in its new aspect (for example that ijk = ij . h), and being assumed it limits the science to certain boundaries, and, along with other assumed laws, furnishes the key to the interpreta- tion of results. The law is by no means a necessary law. Some new forms of the science may possibly modify it hereafter. In the meantime the assumption of the law fixes the limits of the science. The commutative law of multiplication under which order may be deranged, which is assumed as the groundwork of common algebra (we say assumed advisedly) is now no longer tenable. And this being the case it is found that the science of Quaternions breaks down one of the barriers imposed by this law and expands itself into a new field. ij is not equal to ji, it is clearly impossible it should be. A simple inspection of the figure, and a moment's consideration of the definition, will make this plain. The definition imposes on i as an operator on j the duty of turning _; through a right angle as if by a left-handed turn with a cork-screw handle, thus throwing j up from the plane xy ; when, on the other hand, j is the operator ART. 19.] VECTOR MULTIPLICATION AND DIVISION. 35 and i the vector operated on, a similar left-handed turn will bring i down from the plane of xy. In fact ij=k, ji = ~Tc, and so \ 19. • We go on to obtain one or two results of the application of the associative law. 1. Since ij= h, we have i.ij = ik = —j. Now by the law in question, or i^= — 1. Our first result is that the square of the unit vector along Ox is — 1 ; and as Ox may have any direction svhatever, we have, gene- rally, the square of a unit vector = — 1. In other words, the repetition of the operation of turning through a right angle reverses a vector. 2. Again, ijk = i.jk = i .i = i^ = —l.. Similarly it may be proved that jfd = Jcij = —l, ' or no change is produced in the product so long as direct cyclical order is maintained. 3. But ilcj = i.kj = i. — i = — i'= + li . '. ijk = — ikj, or a derangement of cyclical order changes the sign of the product. This last conclusion is also manifest from Art. 18. Vectors generally not at right angles to each other. 20. We have already (Art. 8) laid down the principle of separation of the vector into the product of teasor and unit vector ; and we apply this to multiplication by the considerations given in Art. 18, from whicli it follows at once that if a be a vector along Ox containing a units, )8 a vector along Oy con- taining b units, a = ai, P = hj, and aj8 = a5y. 3—2 36 QUATEEKIONS. [CHAP. III. In the same Way a" = ai .ai'=a'i^ — — a', or the square of a vector is the square of the corresponding line with the negative sign. Seeing therefore the facility -with which -we can introduce tensors whenever wanted, we may direct our principal attention, as far as multiplication is concerned, to unit vectors. 21. We proceed then next to find the product a^, when a and p are vectors not at right angles to one another. ] . Let a, p be unit vectors. Let OA - a, 0B = p. Take OG=y, a unit vector perpen- dicular to OB and in the plane BOA. Take also DO or DO produced = e, a unit vector perpendicular to the plane BOA. Draw AM, AN perpendicular to OB, OC, and let the angle BOA = ; then vector OA = OM+MA = OM+ ON (Art. 1) = part of OB + part of OC (Art. 3). ' Now it is evident that OM as a line is that part of OB which is represented by the multiplier cos ^, or OM = OB cos 0, and similarly that 0N= OG sin : consequently (Art. 3) the same applies to them as vectors ; i. e. vector 0M= /8 cos 6, vector 0N= ysinO ; .: a = I3coa6 + yam0, and a^ = (/3 cos6 + y sin 6) P = P''cos6 + yl3ame. But 13' = -I (19. 1), y^=e (17); [Observe that y, ^ and e of the present Article correspond toj, i and -k of Art. 17.] ,-. a|8=-cos6 + esin^. ABT. 22.] VECTOB MULTIPLICATION AND DIVISION. 37 2. If a, j3 are not unit vectors, but contain Ta and T^ units respectively, we have at once, by the principle laid down in Art. 20, o)3 = TaTfi (- cos e + e sin 6) . 3. It thus appears that the product of two vectors a, )3 not at right angles to each other consists of two distinct parts, a numerical quantity and a vector perpendicular to the plane of o, p. The former of these Sir "William Hamilton terms the scalae part, the latter the vector part. We may now write aP^SajS+ValS, where S is read scalar, V vector : and we find Sal3 = -TaTp cose, ■ • VaP=TaTI3esme. 4. The coefficient of e in Va.(3 ia the area of the parallelogranl whose sides ate equal and parallel to the lines of which a, p are the vectors. 22. To obtain j8a we have, a and ^ being unit vectors, a— 13 C0s6 + y sin ', .: Pa = (3 {13 cos e + y sine) = /3'co3e + /3ysini9 = -cos^-£sin^ (Art. 19. 1 and 18); therefore generally , /3a = TaTl3 (- cos - £ sin e). It is scarcely necessary to remark that whilst y operating on j8 turns it inwards from OB to DO produced, /3 operatiag on y turns it outwards frojn 00 to OD, causing' it to become - e. We have therefore 1. Sal3 = Spa. 2. VaP = ~7pa. 3. aP + j3a=>2Sal3.. L aj3-l3a = 2rap. 38 QUATERNIONS. [CHAP. III. 5. (a + )8)' = (a + )8)(a + /?) 6. {a-Py = a'-2Sal3 + ^\ 7. If a, j3 are at right angles to each other, Sa^= 0, and conversely. J') 8. FajS is a vector in the direction perpendicular to the plane which jjasses through a, jS. yy 9. a^/3^ = 0.^.^0. because /3' is a scalar ; = {Sapy-irapy. Note. a'P' must not be confounded with (a/3)'. 23. Before proceeding further it is desirable we should work out a few simple Examples. Ex. 1. To express the cosine of an angle of a triangle in terms of the sides. Let ABC be a triangle ; and retaining the usual notation of Trigonometry, let CB^a, CA=/3; then (vector ABf = (a - /Sy = a'-2Sa^ + fi' (22. 6), or, changing all the signs to pass from vectors to lines (20) and applying 21. 3, c' = a' -2ab COS G + b'. Ex. 2. To express the relations between the sides and opposite angles of a triangle. Let CB = a, CA = P, BA=y. Then CB + BA = CA gives a + y = p, .•. a* = a (/3 - y) = a/3 - ay. Take the vectors of each side. ART. 23.] VECTOR MULTIPLICATION AND DIVISION, 39 Now Va' = 0, for a? = — a' has no vector part, .-. 7a/3= Vay; i.e. (21. 3) abismG = acesmB, or 6 sin C = cmiB ; , i.e. b : c :: sinB : sin G. Ex. 3. Tlie sum of the squares of the diagonals of a paral- lelogram is equal to the sum of the squares of the sides. Retaining the notation and figure of Ex. 1, Art. 7, CB = a + p, DA=a-P; .: CB' + DA' = 2a' + 2fi', and, changing all the signs, we get (20) for the corresponding lines, CB' + DA' = 2GA'+2AB' = CA' + AB' + BD' + DG\ Ex. 4. Parallelograms upon the same base and between tlie same parallels are equal. It is necessary to remind the reader of what we have already stated, that examples such as this are given for illustration only. We assume that the area of the parallelogram is the product of two adjacent sides and the sine of the contained angle. Adopting the figure of Euclid i. 35 and writing TV(3a as the tensor multiplier of VjSa so as to drop the vector e on both sides ; we have, calling BA, a ; BG, P ; BE = BA+AE = a + xfi ; .: V.p{a + xP)=r{BG.BB), i.e. Vpa=V {BG.BE), remembering that x^," has no vector part. Hence T.Vpa=T {BG . BE), i.e. BG.BABYD.ABG = BG.BEe,m.EBG{2l. 3), which proves the proposition. 40 QUATEENIONS. [CHAP. Ilf. Ex. 5. On the sides A B, AC of a triangle are constructed any two •parallelograms ABDE, ACFG : the sides BE, FG are produced to meet in H. Prove that the sum of the areas of the parallelograms ABDE, ACFG is equal to the area of the parallelogram, whose adjacent sides are respectively equal and parallel to BG and AH. Let BA = a, AE = p, AG = y, GA = S, then AH = p + xa, and AH = — S — yy; .: 7aAH=YaP and VyAE = -Vy^ = VSy (22. 2), hence ^(« + y) ^^= ^"j8 + FSy, i.e. (21. 4), the parallelogram whose sides are parallel and equal to BG, AH, equals the two parallelograms whose sides are parallel and equal to BA, AE ; GA, AG respectively. [The reader is requested to notice that the order GA, AG is the same as the order BA, AE, and BA, AH : so that the vector e is common to all.] Ex. 6. If be any point whatever either in the plane of the triangle ABO or out of that plane, the squares of the sides of the triangle fall slwrt of three times the squares of the distances of the angiilar points from 0, by the, square of three times the distance of the mean point from 0. Let OA=a, OB = p, OG = y, 'then (Art. 14), 0(? = | (a + ^ + y), or o? + p,' + y'+ 2S {aP +Py+ ya) = WG'. Now AB = P-a, BG = y-P, GA=a-y, .-. AB' + BG^ + CA'^2{a:' + ^''+y'')-2S{afi + Py + ya) = 3(a' + ;8^ + y")-90e^ and the lines AB' + BG'' +GA' = Z {OA' + OB' + OG') - (ZOGf. Ex. 7. Tlie sum of the squares of the distances of any point from the angular points of the triangle exceeds the sum of the AET. 23.] VECTOR MULTIPLICATION AND DIYISION. 41 sqiutres of its distances from the middle points of the sides hy tlie sum of tlie squares of half the sides. Retaining the notation of the last example, and the figure of Ex. 4, Art. 7, OD = l{P + y), OE = l{y+a), OF=l{a + P); .: 4 (OB' + OH' + OF') = 2 {a' + 13' + /) + 2S (a^ + /?y + ya) = a' + P' + y' + 90G' = i{a'+P' + y')-{AB' + BC + CA'); A 7i' A. Tir" J. PA' .-.as lines Oiy+OE'+OF'+ , '^^ =OA'+OB'+OC\ Ex. 8. The squares of the sides of any quadrilateral exceed the squares ofilie diagonals hy four times the square of tlie line which joins the middle points of tlie diagonals, Retaining the figure and notation of Ex. 8, Art. 7, we have squares of sides as vectors = a' + {^-aY+{y-PY + y' = 2(a' + P' + y')-2S{ap+Py), and squares of diagonals = l3' + {y-a)' = a' + fi' + y'-2Say', therefore the former sum exceeds the latter by a' + ^' + / - 2;5'a;8 - 2S/iy + 2Say ^{a+y-Pr -C-f'-f)" = i{OQ-opy = iPQ'. Therefore as lines the same is true. N'ote. The points A, B, C, D may be in different planes. 42 QUATERNIONS. " [CHAP. III. Ex. 9. Four times the squares of the distances 0/ any point what- ever /rom the angular points of a quadrilateral are equal to the sum of the squares of the sides, the squares of the diagonals and the square of four times the distance of the point from tlie mean point of the figure. Witli the notation of Art. 14, and the figure of Ex. 7, Art. 7, we have squares of the sides + squares of the diagonals = 3 (a" + )8= + y= + S=) - 2^ (a/3 + ay + qS + ^Sy + /3S + yS). Now (Art. 14) (a + ^ + y + 8)'= = (40Z)'j .•. (iOXy + squares of sides + squares of diagonals = A(OA''+0£' + OC' + OI)''). Ex. 10. T/ie lines which join the mean points of three equila- teral triangles described outwards on the three sides of any triangle form an equilateral triangle whose mean point is the same as that of the given triangle. Let P, Q, B be the mean points of the equilateral triangles on BC, GA, AB; PD=a, DG = P, GE = y, EQ = ^; and let the sides of the triangle ABG be 2a, 25, 2c. PQ' = {a + ^ + y + Zy = a' + /S" 4- y" + 8" + 2SaP + 2Say + 2SaS + 2S^y + 25/38 + 2 »Jo = I (a* + 6- + c') + ■^- ai-ea of ABC, ^'^ which being symmetrical in a, 6, c proves that FQE is equilateral. Again, G being the mean point of ABC, PG=PD + DG = a + ^+^, and line PG^=-^ + 17 + "o" "*" T~"7^ a6 sin C - ^ a5 cos (7 = I (a' + 6' + 0+-^ area ^5(7; .-. PG = QG = RG; and G is the mean point of the equilateral triangle PQE. Ex. 11. In any quadrilateral prism, the SU771 qfiJie squares of (lie edges exceeds tlie sum oftlie squares oftlie diagonals hy eigJU times tlie square of the straight line which joins the points of inter- section of the two pairs of diagonals. „^ Let OA = a, OB = p, OC = y, OD = B; sum of squares of edges = 2{a' + /3'+(y-ay + (y-/3r + 2S=} = 2 {2a' + 2^' + 2y= + 28' - 2^ay - 2SPy}, = l(a + /3-y), 44 QUATERNIONS. [CHAP. III. sum of squares of diagonals = (B + yy + {B-yy+{S + a-py + (8 + l3-ay Also l0G = l{S + y) = vector to the point of bisection of CD, and therefore to the point of intersection of OG, CD, and vector from to the point of bisection of ^i**, as also to that of BE, and therefore to the intersection of AF, BE hence vector which joins the points of intersection of diagonals 1, eight times square of this vector = 2 (a' + ^^ + y^ + 2SaP - 2Say - 2,S'^7), which, added to the sum of the squares of the diagonals, makes up the sum of the squares of the edges. 24. Definition. We define the quotient or fraction - . where a and /3 are unit vectors, to be such that when it operates on a it produces j8 or - . a = y8. This form of the definition enables us to strike out a by a dash made in the direction of ordinary writing, B S thus —.a = B, - is therefore that multiplier which, operating on a, a ' a or on /3 cos + y sin d (21), produces yS. Now cos ^ + e sin operating on j8 cos + y sin 6 produces j3 cos^ ^ + (y + ^P) sin 9 cos 6 + ey sin" 0. But a glance at the figure (Art. 21) will shew that c^ = -y, and «y = i3 i ART. 25.] VECTOR MULTIPLICATION AND DIVISION. 45 .*. cos 6 + e sin 6 operating on /3 cos $ +y sin produces (3 ■ hence — = cos 6 + e sin 6. a It may be worth while to .exhibit another demonstration of this proposition : thus — .al3 = ^ . P (by the associative law) = —1.(19. 1). i.e. (21.1) ^ . (-cos e + csin 6)=- 1. Now (cos 6 + €sm6) { — cos 5 + « sin 6) = - cos' 6 - sin" = -1; u Cor. ^ = - /3a (by 22). 25. 1. Definition. Still retaining a, ^ as unit vectors, since - operating on a causes it to become j8, it may be defined as a versor a acting as if its axis were along OD (Fig. Art. 21). By comparing the result of that article with the definitions of Ai't.17, it is clear that — or cos 6 +i sin 6 is an operator of the same character as — A or c a (as' we have now called the corresponding unit vector) ; with thLs difierence only, that whereas —kai c as an operator would turn o through a right angle, cos 6 + e sin 6 turns it, in the same direction, only through the angle 6 : cos ^ + e sin 6 is then the versor through the angle 6, 2. If a, /3 are not unit vectors, the considerations already advanced render it evident that ^=^(cose+csina a la. Now 7-7- is itself of the nature of a tensor, for it is a numerical la quantity, hence - is the product of a tensor and a versor. 46 QUATERNIONS. [CHAP. III. 26. By comparing the last Article ■with Art. 22 it appears that generally the product or quotient of two vectors may be expressed as the product of a tensor and a veraor. This product Sir W. Hamilton names a Quaternion. Cor. It is evident that a quaternion is also the sum of a scalar and a vector. 27. (1) If "' /3, 7 are unit vectors in the same plane, c a unit vector perpendicular to that plane ; we have seen that - operating on a turns it a round about t as an axis to bring it into the 0/ position /?. If now ^ be a second operator about the same axis in the same direction acting on /3, it will bring it into the position y. But it is evident that - acting on a would at once have brought it into the posi- a y 3 y tion y. This i-s equivalent to the fact that n • = j or in ^i'^ other form (Art. 24) that (cos ^ + 6 sin 4>) (cos $ + esm6)= cos (6+ <}>) + e sin (6 + ^). From this it is evident that the results of Demoivre's Theorem apply to the form cos & + e sin 0. Further, it is evident that since cos 6 + e sin 6 operating with e as its axis, turns a vector through the angle 6, whilst e itself acting in the same direction turns it through a right angle, cos 9 + c sin 6 is part of the operation designated by e, viz. that part which bears to the whole the proportion that bears to a right angle. (2) Eemembering then that the operations are of the nature of multiplication, it becomes evident that cos 6 + ( sin 6 as an e operator may be abbreviated by e^ or t^. And since (cos + e sin 0) (cos <^ + e sin ^) = cos {d + ) + e sin (6 + (jt), ART. 28.] VECTOR MULTIPLICATION AND DIVISION. 47 we shall have or the law of indices is applicable to this operator. (3) Now we have already seen (19. 1) that e° = - 1 ; .-. «* = +l. Conversely, if e" = ± e, w must be an odd number ; if c" = — 1, n must be an odd multiple of 2 ; and if e" = + 1, n must be an even multiple of 2. (4) When a, P are not units, the introduction of the corre- sponding tensor can be at once effected. "We conclude that a quaternion may be expressed as the power of a vector, to which the algebraic definition of an index is applicable. 28. Reciprocals of quaternions — unit vectors. 1. Since a.a = a' = — 1, and -.a = l (Def Art. 24) a = — a. a'j 1 .•. — = — a, or a = — a; a or the reciprocal of a unit vector is a unit vector in the opposite direction. 1 / s , 1 2. Again, a.- = a(— a}=l=-.a; or a vector is commutative with its reciprocal. 3. If y be a versor (say cos ^ + e sin ^, or - j , -. q = \ (Def. extended). Now - = ? j a .-. j8 = qa, by operating on a. 48 QUATERNIONS. [CHAP. III. Also a 1 a = - /?, by operating on |3, and P = qa, = q.- ^ ; .-. q. - = 1=-- .q, q q or a and - are commutative. q This is perhaps better demonstrated by observing that or that if - = cos 5 + e sin 6, then must ^ = cos ^ — e sin ^ j factors which are from their very nature commutative. When the versors are not units the tensors can be introduced as mere multipliers mthout affecting the versor conclusions. 29, We present one or two examples of quaternion division. Ex. 1. To express sin {6 + . a, (3, y being unit vectors in the same plane (Fig. Art. 27), we have - = cos 6 + e sin 6, a ^ = cos ^ + € sin <^, ^=coa{d + ) + e sin {6 + ). ART. 29.] VECTOR MULTIPLICATION AND DIVISION. 49 But l^l.P.. a ft a' . •. COS {e + <^) + e. sin (^ + ^) = (cos 6 + e sin 6) (cos <^ + 6 sin ^) ; whence multiplying out and equating, we have sin {0 + ^)= sin 6 cos <^ + cos 6 sin , cos (Q + <^) = cos ^ cos (^ - sin ^ sin <^. Cor. If the action of the versors be in opposite directions, P lying beyond y, we have (Art. 28) - = cos (5 - ^) - E sin {6 - (j)). But - = cos (^ + e sin , -5 = cos ^ — E sin ^ ; a a B . •■• - =„■- gives cos (5 - ^) - E sin (i9 - ^) = (cos ^ - e sin 6) (cos ^ + e sin cf>), whence sin {6-4,) = sin i9 cos <^ - cos 6 sin ^, cos (9 - cl>) = cos 5 cos ^ + sin 6 sin <^. Ex. 2. ^o j^ftc? 29 2f £"■.£"■. £'fa = €"y = a, so that 2J; 2^ 29 or /^'^^*+^' = l (27.2). Hence (27. 3), 2 -{$ + (!> + \f) is an eve first value is4; .-. e + 4, + ip = 2-n; or the exterior angles of a triangle are equal to four right angles. It will be seen that the demonstration here given is of the nature of that given by Prof. Thomson in the Notes to his Euclid. ABT. 29.] VECTOR MULTIPLICATION AND DIVISION. 51 Ex. 4. In the figure of Euclid I. 47 the three lines AL, BK, CF meet in a point. Let BC= a, GA = /J, AB = y ; the sides being as usual denoted by a, h, c. Let i be the vector which turns another negatively through a right angle in the plane of the paper, so that BD = ia, CK=ip, AG = iy. If BK, AL meet in 0, BO = xBK=x{a + ip), and B0 = BA + A0 = BA+9/BD = -y + yia. ; .: a; (a + 1/3) = — y + yia, XfSa{a + ij3) = — Say, _ Say _ ac cos B Sa (a + ip) a' + ah sin ~ a' + ho' and ■ xSa^ = ySia^ ; _h he which being symmetrical in h and c shews that CF, AL intersect in the same point in which BK, AL intersect. BO Cob. Since BK d' + hc' we have GO W CF a' + hc' also AO be BD a' + bo' AO ■• BD BO GO c'+h' + bc ^BK'^CF a' + bo :1. 4—2 52 QUATEENIONS. [CHAP. Ill, Ex. ,5. If ABGD he a quadrilateral inscribed in a circle ; AB = a., BC = P, CD=y, DA = Z; then apy = ^ 8. Let unit vectors along AB, BG, CD, DA be a', j8', y, 8' ; and let the extciior angles at B and Z> be ^ and +P sin <^), a and j8 being unit vectors ; then qr = TqTr (cos 6 cos <{>- a sin 9 cos - ft cos 6 sin <^ + ap sin sin — (3 sin tf>) (— cos 5 - a sin ) = TqTr (cos 6 cos + a sin 6 cos ^ + /3 cos 6 sin + /3a sin 6 sin ). Now observing that ^a has the same scalar part with afS, but the vector part with a contrary sign, we see that the two ex- pressions for qr and for KrKq likewise have the same scalar part, but that their vector parts have contrary signs. Hence K (qr) = KrKq. (See Tait, § 79 et sq.) 31. ^^0 propose, in this Article, to give and interpret one or two formulje, relating to three or more vectors, wliich are indis- pensable to our progress, reserving to a separate Chapter the demonstration and application of other formulije, the value of which the reader can hardly as yet be expected to understand. 64 QUATERNIONS. [CHAP. IIL 1. To express ^S*. aj8y geometrically. Pirst suppose a, j3, y to be unit vectors OA, OB, 00. Let A0B = 6, and the angle wtich 00 makes witt the plane AOB = (j> ; then since a/3 = - cos e + £ sin 9 (Art. 21), where « is perpendicular to the plane AOB, S. ajSy = S {- COS 6 + e smO) y = Scy sin 6. Now Sey = — cos . angle between e and y = — sia . angle between plane AOB and 0(7 = — siu <}> ; .-. S. afiy = — sin <^ sia ^. O'* Next if a, /8, y are not units, but have re- spectively the lengths Ta, T^, Ty, or a, h, c; we shall have jS' . a/3y = — abc sin 6 sin <^. But ah sin is the area of the parallelogram of which the adjacent sides are a, h ; and c sia ^ is the perpendicular from on the plane of the parallelogram ; .•. — S . a^y = a6 sin ^ . c sin = volume of parallelepiped of which three con- terminous edges are OA, OB, 00. 2. From the nature of the case, no change of order amongst the vectors a, p, y can make any change in the value (apart from the sign) of the scalar of the product of the three vectors ; for it will in every case produce the volume of the same parallelepiped. .-. S.aPy = ^S.yaP = ^S.ayP, &c. Cor. 1. The volume of the triangular pyramid, of which OA, OB, 00 are contermiaous edges is — ^ *S . a/3y. Art. 31.] VECTOR multiplication and division. 55 Cor. 2. If o, j3, y are in the same plane, ^ = ; .-. S.al3y = 0. Conversely, if S . a^y = 0, none of the vectors a, /S, y being themselves 0, -we must have either ^ = or 'f>=0 ; hence in either case the three vectors are co-planar. 3. Since Va^ = y' (21. 3), a vector perpendicular to the plane OAB (Fig. of formula 2) ; F/3y = a', a vector perpendicular to the plane OBG ; and since y, a are both perpendicular to 0£, the line along which is the vector /i ; OB is perpendicular to the plane which passes through y, a, and therefore (21. 3) is in the direction of Vy'a ; hence F(Fa^r/?y)=Fy'a' = wi/?, or the vector of the product of two resultant vectors, one of the constituents of each of which is the same vector, is a multiple of that vector. 4. If OA=a, OB = 13, OB = B, Oi;=e; and if the planes OAB, ODE intersect in OP ; it follows, as in (3), that, Va^ and FSc being both perpendicular to OP, V(Va^VSe) is along OP and is therefore = 9iOf. 5. Connection between the representation of the position of a point by a vector and its representation by Cartesian co-ordinates. If X, y, z be the perpendicular distances of a point P in space from the planes of yz, zx, xy respectively (fig. of Art. 16) j i, j, k unit vectors in the directions of x, y, z; then od is the vector of which the line is x (Art. 3) ; consequently OM along Ox, MN parallel to Oy and iVP parallel to Oz, being x, y, z as co-ordinates, they are xi, yj, zk as vectors. Now vector OP =031+ MN + NP, and is therefore p=od + yj + zk. The same method of representation is evidently applicable when the planes of reference are not mutually at right angles. Tf X, y, z be the co-ordinates of P referred to oblique co-ordinates ; a, P, y unit vectors parallel respectively to x, y, s; then vector OP = xa + y/3+ zy. 56 QUATERNIONS. [CHAP. III. CoE. When X, y, z are at right angles to one another p = xi + yj + eh gives Sip =-x, Sjp = — y, Shp = — z; . : (SipY + {Sjpf + {SkpY = x' + y' + z' = 0P'. Ex. To find the volume of the pyramid of which the vertex is a given point and the base the trianjle formed by joining three given points in the rectangular co-ordinate axes. Let A, B, C be the three given points ; line OA^a, OB=h, OC = c; X, y, z the co-ordinates of the given point P, then vector OA = ai, OB = bj, 00 = ch ; and OP = xi + yj + zh ; .: PA = OA-OP=-{{x-a)i + yj + zk}, PB = - {xi + (y- b)j + zk], PO = -{xi-\- yj + {z-c) h). Now the volume of the pyramid PABC is - I S{PA .PB.PO) (31. 2. Cor. 1) = --^S.{{x-a)i + yj + zk] {xi + {y- b)j + zh] [xi + yh + {z-c) Tc], Multiplying out and observing that only tei'ms which involve all of the three vectors i, j, k produce a scalar in the product, ■we get (+ or -) Vol. =--^{{x — a){bz+cy- be) — cxy — bxz\ 1 , /a; y z D \a -)■ The sign of the result will of course depend on the position of P. Art. 31.] VECTOR mttltiplication and division. 57 Additional Examples to Chap. III. 1. If in the figure of Euclid i. 47 BF, Gil, KE be joined, the sum of the squares of the joining liues is three times the sum of the squares of the sides of the triangle. The same is true -whatever be the angle A. 2. Prove that 4.10" (Art. 7, Ex. 4) = 2 {AB' + AG')~BC\ 3. If P, Q, B, S be points in the sides AB, BC, CD, DA of a rectangle, such that PQ = PS, prove that AR' + OS' = AQ" + CP". 4. The sum of the squares of the three sides of a triangle is equal to three times the sum of the squares of the ILues drawn from the angles to the mean point of the triangle. 5. In any quadrilateral, the product of the two diagonals and the cosine of their contained angle is equal to the sum or difference of the two corresponding products for the pairs of opposite sides. 6. If a, b, c be three conterminous edges of a rectangular parallelepiped ; prove that four times the square of the area of the triangle which joins their extremities is = o'6- + 6V+cV. 7. If two pairs of opposite edges of a tetrahedron be respect- ively at right angles, the third pair will be also at right angles. 8. Given that each edge of a tetrahedron is equal to the edge opposite to it. Prove that the lines which join the points of bisection of opposite edges are at right angles to those edges. 9. If from the vertex of a tetrahedron OABC the straight line OD be drawn to the base making equal angles with the faces OAB, OAO, OBG ; prove that the triangles OAB, OAG, OBO are to one another as the triangles DAB, DAG, DBG. CHAPTER IV. THE STRAIGHT LINE AND PLANE. 32. Equations of a straight line. 1. Let /3 be a vector (unit or otherwise) parallel to or along the straight line ; a the vector to a given point A in the line, p that to any point what- ever P in the line, starting from the same origin ; then AF is a vector parallel to /S = a;/3, say, and OP = OA + AP gives p = a + x^ (1) as the equation of the line. 2. Another form in which the equation of a straight line may be expressed is this : let OA = a, OB = (3 he the vectors to two given points in the line ; then AB = ^-a and AP = x{P-a); .: p = a + x{P-a) (2). Of course the /3 of No. 2 is not that of No. 1. The first form of the equation supposes the direction of the line and the position of one point in it to be given, the second form supposes two points in it to be given. 3. A third form may be exhibited in which the perpendicular on the line from the origin is given. AET. 34.J THE STRAIGHT LINE AND PLANE. 59 Let OD perpendicular io AP = h ; then JDP^p-S and ^S(p-8) = 0, because OD is perpendicular to AF (22. 7) j i. e. SSp = G (3), where C is a, constant. {Note. In addition to this we naust have the equation of the plane of the paper, La which p is tacitly supposed to lie. This may be written as Sep = 0.) ' 33. Equation of a plane. Let P he any point in the plane, OD perpendicular to the plane j and let OX)-S, OP^p; then p - 8 = OP, which is in a direction perpendicular to OD ; .: 5S(p-8) = 0, or Shp = 8', or.S|=l. Cor. 1. If SSp=0 be the equation of a plane, 8 is a vector in the direction perpendicular to the plane. Cor. 2. If the plane pass through 0, p can have the value zero, .-. SSp = is the equation. Cor. 3. Since a vector can be drawn in the plane through D, parallel to any given vector in or parallel to the plane ; if ;8 be any vector in or parallel to the plane, SSp = 0. 34. We proceed to exhibit certain modifications of the equa- tions of a straight line and plane, and one or two results imme- diately deducible from the forms of those equations. 1. To find the equation of a straight line which is perpen- dicular to each of two given straight lines. Let P, y be vectors from a given point A in the required line, and parallel respectively to the given lines. 60 QUATEKNIONS. [CHAP. IV. If OA = a as before, tlien since (22. 8) Vj3y is a vector along the line whose equation is required ; we have or p = a + xVPy, as the equation of the line. 2. To find the length of the perpendicular from the origin on a given line. Equation (1) of Art. 32 is p = a + xj3. If now p = OZ) = 8 ; we get (SS^ = SZa, or -Ol>'=SSa; 1/8 being the unit vector perpendicular to the line. CoE. The same result is true of a plane. 3. To find the length of the perpendicular from a given point on a given plane. Let Sap = C be the equation of the plane, y the vector to the given point. Then if the vector perpendicular be xa (33. Cor. 1), p = y + xa gives Say + xa" — C, and the vector perpendicular is a;a = — a~' (C - Say) ; the square of which with a — sign is the square of the perpendi- cular. 4. To find the length of the common perpendicular to each of two given straight lines. IRT. Si.] THE STRAIGHT LINE AND PLANE. 61 Let /3, ;8, be unit vectors along the lines ; a, a^ vectors to given points in the lines ; p = a + xP, P. = «. + ^fi.' the vectors to the extremities of ttie common perpendicular S. Then since 8 is perpendicular to both lines, it is peipendicular to the plane which passes through two straight lines drawn pa- rallel to them through a given point; .-. (21. 3)8 = 2/ F/3A- But 8 = p-p, = a + K/S-ttj -Xj/?,, hence ^ . 8/3/3, = >S'. (a- a,) j8/3, ; i.e. S{yri3p^.pp^)==S.{a-a;)/3p^, or2/(F/3;8,r = 5.(a-a,)^/3., because SV^ji^SPIB, = ; whence 8 = ?/Fj8^i is known. 5. To find the equation of a plane which passes through three given points. Let a, /3, y be the vectors of the points. Then p — a, a — (3, /3 — y are in the same plane. .-. (Art. 31. 2. Cor. 2) S. (p - a) (a - yS) (^8 - -y) = 0, or Sp{raP+VI3y+rya)-S.aPy = is the equation required. Cor. Va^ + VPy + Vya is a vector in the direction perpen- dicular to the plane; therefore (No. 3) the perpendicular vector from the origin = S.al3y. {VaP + F/3y + Vya)". 6. To find the equation of a plane which shall pass through a given point and be parallel to each of two given straight lines. 62 QUATERNIONS. [CHAP. IV. Leb y be the vector to the gLvea point, p = a + xjS, p = a +x ^ the lines ; then if lines be drawn in the required plane pa,rallel to each of the given straight lines — these lines as vectors will be p, /?! : also p -y is a vector line in the plane ; .-. S.I3l3^{p -y) = (31. 2. Cor. 2), which is the equation required. 7. To find the equation of a plane which shall pass through two given points and be perpendicular to a given plane. Let a, ;8 be the vectors to the given jDoints, SSp = the equa- tion of the plane ; then the three lines p — a, a — /B, S are vectors in the plane ; .-. S.{p-a){a-^)S=0, or S.p(a-P)8+S.aj3S = 0. 8. To find the condition that four points shall he in the same plane. 1. Let OA, OB, 00, OB or a, ^8, y, 8 be the vectors to the four points ; then 8 - a, S — ^, 8 — y are vectors in the same plane ; .-. ^.(8-a)(8-|8)(8-y) = (31. 2. Cor. 2), or S.BPy + S.aSy + S .ajS&^S.afiy (1). 2. Another form of the condition is to be obtained by as- suming that dS + cy + bl3 + aa^0 (2), and substituting in equation (1) the value of 8 deduced from this equation. The result is or a + l + c + d = (3). Equation (1), or the concurrence of equations (2) and (3) is the condition necessary and sufficient for coplanarity. 9. To find the line of intersection of two planes through the origin. ART. 35.] THE STRAIGHT LINE AND PLANE. 63 Let Sap = 0, S^p = be the planes. Since every line in the one plane is perpendicular to a ; and every line in the other perpendicular to /3 ; the line required is perpendicular to both a and ;8, and is therefore parallel to Fa/3, or p = xVaP is the equation. 10. The equation of the plane which passes through and the line of intersection of the planes Sap = a, S^p = 6 is Sp (aP - ba) = 0. For 1° it is a plane through ; 2° if p be such that Sap = a, then must S^p = 6. 11. To find the equation of the line of intersection of two planes. Let p = ma + n^ + xVafi be the equation required. Then Sap = ma' + nSa(3, since Vafi is perpendicular to a, and similaily SI3p = mSal3 + nP''; a^'-lSaP bSal3-aP' a'l3'-{Sal3y {VajSy aSal3-ha' aSal3-ha'' (Art. 22. 9), {SaPY-a'l3' {Val3y ' 35. We ofier a few simple examples. Ex. 1. To find the locus of the middle points of all straight lines which are terminated hy two given straight lines. Let AP, BQ be the two given straight lines, unit vectors parallel to which are /?, y; AB the line which is perpendicular to both AP, BQ. Let be the middle point of AB ; vector OA = a; R the middle point of any line PQ, vector OR = p ; then 64 QUATERNIONS. [chap. IV. OP = p + EP = a + xP, OQ = p + RQ = — a. + yy. But EP + RQ = Q; •■• 2p=a;/3 + 2/y; hence, since Sa^ = 0, Say = 0, Sap = is the equation required ; and the locus is a plane passing through (33. Cor. 2), and perpendicular to OA (33. Cor. 1). Note that, if /3 || y, we have simply 2p = x'l3; and, as there is now but one scalar indeterminate, the locus is a straight line instead of a plane. Ex. 2. Planes cut off", from the three rectangular co-ordinate axes, pyramids of equal volume, to find the locus of the feet of per- pendiculars on tltemfrom the origin. Here the axes are given, so that i, j, h are known unit vectors. Let ai, hj, eh be the portions cut off from the axes by a plane, the perpendicular on which from the origin is p. Then p — ai is perpendicular to p ; .: Sp{p-ai) = 0, or p^ — aSip. Similarly, p' = bSjp, p' - cSkp. Hence p" = abc Sip Sjp Skp = C Sip Sjp Skp, since ahc is by the problem constant. If X, y, z be the co-ordinates of p this equation gives at once (x' ■vy' -V zy = Cxyz as the equation required. ART. 35.] THE STRAIGHT LINE AND PLANE. 65 Ex. 3. To find the locus oftlie middle points of straight lines terminated hy two given straight lines and all parallel to a given plane. Retaining the figure and notation of Ex. 1, let 8 be the vector pei-pendicular to the given plane : we have 2p = a;/3 + yy, 2QP=2a + xj3-yy. Now SBQP=0 (33. Cor. 3); .-. JSS{2a + xl3-yy) = 0; 2SaS SI3S and 2p = x^ + ^y + a=-py = ay + x(J3 + by), where a = -^-^ , b = ~-^ are constants j {Syh for instance is the negative of the cosine of the angle between one of the given lines and the perpendicular to the given plane). Now 13+hy is a known vector lying between ^ and y ; call it e, and 2p = ay + a;e is the equation requii-ed ; which is that of a straight line, not generally passing through (32. 1). Ex. 4. OA, OB are two fixed lines, which are cut hy lines AB, A'B' so that the area AOB is constant; and also the product OA, OA' constant. It is required to find Hie locus oftlie intersec- tions ofAB, A'B'. Let the \init vectors along OA, OB be a, j8 respectively. OA = ma, OA' = 7n'a, OB = nP, OB' = n'P; then the conditions of the problem are mn =>m'n' =C, mm' = a. T. Q. 5 66 QUATERNIONS. [CHAP. IV. Now it AB, A'B inter.sect in P, and OP = p, we have p=OA+AP = ma + X (nj3 — ma), p=OA' + A'P = m'a + x' (n'jS — m'a) ; or p = ma + x( — (3- ma j , t 1 \J n ! p = = ma + x — ,«-ma \m '^ . m — X7n = m' — x'm', X x' m m' ' m X — , m + m m' m'+a' m +a and p = - ;(aa + C/3), m +a and the locus required is a straight line, the diagonal of the parallelogram whose sides are aa, C/J. Ex. 5. To find the locus of a point such that the ratio of its distances from a given point and a given straight line is constant — all in one plane. Let S be the given point, DQ the given straight line, SP = ePQ the given relation. Let vector /S^i) = o, SP==p, DQ = yy, y being the unit vector along DQ, PQ = xa; then Tp = eT{PQ), Art. 35.] the straight line and plane. 67 gives p" = e'PQ'', wliere PQ is a vector, = eVa\ But p + xa. = SQ = SD + DQ = a + yy; . •. Sap + xa? = a', for Say = ; an d x^a* = (a" — Sap)' ; hence a'p' = e' (a' — Sap)', a surface of the second order, whoBe intersection with the plane S. ayp = is the required locus. Ex. 6. The same problem when the points and line are not in the same plane. Retaining the same figure and notation, we see that PQ is no longer a multiple of a ; but PQ = SQ-SP = a + yy-p; ••• p'' = e'(a + 2/7-p)', and because PQ is perpendicular to DQ Sy{a + 7/y-p) = (); ••• {yy', i-e.) -y=Syp, and p° = e' (a — ySyp — p)', a surface of the second order. Cor. If e = 1, and the surface be cut by a plane perpendicular to DQ whose equation is Syp =c, the equation of the section is a' + c'-2»S'ap = 0, another plane, so that the section is a straight line. Ex. 7. To find the locus of the middle points of lines of given length terminated hy each of two given straight lines. 5—2 68 QUATERNIONS. [CHAP. IV. Retaining the figure and notation of Ex. 1, and calling RP c, we have 2p = x^ + l/y (1), and 2EP = RP - RQ=2a + xP-yy (2). From equation (1) we have Sap = (22. 7), 2SPp = -x + ySPy, because /? is a unit vector, 2Syp = xSPy-y. The first of these three equations shews that p lies in a plane through perpendicular to AB (33. Cor. 2). The second and third equations give 2{Spp + SPySyp) 2(.Syp + S^ySPp) -I (-S'^rr-1 • ISTow (2) gives, by squaring, - 4c^ = 4a^ + it-^yS^ + yy - 'ixyS^y, in which, if the values of x and y just obtained be substituted, there results an equation of the second order in p. Hence the locus required is a plane curve of the second order, or a conic section, which by the very nature of the problem must be finite in extent and therefore an ellipse. Ex. 8. If Oi plane he drawn through the points of bisection of tioo opposite edges of a tetrahedron it toill bisect the tetrahedron. Let D, E be the middle points of OB, AC : DFEG the cutting plane : OA, OB, 00 = a, j3, y respectively. OG = my, AF = n{P-a). The portion ODGEA consists of three tetrahedra whose common vertex is 0, and bases the triangles AEF, EFG, FGD. Now OE=-^{y + o.), ART. 35.] THE STRAIGHT LINE AND PLANE. 69 OB = 1/3, OG = my, OF=:a + n{P-a); and 6 times the volume cut off = S.a^{a + y)[a + n{P-a)} + S. -^(a + y)my{a + n{P-a)} + S.{a.+n{P-a)}my 1/3 (31, 2. Cor. 1) = ^ {re + nm + (1 —ii)m}S. ay/3 = -{n+m)S . ay p. But since E, G, D, F are in one 2)lane, and 2m (1 -n)OE- (1 -n) OG + 2mnOI> - mOF = 0, we must have (34. 8) 2/w (1 - w) - (1 - n) + 2mre - m = ; . •. m + n = \ ; and 6 times the whole volume cut off = \s.ayP = ^ of 6 limes the whole volume, hence the plane bisects the tetrahedron. Cob. The pl.;ne cuts other two edges at F and G, so that AF qG_ AB ^ 00 ~ 70 QUATERNIONS. [CHAP. IV. Additional Examples to Chap. IV. 1. Straight lines are drawn terminated by two given straight lines, to find the locus of a point in them whose distances from the extremities have a given ratio. 2. Two lines and a point S are given, not in one plane ; find the locus of a point P such that a perpendicular from it on one of the given lines intersects the other, and the portion of the perpendicular between the point of section and P bears to SP a cou.stant ratio. Prove that the locus of P is a surface of the second order. 3. Prove that the section of this surface by a plane perpen- dicular to the line to which the generating lines are drawn perpen- dicular is a circle. 4. Prove that the locus of a point whose distances from two given straight lines have a constant ratio is a surface of the second order. 5. A. straight line moves parallel to a fixed plane and is ter- minated by two given straight lines not in one plane ; find the locus of the point which divides the line into parts which have a constant ratio. 6. Required the locus of a point P such that the sum of th^ projections of OP on OA and OB is constant. 7. If the sum of the perpendiculars on two given planes from the point A is the same as the sum of the perpendiculars from £, this sum is the same for every point in the line AB. 8. If the sum of the perpendiculars on two given planes from each of three points A, B, G (not in the same straight line) be the same, this sum will remain the same for every point in the plane ABC. 9. A solid angle is contained by four plane angles. Through a given point in one of the edges to draw a plane so that the sec- tion shall be a parallelogram. EX. 10.] THE STEAIGHT LINE AND PLANE. 71 10. Through each of the edges of a tetrahedron a plane is drawn perpendicular to the opposite face. Prove that these planes pass through the same straight line. 11. ABC is a triangle formed by joining points in the rect- angular co-ordinates OA, OB, OC ; OD is perpendicular to ABC. Prove that the triangle AOB is a mean proportional between the triangles ABC, ABD. 12. VapVjSp + {Va^Y— is the equation of a hyperbola in p, the asymptotes being parallel to a, /?. CHAPTER V. THE CIRCLE AND SPHERE. 36. EqiMtitm of the circle. Let AD be tte diameter of the circle, centre G, radius =a, P any point. If vector Ci) = a, GP = p, we have p' = — a' (1). If however AP = p, CP = p-a, we have (p-a)' = -a^ (2). If be any point, OP = p, OG=y, GP = p-y, we have (p-yY = — a' These are the three forms of the vector equation. Form (2) may be written p' - 2Sap = 0. If OG = c, form (3) may be written p'-2Syp = c'-a''. Examples. 37. Ex. 1. The angle in a semicircle is a right angle. Taking the second form p' - 2-S'ap = 0, ,(3). we may again write ifc Sp{p-2a) = Q; ART. 37.] THE CIRCLE AND SPHERE. 73 therefore p, p — 2o are vectors at right angles to one another. But p-2aisDP; .% DP A is a right angle. Ex. 2. If through any point witJtin or vnthout a circle, a straight line be drawn cutting the circle in the points P, Q, the pro- duct OP. OQ is always the same for that point. The third form of the equation inay be written (TpY + 2TpSyUp + c'-a' = 0, ■which shews that Tp has two values corresponding to each value of Up, the product of which is c' — a^. Therefore, ikc. Ex. 3. Jf two circles cut one another, the straight line which joins the points of section is perpendicular to tlie straight line which joins the centres. Let 0, G be the centres, P, Q the points of section ; vector 00= a; a,b the radii ; then (as vectors) OP'^-a', {OP-a.y=-b'; ,'. SaOP = C, a constant. Similarly, SaOQ = C, the same constant ; .-. Sa{OQ-OP)=0, or SaPQ = 0, i. e. PQ is at right angles to OC. Ex. 4. is a fixed point, AB a given straight line. A point Q is taken in tJte line OP drawn to a point P in AB, such that OP.OQ^F; « to find the locus of Q. Let OA perpendicular to AB be a, vector a ; OQ = p, OP = xp; then T{pP.OQ)=k', or xp* = — k'. 74 QUATEENIONS. [CHAP. V. Bat ^a {xp-a) = 0; .-. xSap = — a"; k" hence p' = —. Sap "^ a' is the equation of the locus of Q, which is therefore a circle, passing through 0. Ex. 5. Straight lines are drawn through a fixed point, to find the locus of the fett of •perpendiculars on them from another fixed point. Let 0, A be the points, the lines being drawn through A. Let OA = a, and let p = a + xj3 be the equation of one of the lines through A, S the perpendicular on it from 0. Then S = a + xl3, and ^S'' = SaS, because 8 is perjiendicular to /? ; i.e. h'-SaS=Q, the equation of a circle whose diameter is OA. Ex. 6. A chord QR is drawn parallel to the diameter AB of a circle : -P is any point in AB ; to prove that Pg' + PR''=.PA' + PB\ Let CQ=p, CR = p', PC = a; then PQ' = -(YtctovPQY = - (a + p)' = - (a" + 2Sap + p"), PR' = - (a + p7= - (a= + 2Sap'+ p") ; H .-. PQ'' + PR' = 2PC" + 2AC'-2(Sap+Sap'). But S {p + p) (p — p') = and p — p'=xa, because QR is parallel to AB ; . •. Sap + Sap' = 0, and PQ' + PR' = 2PC' + 2AC' = PA'+PB\ ART. 37.] THE CIRCLE AND SPHERE. 75 Ex. 7. If three given circles he cut hy any other circle, the chords of section will form a triangle, the loci of the angular points of which are three straight lines respectively perpendicular to the lines which join the centres of the given circles ; and these three lines meet in a point. Let A, B, C be the centres of the three given, circles ; a, b, c their radii ; a, /3, y the vectors to A, B, G from the origin ; OA, OB, 00 respectively p, q, r ; 1) the centre of the cutting circle whose radius is R, OB = s, vector OD = S, p the vector to a point of section of circle D with circle A ; then we shall have (p-ay = -a', {p-hy=-2i\ and .-. 2S{8-a)p = B'-a'-s'+p'. Now this is satisfied by the values of p to both points of sec- tion ; and being the equation of a straight line (32. 3) is the equation of the line joining the points of section of circle J) with circle A — call it line 1, and so of the others ; then line 1 is 2S{S-o.)p = B' -a'-s' +p', line 2 is 2S{B- /3) p = B' ~ b' - s' + g\ line 3 is 2S{S-y)p" = B'-c'-s' + r\ If 1 and 2 intersect in P whose vector is p^, 1 and 3 in ^ (p^) ; 2 and 3 in B (p^), we shall have by subtraction atP, 2S (a- P) p^ = a' -b' -p'+ q' ; ■ a.tQ, 2S{y-a)p^ = -a' + c' + p''-r'; at 2?, 2Sifi-y)p, = b'-c^-f + i^; therefore (32. 3) the loci of B, Q, B are straight lines, perpen- dicular respectively to AB, AC, BG. Also at the jDoint of intersection of the first and third of these lines, we have, by addition, 2S(a-y)p = a'-c'-p' + r', which is satisfied by the second : hence the three loci meet in a point. 76 QUATERNIONS. [CHAP. V. Ex. 8. To find the equation of the cissoid. AQ is & chord in a circle whose diameter is AB, ^iV perpen- dicular to AB. AM is taken equal to BW, and MP is drawn perpendicular to A B to meet AQ va. P ; the locus of P is the cissoid. Let vector ^P = x, AC = a, AM=ya, AQ = xir; then 2/ : 1 :: 2 — 2/ : iK, by the construction j 2 ■ ^-l + x- Now kV '-2xSa7r = is the equation of thi i circle 1 2Sa7r Also IT = y = ■AM+MP ya + y; yo^y Soltt 1 /. 2-(y-a')^ c —0 and RA : RB :: a-OR : jS-OR :: 6'-a» : c«-a' :: AI" : 5Z7-'. 41. 5786 Spliere. 1. It is clear that there is nothing in the demonstration of Art. 36 which Kmits the conclusions to one plane ; it follows that the equations there obtained are also equations of a sphere, 2. Further if we assume that the tangent plane to a sphere is perpendicular to the radius to the point of contact, the con- clusion in Art. 38 is applicable also^ 6—2 8^ QUATEENIONS. [CHAP. V. The equation of tie tangent plane to the sphere is therefore Sjrp = — a'. 3. Lastly, the results of Art. 39 are also applicable if we substitute any number of tangent planes passing through a given point for two tangent lines; the equation of the plane which passes through the points of contact is therefore Sp,r = -a'. This plane is the polar plane to the point through which the tangent planes pass. CoE. Since the polar plane is perpendicular to the line which joins the centre with the point through which the tangent planes pass, the perpendicular CD to it from the centre ia along this line and has therefore the same unit vector with it. The eqna- • tion above gives in this case S{C0.Cn{Uj3y} = -a'; .: CO. CD = a' (19). Examples. 42. Ex. 1. Every section of a sphere made hy a plane is a circle. Let p' = — a' he the equation of the sphere, a the vector per- pendicular from the centre on the cutting plane ; c the correspond- ing line. Let p = o + IT ; then the equation becomes tt' + 2/S'air + a' = - a'. But SaTT = ; .-. ■7r' = -{a''-c') is the equation of the section, which is therefore a circle, the square of whose radius is a' - c\ Ex. 2. To find the curve of intersection of two spheres, Let the equations be p'-iSap^C, p^-%Sa'p = C'; Art. 42.] tHE CIRCLE AND SPHERE. 85 .-. 2S{a'-a)p = G-C', & plane perpendicular to the line of ■whicli the vector is a — a, L e. to the line which joins the centres of the two circles. Hence, by Ex. 1, the curve of intersection is a circle. Ex. 3. To find the locus of the feel of perpendicrdars from the origin on planes which pass through a given point. Let a be the vector to the point, S perpendicular on a plane through it ; then SS(p-a) = is the equation of that plane ; therefore for the foot of the per- pendicular ,S'(8'-a8) = 0; or &'-Sa8 = is true for the foot of every perpendicular and is therefore the. equation of the surface required. Hence it is a sphere whose diameter is the line joining the origin with the given point. Ex. 4. Perpendiculars are drawn from a point on the surface of a sphere to all tangent planes, to find the locus of their extremi- ties. Let a be the vector to the given point, Sirp = — a' the equation of a tangent plane. Since the perpendicular is parallel to p, its vector is IT = a + xp ; .: (n--a)' = a;y = a;V = -x'a'; because both p and a are vector radii, But Sirp = — «' gives with xp = ir — a, iStt (tt - a) = - a'x, {ir'-San)' = a*x' = - a" X - a V = -a'(7r-o)'. 86 QUATEENIONa [CHAP. V. Ex. 5. If the point$ from which tangent planes are drawn to a sphere lie always in a straight line, prove that the planes of sec- tion all pass through a given point. Let 0£ be perpendicular to the line in whicli the poiat /3 lies (41), see fig. of Ai-t. 39, CI! = c, vector C^ = 8; then SpB=-c' is the equation of the line. But S^v = -a' is the plane of contact, which is therefore satisfied by 2 i. e. the planes all pass through a point G in CE, such that c' or GE.GG = a\ Ex. 6. If three spheres intersect one another, their three planes of intersection all pass through the same straight line. Let a, 13, y be the vectors to the centres of the three spheres, p" — 2Sap = a, p'-2SI3p=b, p'-2Syp = c, their three equations ; .-. 2S(a-l3)p = b-a, ■ 2S{a — y)p = c — a, 2S{fi-y)p = c-b, are the equations of the three planes of intersection. 'Now the line of intersection of the first and second of these planes is obtained by taking p so as to satisfy both equations, and therefore their difference 2S{^-y)p = c-h, ABT. 42.] THE CIRCUS AND SPHERE. 87 which, being the third equation, proves that the same value of p satisfies it also. The three planes consequejitly all pass through the same straight line. Ex. 7. To find ike locus of a point, the sum of the squares of whose distances from a number of given points has a given value. Let p denote the sought point ; a, (3, ... the given ones ; then (p_a)^+(p_^)» + &C. = S(p-oy = -C. If there be n given points ; this is V-2'S'.p2a+2a'' = -C, (-f)' = ©"-^<^--->- This is the equation of a sphere, the vector to whose centre is -2(a), n ^ ' i. e. the centre of inertia of the n points taken as equal. Transpose the origin to this point, then (36) S.a = 0, and p' = -l{S(aO + (7}. Hence, that there may be a real locus, C must be positive and not less than the sum of the squares of the distances of the given system of points from their centre of inertia. If C have its least value, we have of course p'=0, the sphere having shrunk to a point. Additional Examples to Chap. V. 1. If two circles cut one another, and from one of the points of section diameters be drawn to both circles, their other extre- mities and the other point of section will be in a straight line. . 88 QUATERNIONS. [CHAP. V. 2. If a ctord be drawn parallel to the diameter of a circle, the radii to the points where it meets the circle make equal angles with the diameter. 3. The locus of a point from which two unequal circles sub- tend equal angles is a circle. 4. A line moves so that the sum of the perpendiculars on it from two given points in its plane is constant. Shew that the locus of the middle point between the feet of the perpendiculars is a circle. 5. If 0, C be the centres of two circles, the circumference of the latter of which passes through ; then the point of inter- section A of the circles being joined with 0' and produced to meet the circles in C, D, we shall have AC. AD =210'. 6. If two circles touch one another in 0, and two common chords be drawn through at right angles to one another, the sum of their squares is equal to the square of the sum of the diameters of the circles. 7. A, B, G are three points in the circumference of a circle ; prove that if tangents at B and G meet in D, those at G and A in E, and those at A and .B in ^j then AD, BE, CJ'' will meet in a point. 8. If A, B, G are three points in the circumference of a circle, prove that V{AB. BG . GA) is a vector parallel to the tan- gent at A, 9. A straight line is drawn from a given point to a poiqt i* on a given sphere : a point Q is taken in OP so that OP.OQ = k\ Prove that the locus of § is a sphere. 10. A point moves so that the ratio of its distances from two given points is constant. Prove that its locus is either a plane or a sphere. EX. 11.] ADDITIONAL EXAMPLES. 89 11. A point moves so that the sum of the squares of its distances from a number of given points is constant. Prove that its locus is a sphere. 12. A sphere touches each of two given straight lines which do not meet ; find the locus of its centre. CHAPTER VI. THE ELLIPSE. 43. !• If we define a conic section as "the locus of a point which, moves so that its distance from a fixed point bears a con- stant ratio to its distance from a fixed straight line" (Todhunter, Art. 123), we shall find the equation to be (Ex. 5, Art. 35) a'p' = e'(a'-Sapy (1), where SP = ePQ, vector A^i) = a, SP = p. When e is less than 1, the curve is the ellipse, a few of whose properties we are about to exhibit. 2. SA, SA' are multiples of o : call one of them xa : then, by equation (1), putting xa for p, we get af = e'(l-a:)'j 1+e' e AET. 43.] THE ELLIPSE. 91 i.e. SA=^SJ), 1+e l—e ... AA'=rr^,Sn, l — e the major axis of the ellipse, which \re shall as usual abbreviate by 2a. If C be the centre of the ellipse CS= SA' - OA'^ (^ - ,— ,) Si) = eCA \1— e 1 — e/ = ae, and if vector CS be designated by a, CP by p, we have o- = r: J a and p' = p + a ; whence, by substituting in (1), the equation assumes the form a'p''+(Sa'py = -a'{l-e'); which we may now write, CS being a and CP p, ay + {Sapy = -a'{l-e') (2). 3. This equation might have been obtained at once by re- ferring the ellipse to the two foci, as Newton does in the Prin- cipia. Book I. Prop. 11; the definition then becomes SP+nP=2a, or in vectors, if CP = p, CS = a, T(p + a) + T{p-a) = 2a; i.e. J-{p + ay + J-{p- a)' = 2a ; hence, squaring, a ^— (p — a)' = a' + Sap ; . i. e. ay + {Sap)' = -a*. (1 - e'). 92 QUATERNIONS. [CHAP. VL If now we write Ap for rn iv > where 4>P is * vector ^"^ a (1 — e ) which coincides with p only in the cases in which either a coin- cides with p or when Sap = 0, i. e. in the cases of the priacipal axes ; the equation of the ellipse becomes •^^^^ = 1 : (3). The same equation "is, of course, applicable to the hyperbolaj e being greater than 1. 44. The following properties of p will be very frequently employed. The reader is requested to bear them constantly in mind. 1. (p + a-) = p + ifxr. 2. p. „ „ a'Scrp + SatrSap = Spp a'p + aSap 45, To find the equation of the tangent to the ellipse. The tangent is defiaed to be the limit to which the secant approaches as the points of section approach each other. Let CP-p, CQ = p', then vector PQ = CQ-CP = p'-p = P say; j8 is therefore a vector along the secant. Now Sp'p' = S{p + P)(p+P) ^S(p + p)(^)(U.l) = Spp + Spi>p + S^4,p + s^p. ART. 46.] THE ELLIPSK 93 But Sp'^p'=l=Sp^p; .: Sp^p + Spp + Spp = 0, or (44. 3) 2SP4,p + ;SJ3<^i3 = 0. Now P(f>p involves the first power of /3 whilst )3<^/3 inrolves the second, and the definition requires that the limit of the sum of the two as y3 gets smaller and smaller should be the first only, even if that should be zero : i. e. when (3 is along the tangent, we must have 2Sp<}>p = 0. Let then T be any point in the tangent, CT = rr, then JT = p + cb/?, and Sp<}>p = gives S{7r-p)p = 0; . •. STrp = 1 is the equation of the tangent. Cob. 1. ^p is a vector along the perpendicular to the tangent (32. 3), that is, ^p is a normal vector, or parallel to a normal rector at the point p. Cor. 2. The equation of the tangent may also be written (44. 3) Sp^Tr=l. 46. We may now exhibit the corresponding equations in terms of the Cartesian co-ordinates, as some of the results are best known in that form. Let CM=x, MP = y as usual; then, retaining the notation of Art. 31 with i, j as unit vectors parallel and perpendicular respectively to CA, vector CJf = xi, MP = yj, CS= aei ; .: p = xi + yj, a'p + cuSap ^p = - a' (1 — e') xi + a^yj a* (1 - e') ''~\a' b'J' 94 QtTATEENIONS. where b' = a'{l-e'); and Sp,t,p = -S{^ + yj)(^,+f^ ~a' b'' [chap. VI. is the Cartesian interpretation of Sp(j>p = 1. Again, if x', y' be the co-ordinates of T a point in the tangent, v = v^i + y'3, and S^^p = -S (oo'i + y'j) ^p + p) -^ jL.yy. ~ a' b' ' . ^ A.yjL-\ '■ a" b' is the equation of the tangent. 47. The -values of p and <^/) exhibited in the last Article, p = xi + yj, <^P=-\^>+%) (1)' enable us to write We shall have W = -;i- +-fr (2)- ,. , , iSi^P , M4>P 9 P = 9' ^~'p = a'iSip + b'jSjp, &c. If, further, we write \j/p for /iSip jSjp\ -\ir'--b-)' ART. 48.] THE ELLIPSE. 95 we shall Lave ,s (iSip jSjp\ = -«^P W. i/'~'p = — {aiSip + bjSjp), &c. = -{aiSi^p+ hjSj^p) (5). It is evident that the properties of p (Art. 44) are applicable to all these functions. Now iSp4>P = 1 gives /Spi/' (fp) = — 1 . But since Spij/cr = Scrfp, this becomes ^'I'P'^P = — 1> or Til,p = \; which shews 1. that i^p is a unit vector; 2. that the equation of the ellipse may be expressed in the form of the equation of a circle, the vector which represents the radius being itself of vari- able length, deformed by the function i/r. Lastly, Sa^P = gives Safl3 = S^ail,p = 0; therefore i/'o, i/fj8 are vectors at right angles to one another. 48. To find the locus of the middle points of parallel chords. Let all the chords be parallel to the vector ^ ; ir the vector to the middle point of one of them whose vector length is 2a;)8 ; then ir + xP, rr-x^ are vectors to points in the ellipse ; .-. S{,r + xP){Tr + xP) = l, S{7r~xp){-T-xP) = l, 96 QUATEENIONS. [OHAP. VI. multiplying out, observing that (44. 1), tf>{ir + x^) = ^ir + x^P, &c., we get by subtracting, or, (Art. 44. 3), i. e. the locus required is a straight line perpendicular to <^j3. Now P is the vector perpendicular to the tangent at the extremity of the diameter )3 (Art. 45. Cor. 1). Therefore the locus of the middle points of parallel chords is a diameter parallel to the tangent at the extremity of the diameter to which the chords are parallel. CoE. If a be the diameter which bisects all chords parallel to )3 ; since Sap = 0, we have (Art. 44. 3), SI34,a = 0, which is the equation to the straight line that bisects all chords parallel to a. Moreover j3 is parallel to the tangent at the ex- tremity of o, for it is perpendicular to the normal tfta. Hence the properties of a with respect to /3 are convertible with those of /3 with respect to a : and the diameters which satisfy the equation Sal3 = 0, are said to be conjugate to one another. 49. Our object being simply to illustrate the process, we shall set down in this Article a few of the properties of conjugate diameters without attempting to classify or complete them. 1. If GP, CD are the conjugate semi-diameters a, yS; and if DG be produced to meet the ellipse again in E, and PD, PE be joined ; vector DP = o - )8, vector EP = a + )8. AET. 49.] THK ELLIPSE. 97 Now = Sal3-Saj3 + Spa (44. 1) = 0, because »Sa<^a, (a + Ki /3) = 1, for the two points of contact. Subtracting and applying (44. 1), Stt^jS = : hence tt and )8 i. e. CT, QE are conjugate. 3. The equation of the chord of contact is Sa-^ir = 1. For (Sp^ir = 1 (45. Cor. 2) is satisfied by the values of p at Q and at B, and since Sp(f>ir = 1 or Sir = 1 is the equation T. Q. 98 QUATERNIONS. [CHAP. VI. of a straiglit line, tt being a constant vector (32. 3) it is the line QR. 4. If QR pass through a fixed point E, the locus of T is a straight line. Let cr be the vector to the point E, then .". Sir<7 = 1, or the locus of iT is a straight line perpendicular to (jxr, i.e. parallel to the tangent at the point where CE meets the ellipse. (45. Cor. 1.) The converse is of course true. 5. Let us now take CP=a, GD=P, GN=xa, NQ = y^, GT = za; then the equation of the tangent becomes iSfea<^ (ira + y;8) = 1 ; i. e. xzSa(fia = 1 ; .•. xz= 1, or xa. za= a' ; geometricaUy CJH^ . CT= GP\ 6. The equation of the ellipse gives S{xa + y^)<^{xa + yP) = l, or x'Saa. + y'Spp + 2xySa.<^^ = 1, i. e. a!^+y = l, or, since GN is xa, GP = a, &c., \gp) ^\gb) ~^' the equation of the ellipse referred to conjugate diameters. 7. a = i/f~'i/fa = — (^aiSitj/a + bjSjfa) /3 = ^- V|8 = - {aiSi>pp + hjSjfp) • . : Va/3 = ab Vij (SifaSj^jS - Sif^Sjij/a). AET. 60.] THE ELLIPSE. 99 If now we call k the unit vector perpendicular to the plane of the ellipse, we get Vij=k. And, observing that \^a, i/f/3 are unit vectors at right angles j if the angle between i and i/ra be 6, that between i and i/f/J will be 9 "*" ^' ^'^' ^'^■> 'we shall have (21. 3) Six^ia = — COS 6., Sifp = sme, SjiJ/a = ~Bm6, SjfP = - cose. . •. Sip, be xtftp ; then since Z" is a point in the tangent, /Sir<^p = 1 gives Sx^p^p = 1, or x{p)'= \; .: {x^pYi^pY^l, and Cr=T(xi.py=T^, a* b* (46). 7—2 ST=Tx4>p = -.t'- - Sa(j>p HZ=T- + Sa P 4>P . 3 .: BY.EZ-. .tI —S'atftp JOO QtTATEKNIONS. [CHAP. VI. Ex. 2, The product of the perpendicula/rs from the/od on the tangent is equal to the square of the semi-axis minor. We have ST the vector perpendicular = x^p, and as T" is a point in the tangent, and CY=a + xi^p, S(a + xP (1 - Sap) (1 - Sa(l>p)' a 1 - S'ap = - aV -a' {I- e") (last example) = -a', and the line CY=a. Ex. 4. To find the locus of T when the perpendicular from the centre on the chord of contact is constant. If CT be JT, the equation of QR, the chord of contact, is ScT^v= 1 (Art. 49. 3), and the pei-pendicular (Ex. 1) is T—; or S<^ir .<^Tr = -c' (Art. 44. 8)5 Le. ^.(^+-^^) = c'(47.3), an ellip.s& Ex. 5. ^^, TR are two tangents to an ellipse, and CQ', CR' are drawn to the ellipse parallel respectively to TQ, TR; prove that QR' is parallel to QR. Let GQ = p, GR = p', CT=a, then Sp^a = 1, Sp'a=l. Now since C§' is parallel to TQ, CQ'=xTQ = x(p-a). 102 QUATEBNIONS. Similarly ■CK = y{p'-a), and S.GQ'4,{C^ = \ gives x'S{p-a)^(p-a) = l, i.e. a;'(,Sfa^-l) = l, and and Q'R'=GR'-CQ' = x{p'- = xQR; -P) hence Q'K is parallel to QR. Cor. ^R" : QR' -.-.x'-.l :: 1 : Sa{p + a) = l; wherefore , 2/Sp^o + Saijta = 0. Similarly ^Sp'tjia. + Satjia = 0; .; S (p' - p) tfta = 0, by subtraction, or S^^a = 0, and (48. Cor.) /3, a are parallel to conjugate diameters. EX. 1.] THE ELLIPSE. 103 Additional Examples to Chap. VI. 1. Shew that the locus of the points of bisection of chords to an ellipse, all of ■which pass through a given point, is an ellipse. 2. The locus of the middle points of all straight lines of con- stant length terminated by two fixed straight lines, is an ellipse whose centre bisects the shortest distance between the fixed lines ; and whose axes are equally inclined to them. 3. If chords to an ellipse intersect one another in a given point, the rectangles by their segments are to one another as the squares of semi-diameters parallel to them. 4. If PGF, DOD' are conjugate diameters, then PD, PD' are proportional to the diameters parallel to them. 5. If Q be a point in the focal distance SP of an ellipse, such that SQ is to SP in a constant ratio, the locixs of ^ is a similar ellipse. 6. Diameters which coincide with the diagonals of the paral- lelogram on the axes are equal and conjugate. 7. Also diameters which coincide with the diagonals of any parallelogram formed by tangents at the extremities of conjugate diameters are conjugate. 8. The angular points of these parallelograms lie on an ellipse similar to the given ellipse and of twice its area. 9. If from the extremities of the axes of an ellipse four pa- rallel lines be drawn, the points in which they cut the curve are the extremities of conjugate diameters. 10. If from the extremity of each of two semi-diameters ordinates be drawn to the other, the two triangles so formed will be equal in area. 104 QUATEENIONS. [CHAP. VI. 11. Also if tangents be drawn from the extremity of each to meet the other produced, the two triangles so formed will be equal in area. 12. If on the semi-axes a parallelogram be described, and about it an ellipse similar and similarly situated to the given ellipse be constructed, any chord PQE of the larger ellipse, drawn from the further extremity of the diameter CD of the smaller ellipse, is bisected by the smaller ellipse at Q. 13. If TP, TQ be tangents to an ellipse, and PCF be the diameter through P, then P'Q is parallel to GT, CHAPTER VII. THE PARABOLA AND HYPERBOLA. 51, As already stated, most of the properties of the hyperbola are the same as the corresponding properties of the ellipse, and proved by the same process, e being greater than 1. There are, however, some properties both of it and of the parabola which may be conveniently developed by a process more analogous to that of the Cartesian geometry. This process we shall develope presently. In the meantime we proceed to give a brief outline of the application to the parabola of the method employed ia the preceding Chapter for the ellipse. 52. If S be the focus of a parabola, DQ the directrix, we have SP = FQ, SA=AD = a. ItSP = p, SD = a, we have (Ex. 5, Art. 35) ay={a'-Sapy (1). p — a~^Sap If ^p = ? •(2), to which the properties of p = (6), so that p is a vector perpendicular to the axis. From the same equation Spp = p — a~^Sap, reads by (6), 'vector along N'F = SP - vector along AN', which requires that NP^a'cfip (10), 8^= a~^Sap ] i.e. =aSa-'p (11). For the subtangent AT, put xa for tt in (5), and there results by (6) a! + AS'a-'p= 1, whence (a; - ^ ja = ^o-a»S'a~'p; i. e. vector AT = — vector AN (by 11) ; .-. \meAT=AN; ART. 52.] THE PARA30LA AND HYPERBOLA. 107 and ST = xa gives ST'={a-aSa-'py {a'-SapY a' = p'by(l); .-. line ST =SP, ■whence also the tangent bisects the angle SPQ ; and SQ ia per- pendicular to and bisected by the tangent. From (8) y {<^p + o"') = P(? = pir+N'G = -a'^p + «a (by 10); .-. y = -a', y = za, z = -\, Za. = — a; i. e. NG = -SD, or line NG = SD, whence the subnormal is ( constant And vector GP = - -2/(^p + a-i) = a»(p + a-'); •. yecior SQ = SD + DQ = SD + NP = a + a'^p = GP, and SQGP is a rhombus. Lastly, h = SY =^SA + AY; ^.AY = \a'p = and equation (3) becomes '4a' a~ ' or y" = 4a (a + x) = 4ax' if x' = AN. The locus of the middle points of parallel chords is thus found. Let the chords be parallel to j8, ir the vector of the middle point of one of the chords, then iT + xP = p, and S{ir + xP) {tt + xjS) + 2&-' (ttH- a:/?) = 1 ; which, since the term involving x must disappear, gives aStt^/J + Sa-'P = 0, a straight line perpendicular to <^/3, i. e. (6) parallel to the axis. This equation may be written S^{7r+a-') = 0, which shews (8) that the chords are perpendicular to the normal vector at the point where p — ir, L e. at the point where the locus of the chords meets the curve : in other words, the chords are parallel to the tangent at the extremity of the diameter which bisects them, 54. Examples. Ex. 1. If two chords he drawn always parallel to given lines, and cut one another at points either within or without the parabola, AET. 54.] THE PAEABOLA AND HYPERBOLA. 109 tlie ratio of tlie rectangles of their segments is always the same whatever be their point of section. Let J'Op, QOq be the chords drawn through 0, and always parallel respectively to j8 and y, which we will suppose to be unit vectors. Let S be the vector to 0, then p = 8 + a!j8 gives from equation (3) S{i + xp) (^8 + p + aS'8<^8 + 2,S'a-'8 + Ax=l, the product of the two values of x being OP. Op : OQ.Oq 1 1 Spji ■ Syy a constant ratio whatever be 0. Cor. Let 0, 6' be the angles in which (3 and y cut the axis ; then since /?, y are unit vectors, if p be a vector to the parabola, drawn from S parallel to FOp, which we may now call SP ; p = nP, 4>P = 'I> («j8) = HP (44. 2), will give «/},/, Spp SpP : Sy<{>y :: sin fl— : sinfl'-^ :: sm'^ : sin'^; 1 1 and, OP. Op : OQ.Oq :: -^^ : -r-^. ' /--CI sm^ siQ 5 Ex. 2. Find the locus of the point which divides a system of parallel chords into segments whose product is constant. no QUATERNIONS. [CHAP. VII. By the last example, the equation of the locus is ^8.^8 + 2^a-'8-l _ a parabola similar to the given parabola. Ex. 3. The perpendicular from A on the tangent, and the line PQ are produced to meet in B: find the locus of B. By Ai-t. 52. 8, AB = x {p + a"'), and PB = ya ; Operate by Sp, and X {py = Spp = a^i.py (52.7); .". X = a', and w = s + a' {p + a"') = -;r- + a'p is the equation required ; and, since iS [tt— -^ J a = 0, it is that of a straight line perpendi- cular to the axis, at the distance 3a from S. Ex. 4. To find the locus of the intersection ivith the tangent of the perpendicular on it from the vertex. If IT be the vector perpendicular on the tangent from A, we have by (52. 8) IT = X (4>p + a~') (1), and the equation of the tangent gives, putting tt + ^ in place of ff in (52. 5), and multiplying by 2, 2S-!r4>p + 2Sa-''n-+2Sa-'p = l (2), we have also ^p(S'a7r = X, which gives x, and STrp)', which substituted in (2) gives 2a; {pY for Sp<^p, equation (3) gives a' {^pY + 2Sar'p = 1 ; therefore by subtraction (2a;-a=)(<^p)' + 2>S'a-7r = 0, i. e. {2Sa.Tr - a') (<^p)' + 250" V = 0, which from (1) becomes, multiplying by S^aTv, (2&i7r - o)= (tt - aT'Sairy + iS'avSa-'n = 0. This equation at once reduces to 2;r'/S'a7r-irV + ;SW = 0, an equation which, when 4a is written in place of a, becomes identical with that obtained in Art. 37. Ex. 8. The locus is therefore a cissoid, the diameter of the generating circle being AD. 65. It ■will probably have suggested itself to the reader, that there exists a large class of problems to which the processes we have illustrated are scarcely if at all applicable. Hence there may have arisen a contrast between the Cartesian Geometry and Quaternions unfavourable to the latter. To remove this un- favourable impression, all that is required in a reader familiar with the older Geometry is a little experience in combining the logic of the new analysis with the forms of the old. He will then see how simple and direct are the arguments which he can bring to bear on any individual problem, and consequently how little the memory is taxed. 112 QUATERNIONS. [CHAP. VII, "We propose in this Article to put the reader in the track of employing hia old forms in conjunction with quaternion reasonings. We shall work several examples on the parabola and the hyperbola. Having applied quaternions pretty fully to the ellipse in what has preceded, we will limit ourselves to a single example in this case. 1. The Parabola. If the unit vector along any diameter of the parabola be u., and the unit vector parallel to the tangent at its extremity be /3 ; we may write the equation of the parabola under the form p = xoL + y^ -t'-^y^ • •■(D- Por the particular case in which the diameter in question is the axis, and the tangent at its extremity parallel to the directrix P = £'* + 2/^ (2), where a is AS (Art. 52). This is the most convenient form when the focus is referred to. In other cases a somewhat simpler form may be obtained by supposing a, or if necessary both a and /8 of equation (1) to be other than unit vectors. The equation may then be written under the form P = \<^ + iP (3). To find the equation of the tangent, we have 2 p'=^'a+«';8; .•.p'-p = (i'-i)('-^'a + i8), iii8a3AiNn immi'j ART. 55.] THE PARABOLA AND HYPERBOLA. 113 Now p' — p is a vector along the secant; and its limit is a vector along the tangent : hence any vector along the tangent is a multiple of tu. + ft ; and the equation of the tangent may be written w=^a + t^ + x{ta + P) (4). Examples. Ex. 1. J/ AP, AQ he chords drawn at right angles to one another from A ; PM, QN perpendiculars on the axis, then the latus rectum is a mean proportional between AM and AN; or between PM and QN. liPM=y, QN=y', AP='{-^a^yP,AQ = '-£a-y'p. :SoxfS{AP.AQ) = (22.7); 2 /a therefore also xx — (4a)l Ex. 2. If the rectangle of which AP, AQ are the sides he completed, the further angle will trace out a parabola similar to the given parabola, the distance between the two vertices being equal to twice the latus rectum. p=AP+AQ 4a (y-y'Y Aa <^+(y-y')P a+ (y — y')P+ 8aa. Ex. 3. The circle described on a focal chord as diameter toucJies the directrix; and the circle described on any other chard does not reach the directrix. T. Q. 8 114 QUATERNIONS. [cH. VII. Let PQ be awy chord, centre 0, The equation of the circle with centre 0, radius OP, is AQ + AP\' _ /'A Q -AP \- 1 )' ( AQ + Aiy /. or p'-S{AP + AQ)p+S{AP.AQ} = 0. At the points in which this circle meets the directrix p = -aa + z^ ; Tliis equation is possible only when 2/2/' + 4a'' = ; i. e. when the chord is a focal chord. In this case the two values of z are equal, each being ; and the directrix is a tangent to the circle. Ex. 4. Two parabolas have a common focus and axis; their vertices are turned in opposite directions. A focal chord cuts them in PQ, P'Q, so that PP'SQQ' are in order. Prove (1) that SP . SP' = SQ . SQ'; (2) that SP : SQ' is a constant ratio; and (3) that the tangents at P, P' are at right angles to one another. The equations of the parabolas are '/' the focus being the origin. AET. 55.] THK PARABOLA AND HYPERBOLA. 115 Now since p, p' are in the same straight line when the common chord is the focal chord, we have y' = py, ••• {yy' - ^0.0!) {a'y + «2/') = O- Taking the former factor, we must have y, y' on the same side of the axis with a constant product ; therefore SP.SP^SQ.SQ'. The second factor gives SP : SQ' a constant ratio a : a'. Lastly, by Equation (4), the tangent vectors at P and P are parallel to therefore the tangents are at right angles to one another. Ex. 5. If a triangle he inscribed in a parabola, the three points in which the sides are met by the tangents at the angles lie in a straight line. Let OPQ be the triangle. Take as the origin, then p'='-'a + «'A w=^a + tp + x{fa. + P}, tr' = %a. + t'fi + x'{l'a + P), 8—2 116 QUATERNIONS. [cH. VII. are the vectors OP, OQ, and the equations of the tangents at P and Q. If QO meet in A the tangent at P, OA = *^a + tp + x{ta + [i) = yOQ ft'' a + t'^; Z t" = 2 2/. t + x = = «y, t' tu'-t" and ^^ = ^^^(1'"-''^) Similarly if the tangent at Q meets PO in B, t 2t-t If the tangent at meets PQ in G, OG=OP + z{PQ) = OP + z{OQ-OP) But 00 = vl3; t' t'^-e „ ART. 55.] THE PARABOLA AND HYPERBOLA. 117 and 00=— -,p. V t to , , 2t-t' ll'-t f-t" „ and also ; r— = ; t t tt ' therefore (Art. 13) ^, B, G are in a straight line. 2. The ellipse. If a, /3 are unit vectors along the axes, the equation of the ellipse may be written .p = xa + yp, where y' = —2 ifl' -o(?)=m (a' - a;") ; and the equation of the tangent will be readily seen to be TT = xa + yP + X{ya. — mx^). A single example will suffice. Ex. If tangents be drawn at three points P, Q, II q/" an ellipse intersecting in B,', Q', P', prove that PR'. QP. RQ' = PQ'. QR'. RF. I{ X, y; X, y ; x", y" are respectively the co-ordinates of P, Q, R; we shall have GR' = xa.+ yji + X {ya- mx/3) = x'a + y'P + X' (y'a - rnx'/S) ; . •. x+ Xy = x' + X'y, y — mXx = y' — mX'x ; . •. mX (s/y — y'x) = mx' + y' — mxx' — yy' = b' — mxx' — yy'. Hence mX' (xy' — x'y) = ¥ — mxx' — yy' = -mX{xy'-x'y); .: X = -X', r=-r for §', .Z^ = -.^'forP', and XTX^-XYZ". 118 QUATEKNIONS. [CH. VII. Now y = -pQ, , «BC, hence the proposition. 3. The hyperbola. If a, p are unit Tectors parallel to the asymptotes CX, CY, the equation of the hyperbola may be written Go = xa+ - B, X Since xy = — -r — = 0. If o, /3 be not" both units we may write the equation under the simpler form P = <<^ + f (1). To find the equation of the tangent, we have as usual a vector parallel to the secant /8> = (^(-f)- and a vector parallel to the tangent will be ta-- (2). ART. 55.] THE PARABOLA AND HYPERBOLA. 119 Hence the equation of the tangent is n^ta + ^ + xfta-^') (3). Cor. It is evident that t' t' are conjugate semi-diameters. Examples. Ex. 1. One diagonal of a parallelogram whose sides are the co-ordinates being the radius vector, the. other diagonal is parallel to the tangent. We have €2^= ta, NQ = ^ , and the other diagonal is td-^, t ' which, equation (2), is parallel to the tangent at Q. Ex. 2. Any diameter CF bisects all the cliords which are pa- rallel to the tangent at P. Let CP be ta + t' Q the tangent at P is parallel to ta- P. ' t' CQ: .CV+VQ = x(ta->r^ + T(ta-^\ But as Q is a point in the hyperbola, this equation must have the form 120 QUATERNIONS. [CH. VII. .-. {X+T)t=T, and , X^-J' = \, an equation which gives two equal values of T with opposite signs, for every value of X. Hence all chords are bisected. Cor. X'-T^l is \CP) \C1)J ~ ^' CD being ia-^ = PO. This is the ordinary equation of the hyperbola referred to conjugate diameters. Ex. 3. If TQ, TQ' he two tangents to the hyperbola intersect- ing in R and terminated at T, T', Q, Q' by the asymptotes ; then (1) TQ' is parallel to T'Q; (2) area of triangle TUT' = area of triangle QRQ', and (3) CR bisects TQ' and T'Q. The equation of the tangent 3 Tr = ta-i h X gives CT=2ta, T=2t, (the coefficient of )3 being 0), CT'=1t'a, 23 CQ'^f; .: Q'T=2at-^f- = ^{att'-/3), QT'J{att'-p): therefore Q'T is parallel to QT'. ART. 5o.] THE PARABOLA AND HYPERBOLA, Again, CR=CQ + QE=CQ+x{CT-CQ) ■f-K-f)- Also cn-J^...(.^-§): .: xt = x't', 1 a; 1 a;' t ■t~ t' t" t' t + t" ^-h- 121 and ica;' = (1 - a;) (1 — x'), i.e. QR. Q'R = RT .R'T, and the triangles TRT, QRQ' are equal. Lastly, (7i? = f+^,(a.-f) < + «' \ «' / or OR is in the direction of the dingonal of the parallelogram of which the sides are CT, CQ'; and therefore CR bisects TQ' and T'Q. Ex. 4. If through Q, P, Q' j^di'allels he drawn to CX meeting CY in E, F, G ; GE, CF, CG are in continued jrroportion. CP = ta + ^; Q'Q = m{tai-^^; = GV+ YQ = x(,..f).r(,.-f). 122 QUATERNIONS. [CH. VII. CE-- = (X- ■^f. CF-. ' t' CG-- -iX.Y)\, CE. CG = CF' > X ,_Y> ' = 1 (Ex. 2). and because Ex. 5. If a chord of a hyperbola he one diagonal of a parallelogram whose sides are parallel to the asymptotes, i/ie ot/ier diagonal passes through the centre. Let the chord be PQ ; p, p the vectors to P and Q ; then QP = p-p' = at + ^^-{at' + ^)j. Now when one diagonal of a parallelogram is ma. + n^, the other will be ma—np. Therefore in the case before us, the other diagonal is And it is therefore in the same straight line with the line which joins the centre of the hyperbola with the middle point of PQ ; whence the truth of the proposition. ART. 55.] THE PARABOLA AND HYPERBOLA, 12.3 Ex. 6. If two tangents to a hyperbola at the extremities Q> Q' of a diameter, meet a tangent at P in the points T, T' ; and if CD, CD' are tJie semirdiameters conjugate to CP, CQ ; then (1) PT : QT :: PT' : Q'T' :: CD : CD'; and (2) FT .PT'=CD\ If t, t', — t', correspond to P, Q, Q', then CT=at + "+x (-f) gives t + xt = t' + x't'. 1 x_l x' t t~ t' 7 ' t'-t ''^t' + t^ ''■ Similarly cr = atJ^y{at-l) gives t + yt = -t'-y't', 1 y 1 y' t t~~t'^l' whence «' + < y-t'-t=-y' Now X : y :: x' : y' gives PT : QT :: PT' : QT' :: CD : CD'. And a-y = l gives PT .Pr=G'D\ Cob. gives ^y'=\, QT . Q'T' = CD'\ 124! QUATERNIONS. [CH. VH. Ex. 7. Straight lines move so that the triangular area which they cut off from two given straight lines which meet one another is constant: to find the locus of their ultimate intersections. Let OAA', OBB' be the fixed lines, AB, A'B' two of tlie moving lines with the condition that oa.ob = oa'.ob: If a, yS be unit vectors along OA, OB, OA = ta, 0B = ul3; OA' = t'a, OB' = u'P, the point of intersection oi AB, AB' gives p = ta + x {u^ — to) = t'a + x (w'/3 — t'a), .'. xu = a^u', and t{\~x) = t'{l-x') -''(■-?)■ Now tu = t'u' = c because the triangle has a constant area ; .-. a; =^^,= 2 ultimately; the equation of a hyperbola. Additional Examples to Chap. VII. 1 . In the parabola SY'^SP.SA. 2. If the tangent to a parabola cut the directrix in R, SR is perpendicular to SP- 3. A circle has its centre at the vertex A oi & parabola whose focus is S, and the diameter of the circle is ZAS. Prove that the common chord bisects AS. 4. The tangent at any point of a parabola meets the directrix and latus rectum in two points equally distant from the focus. ART. 55.] THE PARABOLA AND HYPERBOLA. 125 5. The circle described on SP as diameter is touched by- tlie tangent at the vertex. 6. Parabolas have their axes parallel and all pass through two given points. Prove that their foci lie in a conic section. 7. Two parabolas have a common directrix. Prove tliat their common chord bisects at right angles the line joining their foci. 8. The portion of any tangent to the parabola between tan- gents which meet in the directrix subtends a right angle at the focus. 9. If from the point of contact of a tangent to a parabola a chord be drawn, and another line be drawn parallel to the axis meeting the chord, tangent and curve ; tliis line will be divided by them in the same ratio as it divides the chord. 10. The middle points of focal chords describe a parabola whose latus rectum is half that of the given jarabola. 11. FSQ is a focal chord of a parabola : PA, QA meet the directrix in y, z. Prove that Pz, Qy are parallel to the axis. 12. The tangent at B to the conjugate hyperbola is parallel iaCP. 13. The portion of the tangent to a hyperbola which is in- tercepted by the asymptotes is bisected at the point of contact. 14. The locus of a point which divides in a given ratio lines which cut off equal areas from the space enclosed by two given straight lines is a hyperbola of which these lines are the asymp- totes. 15. The tangent to a hyperbola at P meets an asymptote in T and TQ is drawn to the curve parallel to the other asymp- tote. PQ produced both ways meets the asymptotes in R, R' : BR' is trisected in P, Q. 12G QUATERNIONS. [CHAP. VIT. 16. From any point H of an asymptote, E^, EM axe drawn parallel to conjugate diameters intersecting the hyperbola and its conjugate in F and D. Prove that CF and CD are conjugate. 17. The intercepts on any straight line between the hyper- bola and its asymptotes are equal. 18. If QQ' meet the asymptotes in F, r, FQ.Qr = FO\ 19. If the tangent at any point meet the asymptotes in X and Y, the area of the triangle XCY is constant. CHAPTER VIII. CENTEAL SUEFACES OF THE SECOND OEDER, PAETICULAELY THE ELLIPSOID AND CONE. 56." The Ellipsoid. In discussing central surfaces of the second order, we shall speak as if our results were limited to the ellipsoid. That such limitation is not, in most cases, necessarily imposed on us, will be apparent to any one who has a slender acquaintance with ordinary Analytical Geometry. We adopt it in order that our language may have more precision, and that, in some instances, our analysis may have greater simplicity. If the centre be made the origin it is clear that the scalar equation can contain no such term as ASap, for the definition of a central sur- face requires that the equation shall be. satisfied both by +p and by -p. If we turn to the equation of the ellipse (Art. 43), we shall see at once that the equation of the ellipsoid must have the form ap' + hS'ap + 2cSapSPp + . . . = 1. Now if, as in the Article referred to, we put p in its present form. 57. To find tlie equation of the tangent plane. Let a secant plane pass through the point whose vector is p ; and let p' be the vector to any point of section. Put p' = p + p, ■where j8 is a vector along the secant plane ; then Sp'i,p' = S{p + P)cl>{p + j3). Hence, observing that (44) {p + l3) = i>p + ct>p, and Sp(l>^ = S^p, we have Sp'p + 2SP4>p + 8/3^/3 ; i.e. 2S/Scj>p + ,Sl3cl>p=0. Now (45), as the secant plane approaches the tangent plane, the sum of these two exjjressions approaches in value to the first alone : that is, for the tangent plaae, Sfi(f>p = 0, where /3 is a vector along that plane. If TT be the vector to a point in the tangent plane, ■n- = p + xj3 ; . ". S{Tr — p)(l>p = xS^p = Sp^p = 1 is the equation of the tangent plane. Cor. (f>p is a vector perpendicular to the tangent plane at the extremity of the vector p. 58. If OF be perpendicular from the centre on the tangent plane; then, since 4>p is a vector perpendicular to that plane, OY=X(f>p and Sx{cj)py=l, giving 0Y=T{x4>p) = T—. Sir W. Hamilton terms <^p the vector of proximity. ART. 61.] CENTRAL SURFACES OF THE SECOND ORDER. 123 59. If tanj»ent planes all pass through a fixed point, the curve of contact is a plane curve. Let T be the fixed point ; vector a ; p the vector to a point of contact. Then (Art. 57) Sap = 0, or Spa. = 0, the equation of a plane through the origin perpendicular to <^a r that is, the curve of contnct lies in a plane through the centre parallel to the tangent pLtne at the extremity of the diameter which is parallel to the given line. 61. To find the locus of the middle points of parallel chords. Let each of the chords be parallel to o, ir the vector to th*; middle point of one of them; then ir+cca, ir — aki are points in the ellipsoid. From the first, a + x'Safpa = 1. T. Q. 9 130 QUATERNIONS. [CHAP. VIII. rrom the second, S-irTr — IxSir^a + a?Sa<^a = 1 ; ',•. subtracting, STr(j)a=0 (1), i. e. the locus is a plane through the centre perpendicular to a, or parallel to the tangent plane at the extremity A of the diameter which is drawn parallel to a. If we call this the plane BOC, B and G being any points in which it cuts the ellipsoid ; and if OB = P, 00 = y, we shall have and therefore So.<^P = 0, or a satisfies the equation Sit^P = of the plane which bisects all chords parallel to OB (Equation 1). Let AOO be this plane which bisects all chords parallel to OB. Then, since 00 or y is a vector in it, SyP = 0, i.e. Sfty = 0. But we have already proved that Syy = 0, which is the plane bisecting all chords parallel to y ; that plane is therefore the plane AOB: we are thus presented with three lines OA, OB, OG such that all chords parallel to any one of them are bisected by the diametral plane which passes through the other two. We may term these lines conjugate semi-diameters, and the corresponding diametral planes conjugate diametral planes. It is evident that the number of conjugate diameters is un- limited. Cor. We have the following equations : Sa<\>p = = Spa, SI3y = = Sy4.(i, ,?a^y = = >S'y<^a (2). ART. 62.] CENTRAL SURFACES OF THE SECOND ORDER. 131 They shew that y is perpendicular to both <^a and <^j3, and is therefore a vector perpendicular to their plane ; hence, as in 34. 4, y = xVp. In the same way, since <^y is perpendicular to both a and )8, we have y = yVaP; or, neglecting tensors, we have the following vector equalities : y = F^a<^/3, j8 = ry, a = F<^y3.^y, y=ral3, p=Vay, ,^a = F^y (3). Note also upon which Hamilton founded his solution of linear equations. 62. If as iu Art. 47 we write — ^t^p for <^p, x^/p being still a vector, the equation of the ellipsoid assumes the form i.e. (44) Stjfpfp = -1 (^py = -T{rl.py = -l.. (1), which, if we put (r= ijrp, becomes Tcr= 1, the equation of a sphere. Hence the ellipsoid can be changed into the sphere and vice versd, by a linear deformation of each vector, the operator being the function xj/ or its inverse. The equations Sacl>p = 0, (tc. now become Satl/'B = 0, i.e. >S'i/'ai///3 = 0, &c., &c (2). (1) and (2) shew that i/fa, i/^/S, ^y are unit vectors at right angle to one another. If we term the sphere Tu = \ the unit-sphere, we may enunciate this result by saying that the vectors of the unit-sphere which correspond to semi-conjugate diameters form a rectangular system. 9—2 132 QUATERNIONS. [CHAP. VIII. 63. Let VIS now take i, j, k unit vectors along the principal axes of X, 1/, z; then we shall have p=^xi + yj + zk (1), .•. Sip = — x, &c. 80 that for the sake of transformations in ■which it is desirable that the form of p should be retained, we may write p = -{iSip+jSjp + kSkp) (2); and as p is a linear and vector function of p, its vector portions along the principal axes will be multiples of iSip, jSjp, kSkp ; we may therefore write iSip jSjp kSkp . 't'p^ir^'w'^^^ ^^'' the form a' having been assumed in order to make the equation Spp = 1 coincide with the Cartesian equation x' y' z' , 1- — + 1 a' b' c'~ As (}>p = -ip>j;p (4), we require to take ij/p so that performing the operation tj/ twice on p shall give the same result (with a — sign) as performing the operation -'p = aSSip + b'jSjp + c''kSkp (7), because 4"K^9 produces p. \lf~^p= — {aiSip + bjSjp + ckSkp) (8), p = \j/-'}J/p = - (aiSiij/p + bjSjipp + ckSktj/p) (9). It is evideDt that the properties of Art. 44 apply to all these functions. » 64. Examples. Ex. 1. Find the point on an ellipsoid, the tangent plane at which cuts off eqiml portions from the axes. Let X, y, % be the co-ordinates of the point, p the portion cut oflF, then p-=xi + yj + zk. Now pi, pj, pk are points on the tangent plane ; .*. Spip= 1, which gives 134 QUATERNIONS. [CHAP. VIII. or py 1 Similarly 1J~ ' X y z \ a' b' c' p Ja'+b' + c'' Ex. 2. To find the perpendicular from the centre of the ellipsoid on a tangent plane. 0Y'^{t1>J; (Art. 58) .-. ±,=(T.^p)^ = -(<^py = ^ + |;+ i' (Art. 63, 1. 3). Ex. 3. To find the locus of the points of contact of tangent planes vihich make a given angle with the axis of z. We have SkV{4>p)=p, Skp=pTTr = SirS, if S be ir, Sir = Sir'ir ; .'. Sir'ir = — C is the equation required ; hence the Cartesian equation is (63. 6) 3 3 3 a o c Ex. 5. The sum of the squares of three conjugate semi-dia- meters is constant. Let a, p, y be the semi-diameters ; if/a, >/f/3, i/ry are rectangular unit vectors (Art. 62). Now a = — (aiSifa + bjSjfa + ckSkij/a.) (63. 9) ; .-. (TaY = -a'=a' (Si^^a)' + ¥ (-S)>a)= + c' {SkfaY, {TPY = a' (Si^l^PY + S' i'SjfPy + o' i^Hl^f, {TyY = a' {SixlryY + V (Sj^PyY + c' (Sk^PyY : adding, and observing that {Si4,-'p=-a''iSip + b^jSjp-hc''kSkp (63. 7), and <^p = - {iSip +jSj^p + hSk^p) (63. 2) ; .-. Sp4>p = 1 = o' {Si<^pY + V {SjpY + c- {Skcl>pY = {t^pY {a" {Si upY + V {Sju^pY + e {Sk up'' are at right angles to each other ; that is, so that the tangent j)laiies at their extremities are at right angles to one another (57. Cor.). 1 1_ _ 1 (Tpr-{T^p'y-(Tp"y = a' {{Siup"y\ + ¥{{SjU4,py + ...} + ... = a? + 1^ + 0^ (31. Cor.). But 7^-7— Tj , «tc. are the perpendiculars from the centre on the (■' W tangent planes at p, p, p" (58). Hence the proposition. Ex. 7. The sum of the squares of the projections of three con- jugate diameters on any of the principal axes is equal to the square oj that axis. Let a, /?, y be conjugate semi-diameters ; then, since o = - [aiSixf/a + bjSjipa + ckSkfa) (63. 9), Sia = aSiij/a. Similarly, Si/i = aSiif/P, Siy = a&'i/f-y ; .-. {Siay + (Si^y + {siyy = a' {(Si,j,i.y + (Siii^py + (&Vy)'} = a' (31. Cor.), because ipa, xj/ji, KJiy are at right angles to one another (62). But — /Sia is the projection of 7a along the axis of x ; and similarly of the others. Hence the proposition. Ex. 8. The sum of the reciprocals of the squares of the three perpendiculars from the centre on tanyent planes at the extremities of conjugate diameters is constant. Let Oy^, Oy^, Oy^ be the perpendiculars, i^ = -(,^a)= (58) ^ = {Sif,(3f^(^ a* b* c* ^ ' ART. p4,] CENTRAL SURFACES OF THE SECOND ORDER. 137 1 JSiPY (SjPY {S/clSY 1 _(SiyY ASjyY (Sky)\ Oy^ a* b' ^ c* ' 111/,.-, -TV -a-'-b'^? (^"•^)- Ex. 9. If through a fixed point within an ellipsoid three chords he drawn mutually at right angles, t/ie sum of the recipro- ods of tlie products of their segments will be constant. Let 6 be the vector to the given point ; a, ^, y unit vectors ])arallel to three chords at right angles to each other. Then 6 + xa = p gives S(0+xa){e + xa) = l ii quadratic equation in x, the product of whose roots is se^e - 1 . Saa . •. tlie product of the reciprocals of the segments of the chord is 1 Satba 1 x,ax^a^ ^0e-l ' {Toy' and the sum of the reciprocals of the products of the segments is 1 ( Sa^a S^P Sy^ ISe^e - 1 • {{Taf "^ {T^f'^ [Tyf]' ^ . „ , (Sia)" (Sja)' (Ska)' Now since Saa = — ^— + ^ ^ + ' — j — (63. 2, 3), the sum of the reciprocals of the products + l[{Sjay+{Sjpy+{Sjyy] 13S QUATERNIONS. [CHAP. VIII. + l{{Sf^-)'+ }] Cor. If 6 be not constant, but S60 be so, i. e. if the given point be situated on an ellipsoid concentric with and similar to the given ellipsoid, the same is true. Ex. 10. If the poles lie in a plane parallel to yz, the polar planes cut the axis of x always in the same point. Let pi be the distance from the origin of the plane in which the poles lie, 8 any line in that plane, then ir =pi + 8 is the vector to a pole, and Sp4, {pi + 8) = 1 (59) the equation of the corresponding polar plan^. At the point where this plane cuts the axis of x, p = xi; .: Spod<^i + xSi<^h = 1. Now 8 is a vector in a plane perpendicular to <^t, .-. Sil = SZp = a the equation of A, Sp(j>p = b B, S{p-a){p-a)^C C. AKT. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 139 No-w at the intersection of A and G, p is the same for both ; therefore the equation of the plane of intersection is to be found by subtracting the one from the other. It is therefore 2Spp=\ (1), S{p-a)a; and, since by the question, the former is perpen- dicular to the latter, ^tt is perpendicular to ^n- - ^a, .•. iS-ir (tf}ir — aj = 0, the equation of the locus of the centres of the cutting ellipsoids. This equation will be reduced to the requisite form by ob- serving that Scj>Tr(j)Tr = SirTra = &'a'ir ; .-. S (w - a) '7r = 0, the equation of an ellipsoid of which the semi-axes are propor- tional to a', b', c' (63. 6). The Cartesian equation is or b* c* \a* 6* c* / ART. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 141 Ex. 14. If p =1 of the outer ellipsoid. The tangent plane is Sirp = 1. Now if o- be the vector to the elliptic section measured from the point of contact, ir = p + o- is a point in the outer ellipsoid ; .'. a'S (p + 0-) (^ (p + p = (57. Cor.); .•. a' + a'SiTfjxr = 1, T — -5'S'o-^o-= 1, the equation of an elliiise of which the centre is the point of contact. Ex. 15. Fiiid tlie equation of the curve described by a given point in a line of given length whose extremities move in fixed straight lines. Eii-st, let the straight lines lie in one plane. Let unit vectors parallel to them be a, y3. Let the vectors of the extremities of the moving line be xa, y/J, and its length I. Then the condition is (y/3-CDa)' = -r, or x' + y' + 2xySa^ = l' (1). The vector to a point which divides this line in the ratio e : 1 is p = xaA- e {yP - xa) = a;o(l —e) + eyP; . : Sap = — (1 - e) a; + eySa^, SPp = {l-e)xSaP~ey; 142 QUATERNIONS. [CHAP. VIII. Sap + Sa^SPp _Sj3p+SaPSap which values being substituted in equation (1) give the required equation, viz. : (Sap-J-SaPSPp)' (SI3p + SaPSap)' ■ {l-ey + H 2 f°^ (^ap + SapS^p) {Spp + SapSap) e(l e) But p is subject to the additional condition (31. 2. Cor. 2) S.a/ip = ; and the locus is a plane ellipse. When the given straight lines are at right angles to one another, the equation is much simplified, for and our equations are Sap = -{l-e}x, SPp = -ey; whence (T^^ + "1^ = ' ' an, ellipse of which the semi-axes are le and ? (1 — e). Generally, if the given lines do not meet, let the origin be chosen midway along the line perpendicular to both j then we have y and — y being the vectors perpendicular to the lines, p = (y + aa) (1 - e) + e (- y + 2//3). The first gives and the second gives, as in the simpler case above, Sap = — (1 - e) as + eySaP, SPp = Q.-e)xSa^~ey. ART. 65.] CENTRAL SURFACES OF THE SECOND ORDER. 143 Hence the elimination of x and y again leads to the equation of an ellipsoid, the only difference being that I' is diminished by the square of the shortest distance between the lines; i.e. the axes are less than in the former case. In the extreme case, where l=2Ty, the equation cannot be satisfied except by x = 0, y = 0, (i. e. the locus is reduced to a single point), unless indeed -we have for then a; = ± y, and the locus is a straight line parallel to each of the preceding lines. 65. !r/ie cone. 1. To find the equation of a cone of revolution ■whose vertex is the origin 0. Let a be a unit vector along the axis OA, p the vector to a point P on the surface of the cone ; then Sap = —Tp cos 9, 6 being the angle POA. But this angle is constant, .■. S'ap = p = 2a'p + aSI3p + pSap, the equation of the cone is reduced to Spp = 0. ABT. 67.] CENTRAL SURFACES OF THE SECOND ORDER. 145 It is evident that all the properties of p, Art. 44, are appli- cable here. As in Art. 57, the equation of the tangent plane is Svp = 0. 67. Examples. Ex. 1. Tangent planes are drawn to an ellipsoid from a given external point, to find the cone which has its vertex at the origin, and which passes through all the points of contact of the tangent planes with the ellipsoid. Let a be the vector to the external point, p a point in the ellipsoid where a tangent plane through a touches it. Then the equation of the ellipsoid is Spp=\, and the equation of the tangent plane *Sa<^p =1, i. e. Spf\>a = 1. The equation Sp<^p = {Spi^af, tc' y* «* /xx yy' sa'y '*'' a^'^W^c'^W^W'^'^)' represents a surface passing through the points of contact; and is the cone required. Ex. 2. Of a system of three rectangular vectors two are con- fined to given planes, to find the surface traced out by the third. Let TT, p, a- be the three vectors, of which two are confined to given planes whose equations are Sair = 0, Sjip^O, to find the locus of tr. Since the vectors are at right angles, we have iSirp = 0, >S'7ro- = 0, S(Tp = 0, and we have five equations from which to eliminate ir and p. Since Sair = 0, S(nr = 0, IT is at right angles to both a and S'<7p = 0, jo is at right angles to the plane /3cr ; therefore p = 2/F;8cr, and irp = xt/ Vaa- Vpcr. Now Sirp = 0, therefore S . Vaa V/Sfr = 0, or ^ (ao- - Saa) (^a - S^a) = 0, or (T'Sa^-Sa(rSI3(T = 0, the equation of a cone of the second order, which has circular sections (65. 2). Cor. The circular sections are parallel to the two planes to which the two vectors are confined. Ex. 3. The equation p — t'a + u'P + (t +uy y=0 is that of a cone of the second order touched hy each of the three planes through OAB, OBG, OCA ; aiwL the section ABC through the extremities of a, P, y is an ellipse touched at their middle points hy AB, BC, CA. 1. If the surface be referred to oblique co-ordinates parallel to a, p, y respectively, we shall have p = xa + yP + zy, therefore x-t, y-u', z = {t+ uy, or fi=Uoc + JyY=x + y+2>Jxy, which gives {z—x — yY = 4a;y, a cone of the second order. 2. If i = — M, the equation becomes P = t'{a + I3), the equation of a straight liije bisecting the base AB, which since it satisfies the equation relative to t, shews that this line coincides with the cone in all its length j i. e. the cone is touched in this line by the plane OAB. Similarly, by putting t = 0, u = respectively, we can shew that the cone is touched by the plane BOO, COA in the lines which bisect AO, OA. ABT. 67.] CENTRAL SURFACES OF THE SECOND ORDER. 147 3. Restricting ourselves to the plane ABC, we Lave the section of a cone of the second order enclosed by the triangle! ABC, ■which triangle is itself the section of three planes each of ■which touches the cone. Ex. 4. The equation p = aa + h^ + ey with the condition ah + bc + ca = is a cone of the second order, and the lines OA, OB, OC coincide throughout their length with the surface. 1. It is evident that the equation gives xy ■'eyz-v zx= 0. 2. That if 6 = 0, c = 0, the question is satisfied by p = aa, ■whatever be a, therefore &c. Ex. 5. Find the locus of a point, the sum of the squares of whose distances from a number of given planes is constant. Let (S^SjP, = C„ S^J)^ = Cj, ^°-> therefore /S'8, (p + a;,8,) = C„ &c. and a!,8,» = C,-,SS,p, &c., i.e. the square of the line perpendicular to the first plane from the given point and, by the question, f 'yg ' j + ( 'yg • j +&C- IS constant* The locus is therefore a surface of the second order. 10—2 148 QUATERNIONS. [CH. VIII. Ex. 6. The lines which divide proportionally the pairs of opposite sides of a gauche quadrilateral, are the generating lines of a hyperbolic paraboloid. Let ABGD be the quadrilateral. AD, BG are divided proportionally in P and R. Let GA = a, GB = p, Ci) = y; CR = m^, DP=-mDA; ie. CP-y = m{a.-y); therefore BP=GP -GR = y + m{a-y)-mP, p = GQ=GR+pRP = mp + p{y+m(a — y) — mfi} = xa + yP + zy, say; therefore x = pm, y = m- pm, s =^ (1 - m) ; X therefore m = x + y, p = x + y' x + y or {x + s)(x + y)= x, the equation referred to oblique co-ordinates parallel to a, /5, y. Pascal's Hexagram. 68. Let be the origin, OA, OB, OC, OB, OB five given vectors lying on the surface of a cone, and terminated in a plane section of the cone ABGDEF, not passing through ; OX any vector lying on the same surface. Let OA = a, OB = p, Oq = y, OD = h, 0E = €, OX=p. The equation S. F(Fa;8F8£) V{7Py7(p) F(Fy8Fpa)=0 (1) is the equation of a cone of the second order whose vertex is and vector p along the surface. For AKT. 68.] CENTRAL SURFACES OP THIJ SECOND ORDER. 149 1. It is a cone whose vertex is because it is not altered by writing xp for p. Also it is of the second order in p, since p occurs in it twice and twice only. 2. All the vectors OA, OB, 00, OD, OE lie on its surface. This we shall prove by shewing that if p coincide with any one of them the equation (1) is satisfied. If p coincide with a, the last term of the left-hand side of the equation, viz. Vpa, becomes Foa = Fo" = 0, and the equation is satisfied. If p coincide with )8, the left-hand side of the equation be- comes s. viVaprSi) r(rpyVep) r{rysr^a) (2). Now F(F/3yF£;8) =- riVcpvPy), (22. 2), is a vector parallel to P (31. 3), call it «i/3 j and F.{F(Fay3FS€) F(Fy8F/?a)}= F.{F(Fa;8FS£)F(Faj8FyS)}, (22. 2), = a multiple of Vafi, (31. 3), = wFo/8, say. 150 QUATEENIONS, [CH. VIII; Hence the product of tlie first and third vectors in expression (2) becomes scalar + n VaP, . and the second is mP; therefore expression (2) becomes, by 31. 2, S . (scalar + n Va^) m^ = Smn/BVaP = 0, because Fa;8 is a vector perpendicular to /3. Equation (1) is therefore satisfied when p coincides with yS. If p coincide with y both the second and third vectors are parallel to P (31. 3)j therefore their product is a scalar, and equa- tion (1) is satisfied. The other cases are but repetitions of these. Hence equation (1) is satisfied if p coincide with any one of the five vectors a, j3, y, 8, c; ie. OA, OB, OG, OD, OE are vectors on the surface of the cone. 3. Let F be the point in which OX cuts the plane ABODE ; then ABODEF are the angular points of a hexagon inscribed in a conic section. 4. Let the planes OAB, ODE intersect in OP; OBC, OEF in OQ; OCD, OF A in OR; then V. ral3r8e = mOP, (31. 4), F. VPy7ip = nOQ, V.YyWpa. = pOR; therefore >S'. F(Faj8FS£) V{VPyV€p) 7{7yZ7pa) = mnpS(pP. OQ.OR); hence equation (1) gives S{OP.OQ.OR) = 0, or (31. 2. Cor. 2) OP, OQ, OR are in the same plane. Hence PQR, the intersection of this plane with the plane ABCDEF is a straight line. But P is the point of intersection of AB, ED, &c. ART. 68.] CENTRAL SURFACES OF THE SECOND ORDER. 131 Therefore, the opposite sides (1st and 4th, 2nd and 5th, 3rd and 6th) of a hexagon inscribed in a conic section being produced meet in the same sti-aight line. Cor. It is evident that the demonstration applies to any six points in the conic, whether the lines ■which join them form a hexagon or not. Additional Examples to Chap. VIII. 1. Find the locus of a point, the ratio of whose distances from two given straight lines is constant. 2. Find the locus of a point the square of whose distance from a given line is proportional to its distance from a given plane. 3. Prove that the locus of the foot of the perpendicular from the centre on the tangent plane of an ellipsoid is {axy + (hyf + {czf = {x' + y' + zj. 4. The sum of the squares of the reciprocals of any three radii at right angles to one another is constant. 5. If Oy,, Oy^, Oy^ be perpendiculars from the centre on tangent planes at the extremities of conjugate diameters, and if Gi> Qii Qa ^^ tli^ points where they meet the ellipsoid; then 1.1 1 _1 i 1 OY^'. OQ,' OYJ'. OQ,' 0Y^\ OQ^^ a' b* c* G. If tangent planes to an ellipsoid be drawn from points in a plane parallel to that of xy, the curves which contain all the points of contact will lie in planes which all cut the axis of z in the same point. 7. Two similar and similarly situated ellipsoids intersect in a plane curve whose plane is conjugate to the line which joins the centres of the elKpsoids. 8. If points be taken in conjugate semi-diameters produced, at distances from the centre equal to p times those semi-diameters respectively; the sum of the squares of the reciprocals of the 152 QUATERNIONS. -[CH. Vlrf. perpendiculars from the centre on their polar planes is equal to jp' times the sum of the squares of the perpendiculars from the centre on tangent planes at the extremities of those diameters. 9. If P be a point on the surface of an ellipsoid, PA, PB, PC any three chords at right angles to each other, the plane ABO -will pass through a fixed point, which is in the normal to the ellipsoid at Pj and distant from P by 2 P i. i i' ■where p is the perpendicular from the centre on the tangent plane at P. 10. Knd the equation of the cone which has its vertex in a given point, and which touches and envelopes a given ellipsoid. CHAPTER IX. FOHMtTL^ AND THEIR APPLICATION. 69. Products of two or more vectors. 1. Two vectors. The relations which exist between the scalars and vectors of the product of two vectors have already been exhibited in Art. 22. We simply extract them : (a) SaP = SPa. (b) Va.p = -7l3a. (c) aP + Pa=2Sap. (d) a^-Pa^^Va^. These we shall quote as formulae (1). 2. "We may here add a single conclusion for quaternion products. Any quaternion, such as a/3, may be written as the sum of a scalar and a vector. If therefore q and r be quaternions, we may write q = Sq+Vq, r = Sr +Yr; therefore qr = SqSr + SqVr + Sr Vq + Vq Vr, and S . qr = SqSr + S . VqVr, V. qr = SqVr + SrVq + V. VqVr, where S . VqVr is the scalar part, and V. VqVr the vector part of the product of the two vectors Vq, Vr. If now we transpose q and r, and apply (a) and (6) of for- mulae 1, we get S.qr = S.rq 1 V.qr + V.rq = 2 (SqVr +SrVq)) ^ '' 154 QUATERNIONS. [CH. IX. 3. Three vectors. By observing that ^S*. ySa^ is simply the scalar of a vector, and is consequently zero, we may insert or omit such an expression at pleasure. By bearing this in mind the reader will readily apprehend the demonstrations which, follow, even in cases where we have studied brevity. S.aPy = S.{SaP+Va^)y = S.yVo.p, (byl. a), = S.y{SaP+VaP) = S.ya^ (3). Again, S.apy = ,S . a (SI3y + F/3y) = S{r/iy.a),(hyl.a), = S{SPy+r^y)a = '^-;8ya (3). The formulae marked (3) shew that a change of order amongst three vectors produces no change in the pcalar of their product, provided the cyclical order remain unchanged. This conclusion might have been obtained by a different pro- cess, thus : In (2) let q = a)3, r = y, there results at once S.a^y = S.yal3. Again in (2) let q = ya, r = p, there results S.yaP=S.Pya. We have therefore, as before, S.a^y = S.yaP = S.pya. (3). 4. S.apy = S.aVPy = -S.aVy^, (by 1.6), = -S.ayl3 (4). Similarly S.a^y^-S.^ay (4), or a cyclical change of order amongst three vectors changes the sign of the scalar of their product. ABT, 69.] FOrvMUL.E AND TEEIR APPLICATION'. 155 5. It has already been seen (Art. 31. 1) that — *S'. a/3y is the volume of the parallelepiped of •which the three edges which terminate in the point are the lines OA, OB, 00 "svhose vectore are a, j8, y respectively. We may express t.liis volume in the form of a determinant, thns : Iiet a, )S, y be replaced by xi-vyj^zk, xi + y'j + z'k, x"i + y"j+z"k {Axt. Z\. 5); X, y, z being the rectangnlar co-ordinates oi A, x, y', z those oi B, a:", y", z" those of G, measured from as the origin ; then S . aj3y = S . (xi + yj + sJc) X {x'i + yj + z'h) x{x"i + y"j + z"k). Isovr if -we observe first that the scalar part of this product is cxjnfined to those terms in ■which all the three vectors i, j, k appear ; and secondly that the sign of any term in the product ■will by formulse (3) and (4) be — or + according as cyclical order is or is not retained, ■we perceive that ■we have the exact con- ditions ■which apply to a determinant ; therefore S.aPy = - x, y, s \ *', y', z' (5). rf If It x , y , z The volume of the pyramid OABC is one-sixth of the above. Xote relative to the siffii of tJie scalar. Since ijJc = —l (19), it is clear that if OA, OB, 00 assume the positions of Ox, Oy, Oz in the figure of Art. 16, S'y8 + VySSafi - 8S. aj3y + yS . afiS, by (12). The two expressions being equated, and the common terms deleted, there results BS.a^y= rapSyB+ V^ySaB + VyaS/SB (14). 15. S.al3yS = S.{S.al3y+r. aj3y) 8 = S.{r.aPy)8 = S. (aSjSy - pSay + ySa^) 8, by (10), = Sa^SyS-SayS^S + Sa8S^y (15), 16. SiVapry8) = S.{aP-Sal3){yB-SyB) = S.a^yS-Sa^SyS = Sa8SI3y-SaySpS, by (15) (16). 17. S.apyS = S.(Va^y)B = S.Bra^y = S.Sal3y (17). - ART. 70.] FORMULA AND THEIR APPLICATIOK. 159 18. Five vectors. As we do not purpose to exhibit any applications of the relations which exist among five or more vectors, we shall confine ourselves to simply writing down the two following expressions. S.aPySi = -S.€Syl3a, r.al3ySe=V.€Byl3a (18). 70. Many of these formulte might have been proved difier- ently, and some of them more directly, by assuming for instance that a, /3, y are not in the same plane. In this case anj/ other vector 8 may be expressed in terms of u, jS, y, by the equation S = JBa + yj8 + «y, (31. 5); therefore S. PyS = xS . ^ya = xS.a/3y, (3), S.ySa=7/S. yl3a = -yS. a/Sy, (4), S.SaP = /i;S.ya.l3 = zS. ajSy, (3) ; therefore SS . aySy = xaS . a/Sy + y^S . a/Sy + zyS . a/3y = OaS. /8yS - ^S. ySa + yS . 8aj8 which is formula 13. 71. Examples. Ex. 1. To express the relation between the sides of a spherical triangle and the angles opposite to tltem. Retaining the notation and figure of Ex. 2, Art. 29, we shall have Fa)8 F/?y = y sin c . a' sin a, where y, a are unit vectors perpendicular respectively to the planes OAB, OBC. Tlierefore F. Fa/S VPy = sin c sin a . /3 sin B. Also -pS.aPy = ^ sine sin<^, (31. 1), where ^ is the angle laetween OC and the plane OAB. Now these results are equal (formula 11), therefore sin <^ = sin a sin B, 160 QtTATEENIONS, [CH. IX. Similarly sin = sin h sin A ; therefore sin a sin 5 = sin 6 sin A, or sin a : sin 6 :: sin^ : sin 5. Ex. 2. To Jmd the condition that the perpendiculars Jrom the angles of a tetrahedron on the opposite faces shall intersect one another. Let OA, OB, 00 be the edges of the tetrahedron (Fig. of Art. 31), o, j3, y the corresponding vectors. Vector perpendiculars from A and\B on the opposite faces are Vj3y, Vya respectively (22. 8). If these perpendiculars intersect in 'i3 \n q) \q m/ \m n/ that o{ ACB, A'C'B' is V^ 9'/ W "»/ ^ W pJ ' and that of BCD, B'C'D' \p qJ \q nj ' \n pj Now to prove that any three of these lines lie in the same plane, all that is necessary is to prove (31. 2. Cor. 2) that the scalar of the product of their vectors equals 0. If we take the vectors of the first three, we may write them under the form aa + bp + cy, a' a + 6'/3 + cS, a" a + h'y - bS, respectively ; so that the scalar of their product is iS. {aa + 6/3 + cy) (a'a + 6'y3 + c8) (a"a. + b'y - 58). Now the coefficient of every difi"erent scalar in this product is separately equal to 0. That of S. afiy for instance is, omitting the common factor b', \n p/ \q m) \m nj \p q) \p mj \n q) ' in which every term vanishes. That again of S. /8yS is , -bcb' + cb'b, which is ; and so of the rest. Hence the intersections, two and two, of the first three pairs of planes lie in the same plane ; and the same may be proved in like manner of any other three: whence the truth of the pro- position Ex. 6. CP, CD are conjugate semi-diameters of an ellipse, ART. 71.] FORMULA AND THEIR AtPLICATION. 16-5 as also OP, GD' ; PP, DD' are joined ; to prove that tlie area of the triangle POP equals that of the triangle DGD'. Let a, P, a', P' be the vectors GP, GD, GP„ CD' ; k a. unit vector perpendicular to the plane of the ellipse. Since a==ij/~'tj/a = — (aiSi\j/a + bjSjij/a.), &o., &c. (47. 5), therefore Vaa = V. {aiSiil/a + hjSj^a) (aiSi\j/a' + hjSjfa) = ahk {SifaSjfa - Sjij/aSiij/a) = aJbkS . k V (^af a'). (Formula Ifi.) Similarly V^^ = alkS . k V Wf/B'). Now \j/a, \j/P are unit vectors at right angles to one another ; as are also \j/a', xfifi' ; therefore the angle between \j/a, and i/fa' is the same as that between ffi and i/fy3'. Hence S.kV (ij/aipa') =S.kV (i/'/?i/')8'), and Vaa:=Vpp', i. e. area of triangle PGP' = that of triangle DGB'. Ex. 7. If a parallelejnped he constructed on the semi-con- jugate diameters of an ellipsoid, the sum of the squares of the areas of the faces of the parallelepiped is equal to the sum of the squares of the faces of the rectangular parallelepiped constructed on the semi-axes. By 63. 9, a = - {aiSi\(/a + bjSjtj/a + ckSkij/a) j8 = - (aiSifP + bjSjtpP + ckSkfP) ; therefore FajS = abk (Siif/aSjij/P - Si^j/pSjfa) + acj (SifaSkfP - Sixl/^Skipa) + bci {Sj>l/aSkfl3 - Sjxj/liSkfa). Now Si>paSJi>l3-Sifl3Sj^y + (S{>l>yy = 1, we shall have ( Vapy + ( Vayy + ( F/Sy)' = - {(a&)» + (ac)' + (6c)'}, which (21. 4) is the proposition to be proved. Ex. 8. To find the locus of the intersections of tangent planes at the extremities of conjugate diameters of an ellipsoid. Let TT be the vector to the point of intersection of tangent planes at the extremities of a, ji, y : then STrcl>a= 1, (57), gives Ari/fV = — 1, or Sx(/Tn{'a = — 1, StJ/iTij/y = — 1 . From these three equations we extricate ij/v by means of for- mula (14), which gives xj/irSij/a^l/litl/y = Fi/'a^/3*S'i/'jr^y + Yij/PilrySij/Tr^a therefore ^tt = Y<^a.\l/ji + Vxj/Ptj/y + V\jr/\j/a = '/'V + = -(1 + 1 + 1) = -3, ^■*.^+^=l- 3a' 36" 3c' ' an ellipsoid similar to the given ellipsoid. ART. 71.] FORMULjE and THEIR APPLICATIdN. 167 Ex. 9. If 0, A, B, 0, JD, E are any six points in space, OX any given direction, OA', OB', 00', OD', OE' the projections ofOA, OB, 00, OD, OE on OX; BODE, ODEA, DEAB, EABC, ABCD the volumes of the pyramids whose vertices are B, G, D, E, A, with a positive or negative sign in accordance with the law given in tlie note to 69. 5 j then OA'.BCDE + OB'. CDEA + 00'. DEAB + OD'.EABO + OE'.ABCD = 0. Let OA, OB, 00, OD, OE be o, p, y, 8, e respectively. Write for aS (y - 13) (S - jS) (e - /3) its value a {S. ySe - S. Se/3 + S.€Py-S. jSyB), and similar expressions for pS{a - y) (S - y) (e - y), &o., and there will result, by addition, aS{y-P)(S-p){.-P) + pS(a-y){S-y){.-y) + yS{a-B)(P-S){e-B) + SS(a-i)(fi-€)(y-€) + eS{P -a) (y- a) (8- a) = 0, i. e. retainiug tbe notation adopted in the Note referred to, OA . BODE + OB. CDEA + OG. DEAB + OD.EABG + OE.ABGD = 0. Now let ir be a vector along OX ; then the operation hj S.ir on the above expression gives the result required. In some of the examples which follow, we will endeavour to show how a problem should not, as well as how it should, be attacked. Ex. 10. Given any three planes, and the direction of the vector perpendicular to a fourth, to find its length so that they may meet in one point. Let Sap = a, Sj3p = h, Syp = c be the three, and let S be the vector perpendicular to the new plane. Then, if its equation be Shp = d, we must find the value of d that these four equations may all be satisfied by one value of p. ■168 QUATERNIONS. [CH. IX. Formula (14) gives pS. 0(8y = Va/SSyp + V/3ySap + VyaS^p = cVaP + aV^y + hVya, by the equations of the first three. Operate by S.h, and use the fourth equation, and we have the required value dS. a/3y = aS. /3yS + bS. yaS + cS. a^S. Ex. 11. The sum of the (vector) areas of the faces of any tetrahedron, and therefore of any polyhedron, is zero. Take one corner as origin, and let a, /?, y be the vectors of the other three. Then the vector areas of the three faces meeting in the origin are 111 That of the fourth may be expressed in any of the forms lr(y-a)(^_a), ^ r(a-^) (y-^), I ^(jS-y) (a-y). But all of these have the common value ^r{y^ + ^a + ay). which is obviously the sum of the three other vector-areas taken negatively. Hence the proposition, which is an elementary one in Hydrostatics. Now any polyhedron may be cut up by planes into tetrahedra, and the faces exposed by such treatment have vector-areas equal and opposite in sign. Hence the extension. Ex. 12. If tlie pressure he uniform throughout a fluid mass, an immersed tetrahedron (and therefore any polyhedron) experiences no couple tending to m,ake it rotate. This is sTipplementary to the last example. The pressures on the faces are fully expressed by the vector areas above given, ancj ART. 71.] FORMULA AND THEIR APPLICATION. 1G9 their points of application are the centres of inertia of the areas of the faces. The coordinates of these points are ^{a + p), 3(/3 + y), ^(y+a), ^(a + ^ + y), and the sum of the couples is I V. { Vap. (a+/?) + Vpy . (/3 + y) + Fya . (y + a) + V{yP + l3a + ay).{p.+P + y)) = -| F(Fa/3.y+ F/3y.a+ Fya.;8) = 0, by applying formula (9). Ex. 13. What are the conditions that the three planes Sap = a, Sj3p = h, Syp = c, shall intersect in a straight line t There are many ways of attacking such a question, so we will give a few for practice. (a) pS. a/?y = YoL^Syp + Y^ySap + VyaS^p = cFaj8 + aF/8y + 6Fya by the given equations. But this gives a single definite value of p unless both sides vanish, so that the conditions are /S'.a^y = 0, an d e Va/S + a F/3y + 6 P y a = 0, which includes the preceding. (b) S {la — mji) p = al- hm is the equation of any plane passing through the intersection of the first two given planes. Hence, if the three intersect in a straight line there must be values of I, m such that la - mj3 = y, la - mb = c. The first of these gives, as before, AS'.a/3y=0, 170 and it also gives QUATERNIONS. [CH. IX. Fya = m Va^, 7Py = -l Va^, SO that if -we multiply the second by VaP, la Va^ -mbraP = vYaP becomes — o FySy — 6 Vya = c Va^ ; the second condition of (a). (c) Again, suppose p to be given by the first two in the form p=pa + q/3 +x FajS, we find a =pa' + qSa/S, because Sa Va^ = 0, b = pSaP+qP'; + a;Faj3, + xS. aj3y. Now a determinate value of x would mean intersection in one point only ; so, as before, S.aPy = 0, c {a' 13' - ^a/3) = a (P'Say - SafiSISy) - b (SaPSay - a»>S'y3y). The latter may be written S.a[c (afi' - l3Sal3) - a (y|8» - ^S^y) - b {aS/3y - ySa^)] = 0. Now S.a{aiP''-liSaP) = Sa{P.lia-pSPa) = S.a(prpa) = -S.a iPValS) =^-S {apVaP). Similarly, S . a (y/3' - ^SPy) = S (afi V/Sy), and S.a{aSPy-ySaP)=S. a (F. ^Vya), (formula 8), = ^(a;8F7a). The equation now becomes S . ap {cVaP + aVPy + bVya) = 0. therefore P a' , Sap = a 1 a, SaP +p + Sap, P' b, P' SaP, b so that the third equation gives, operating by S. y. c a' , Sap = Say a, SaP + Spy a" , a S aP, P' b, p^ Sap, b 1 ABT. 71.] TOKMUL^ AKD THEIR APPLICATION. 171 Now since *S' . a^y = 0, a, P, y are vectors in the same plane ; therefore y may be written ma + myS, and c Va^ + a ) "jffy + Vya assumes the form eVaP, ■which, unless e = 0, gives S{a/3ral3) = 0, or VaP is in the same plane with a, P ; but it is also perpendicular to the plane, which is absurd ; therefore e = 0, or cFa/S + aVPy+hrya=Q ; thus the third and prolix method leads to the same conclusion as the fii-st. Ex. 14. FiTid iJie surface traced out hi/ a strai{/?it line which remains always perpendicular to a given line while intersecting each of two fixed lines. Let the equations of the fixed lines be •BT = a + a;/?, ■or, = a^ + ajj/Jj. Then if p be the vector of the new line in any position p = i!T + y (xa-j — ■ar) = (1 - y) (a + xfi) + y (a, + a;,/?,). This is not, as yet, the equation required. For it involves essentially three independent constants, x, a;,, y ; and may there- fore in general be made to represent any point whatever of infinite space. The reader may easily see this if he reflects that two lines which are not parallel must appear, from every point of space, to intersect one another. We have still to introduce the condition that the new line is perpendicular to a fixed vector, y suppose, which gives S.y{^,-^)^0 = S.y [(a. - a) + a!,)8, - a;^]. This gives a;, in terms of x, so that there are now but two indeterminates in the equation for p, which therefore represents a surface, which, it is not difficult to see, is one of the second order. 172 QUATERNIONS. [CH. IX. Ex. 15. Find the condition that the equation S. p(j>p = 1 may represent a surface o/ revolution. The expression p here stands for something more general than that employed in Chap. VIII. above, in fact it may be written <{,p = aSa^p + l3Sj3^p + ySy^p, where a, a^, 13, /?„ y, y, are any six vectors whatever. This will be more carefully examined in tlie next chapter. If the surface be one of revolution then, since it is central and of the second degree, it is obvious that any sphere whose centre is at the origin will cut it in two equal circles in planes perpendicular to the axis, and that these will be equidistant from the origin. Hence, if r be the radius of one of these circles, ^ the vector to its centre, p the vector to any point in its circumfei'ence, it is evident that we have the following equation Spp -l-C{p' + r') = {Sep)' - e', where G and e are constants. This, being an identity, gives l-e'' + Cr' = } Spp-Cp' = {Sepyj- The foi-m of these equations shows that is an absolute con- stant, while r and e are related to one another by the first; and the second gives p = Cp+ eSep. This shows simply that S. €prj>p = 0, i. e. £, p, and <;()p are coplanar, i. e. all the normals pass through a given straight line ; or that the expression Vp(t>p, whatever be p, expresses always a vector parallel to a particular plane. Ex. 16. If three mutually perpendicular vectors be drawn from a point to a plane, the sum of the reciprocals of the squares of their lengths is independent of their directions. ART. 71.] FORMFL^ AND THEIR APPLICATION. 173 Let Sfp = 1 be the equation of tlie plane, and let u, /8, y he any set of mutually perpendicular unit-vectors. Then, if xa, y^, zy be points in the plane, we have xSat = 1, ySfii = 1, zSyi = 1, .-(vhence - c = aSae + /35/3e + ySyi (63. 2) = - + ^ + ^ . X y z Taking the tensor, we have x' y^ z' Ex. 17. Find the equation of the straight line which meets, at right angles, two given straight lines. Let ■BT = a + xj3, ■nr = a, + aj^jS, , be the two lines; then the equation of the required Une must be of the form and nothing is undetermined but o^. Since the first and third equations denote lines having one point in common, we have SimDarly S. )3, V^fi^ (a, - a,) = 0. Let o, = y^ + yi^,, (it is obviously superfluous to add a term in TySyS,), then S.ap}W, = y^T'rfiP^, S.a^PJ^p,= -yT'JW,, and, finally, Ex.18. I/Tp = Ta=^Tp=l, and S.a^p = 0, show that S.U(p-a)r{p-fi)=^l{l-SaP). Interpret this theorem geometriccdly. 174 QUATERNIONS. [CH. IX. We have, from the given equations, the following, which are equivalent to them, p = xa + y^) ' Hence -x^-y*+ ixySafi = - 1 , {x- 1) a + j//3 U(p-a) = J(x-iy-2(xy-y)Sap + y'' xa + (y-l)P Jx" - 2 {xy - x) Hafi + (y - if S.U{p-a)U{p-^) = -a^('c-l) + [a:y + (a'-^)(y-l) ]'S'a/?- y(y-l) Jx'+y'--2x+ I - 2(xy-y)Safi Jx?-i-y'-2y+l-2{xy-x}Sa^ _ x + y-{x + y-l) Safi - 1 ~ j2-2x + 2ySa$ J 2^ 2y + 2xSafi {x + y-\)(l-Sap) 2 J{l-x-y){l - ^afi) + xy {1 - (Hafiyi _x + y-\ I 1 - ABC = OABC-OBCD + OCDA-ODAB. 14. When A, B, C, D are in the same plane, a.BCI)-l3.CI)A + y.DAB-S.ABG=0, where BCD, &c. are the areas of the triangles. 15. 8F.a;8y + aF,/3yS + /3F.78a + yF,8a^ = 4.S.aj8y8. 12—? CHAPTER X. VECTOR EQUATIONS OF THE FIRST DEGREE. With the object of giving the student an idea of one of the physical applications of Quaternions, we will treat the solution of linear and vector equations from an elementary kinematical point of view. Def. Homogeneous Strain is such that portions of a body, originally equal, similar, and similarly placed, remain after the strain equal, similar, and similarly placed. Thus straight lines remain straight lines, parallel lines i-emain parallel, equal parallel lines remain equal, planes remain planes, parallel planes remain parallel, and equal areas on parallel planes remain equal. Also the volumes of all portions of the body are increased or diminished in the same proportion, as is easily seen by su])posing the body originally divided into small equal cubes by series of planes perpendicular to each other. It is thus obvious that a homogeneous strain is entirelv deter- mined if we know into what vectors three given (non-coplanar) vectors are changed by it. Thus if a, /3, y become a', ^, y re- spectively, any other vector which may be expressed as is changed to p=-s:^y('''^-i^yp+^'^-y''p+y'^-''M- CHAP. X.J VECTOR EQUATIONS OF THE FIRST DEGREE. 181 No needful generality is lost, while much simplification is gained, bj taking a, /J, y as unit vectors at right angles to one another. We thus have p = - (aSap + j8(S'/3p + ySyp), p' = — (a! Sap + PSjip + y'Syp). Comparing these expressions we see that Homogeneous Strain alters a vector into a definite linear and vector function of its original value. In abbreviated notation, we may write (as in Art. 63, though our symbol, as will soon be seen, is more general than that there employed) <^p = — {a Sap + PfSPp + y'Syp), where <^ itself depends upon nitie independeat constants involved ia the three equations <^a = a'] <^y=y'J For a, /?', y may of course be expressed in terms of a, /3, y : and, as they are quite independent of one another, the nine co- efficients in the following equations may have absolutely any values whatever; if>a = a' =Aa+ cp + b'y'\ .^^ = /r = c'a +/?/? + ay i: (a). ^y = y' = 6a + a'j3 + Cy) In discussing the particular form of (j> which occurs in the treatment of central surfaces of the second order we found, Art 44, tliat it possessed the property S.(rcl>p = S. p(T (6), whatever vectors are represented by p and tr. Remembering that a, j3, y form a rectangular unit system, we find from (a) S.pa = -c)^ S.a4,p = -cV 182 QUATERNIONS. , ; [CHAP, ■)vith Other similar pairs ; so that our new value, of. ^ satisfies (6) if, and only if, we have in (a) a = a''\ '■'•' ■• l=V ■ («). c = c' The physical meaning of this condition will be -seen imme- diately. But, although (6) is not generally true, we have S.,T4,p = - {Sa'aSap + S^'crS^p + Sy'aSyp) = -jS.p{aSo:,T + pSl3:'cr = - {aSacr +■ pSp'(T, + ySy'a) ..{d). And with this we have obviously S . a-tf>p =S . p'<^ («)) which is the general relation, of which (b) is a mere particular case. By putting a, (3, y in succession for a in (d) and referring to (a) we have fj>'a =Aa + c'/3 + by\ ^'I3=ca +£p+a'yl..... (/), 'y = Va + ap + Cyi Comparing [f) with (a) we see that p = 4>'p, ... I whatever be p, provided the conditions (c) be fulfilled. This agrees with the result already obtained. Either of the functions ^ and ^', thus, defined together, is called the Conjugate of the other : and when they are equal (i. e. when ((■) is satisfied) i^ is called a Self-Corijiigate function. As we X.J VECTOR EQUATIONS OF THE FIRST DEGREE. 183 employed it in Chap. VI, <^ was self-conjngate J and, even Lad it not been so, it was involved (as we shall presently see) in such a manner that its non-conjugate part was necessarily absent. We may now write, as before, p = - {a'Sap + ^SPp + y'Syp), and, by {d), tj>'p = — {aSa'p + I^Sji'p + ySy'p). From these we have by subtraction, ((^ — (fi) p = p — 'p = aSa'p — a'Sap + PSpp — ^S(ip + ySy'p — y'Syp = V. 7aa.'p+ V. VPfi'p+V. Vyyp = 27. ^P (9); if we agree to write 2€ = r {aa' + 13^' + yy') {h}. We may now express that is self-conjugate by writing . = 0, the physical interpretation of which equation is of the highest importance, as will soon appear. If we form by means of (a) the value of c as in (h) we get 2£ = (cy - b'P) + (aa - c'y) + (6/3 - a' a) = {a-a')a + {b-h')^ + {c-c')y, which obviously cannot vanish unless (as before) the three con- ditions (c) are satisfied. By adding the values of p and ')p=ij>p + 'p = -{aSa'p+ a'Sap + ^SP'p + ^Sftp + ySy'p + y'Syp) = - F (apa' + PpP' + ypy') - p {Saa' + S^^' + Syy'). As we have (by 69. 6) V .apa' =V. a' pa, &C. this new function of p is self-conjugate; as will easily be seen by putting it for <^ in (b) and remembering that (by 69. 17) we have *S^ . a-apa' = S . pa'tra = S . paxra', &o., &c. 184 QUATERNIONS. [CHAP. Hence we may write { + ')p = 2^p (i), where the bar over ct signifies that it is self-conjugate, and the factor 2 is introduced for convenience. Prom (ff) and (i) we have ^p ='!3-p+ Fepj ^jy 'p = CTp — Vep) If instead of tj>p in any of the above investigations we write ( + g)p, it is obvious that 'p becomes (tf>' + g)p : and the only change in the coeflBcients in (a) and (/) is the addition of g to each of the main series A, B, C. We now come to Hamilton's grand proposition with regard to linear and vector functions. If ^ be such that, in general, the vectors p, <^p, ^'p (where <^'p is an abbreviation for <^ (<^p)) are not in one plane, then any fourth vector such as ^p (a contraction for ^ (^ {4'p))) '^^^ ^^ expressed in terms of them as in 31. 5. Thus <^'p = mj,4t'p — m^<^p + mp {k), where m, m,, m^ are scalars whose values will be found immedi- ately. That they are independent of p is obvious, for we may put a, p, y in succession for p and thus obtain three equations of the form ^V = m^^'a - m,<^a + ma (I), from which their values can be found. Conversely, if quantities m, »i|, wij can be found which satisfy (?), we may reproduce (k) by putting p=xa + y^ + ey and adding together the three expressions (I) multiplied by x, y, z respectively. For it is obvious from the expression for ^ that x^p = (xp), x'p = ^ (xp), &c., whatever scalar be represented by x. X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 185 If p, (f>p, and <}>'p are in the same plane, then applying the Btraiu fj) again we find <}>p, ify'p, ^p in one plane; and thus equa- tion (^) holds for this case also. And it of course holds if 4>P ^ parallel to p, for then 'p and 'p are also parallel to p. We will prove that scalars can be found which satisfy the three equations (l) (equivalent to nine scalar equations, of which, however, six depend upon the other three) by actually determining their values. The volume of the parallelepiped whose three conterminous edges are X, p., v is (31. 1) , S . Xp-v. After the strain its volume is S . Xp(l>v, so that the ratio ' ^ o . Xpv is the same whatever vectors X, pi, v may be ; and depends there- fore on the constants of ^ alone. We may therefore assume >^ = P, I and by inspection of {k) we find S . 4>X'f>p<^v S.p4>'p4?p , . ti . Xpv , S . p(l>p<]>'p which gives the physical meaning of this constant in (k). As we may put if we please we see by (a) that m =- IS.aliy A, c, h' o', B, a h, a', C 18G QUATERNIONS, • [chap, ■which is the expression for the ratio in which the volume of each portion has been increased. This is unchanged by putting ' for , for it becomes, by (/), A o', b c, B, a' v, a, C Eecurring to (w) we may write it by (e) as from which, as X is absolutely any vector, we have fjL(f>v =TnVfjiv'\ .(n). or 'iJi'v = mVfiivj In passing we may notice that (n) gives us the complete solution of a linear and vector equation such as ^o- = S, ;, where 8 and <^ are given and cr is to be found. We have in fact only to take any two vectors ^ and v which are perpendicular to 8, and such that Yl^v = 8, and we have for the unknown vector- , cr = — Y4>ii.v, m which can be calculated, as <^ is given. If in (n) we put 4> +9 for <}> we must do go for the value of m ;n (m). Calling the latter M^ we have iS . X.fj.v S , X^jx^v + S . iJiX + S . vX +9)[m^-' F/*v + jr ( F^> V + fVf v) + g' F/^v] = M, Viwi'"^^'' From the latter of these equations it is obvious that F<^V '' + yi^'v must be a linear and vector function of F/iv, since all the other teims of the equation are such functions. As practice in the use of these functions we wiU solve a problem of a little greater generality. The vectors Vft-v, Vii V, and Vfi^'v are not generally coplanar. In terms of these (31. 5), let us express <^F/ii«. Let Fixv = xViJ.v + 2/V'iJi.v + zVix'v. Operate by S.X, S.fi, S.v successively, then S . fiv(l>'X = xS . X/xv + yS . vX^'/i + zS . Xfi'v, S . IXV'v = zS . VIX(f>'l'. The two last equations give (by 69. 4) y-.-!, »=-l, and therefore the first gives S .iJivX + S .vXn + S .X/i-ff/v iS . Xfiv = h-1, V (?)• Hence, finally, 'iJ.v— Vfx'jt'v (r). Substituting this in (5), and putting a for F/tv, which is any vector whatever, we have ( + g)[rii-' + g{ii.,- ) +9']'^=im + iJ;g+[i^' + g')ar, or, multiplying out, (m - g r- g"^ + gnitfr^ + g",^ + g'fit + g") '- /tj,^'' + /i,<^ - m)(r = 0. Comparing this with (k) we see that [chap. =/^.= = /^,= S . kfJ-V and thus the determinatioii is complete. We may write (k), if we please, in the form miji'^p = m^p — tn^tjip + 'p, . •(«), .(k'), which gives another, and more direct, solution of the equation (above mentioned) ffxT = S. Physically, the result we have arrived at is the solution of the problem, " By adding together .scalar multiples of any vector of a body, of the corresponding vector of the same strained homo- geneously, and of that of the same twice over strained, to repre- sent the state of the body which would be produced by supposing the strain to be reversed or inverted." These properties of the function <^ are sufficient for many applications, and we proceed to give a few. I. Homogeneous strain converts an originally spherical por- tion of a body into an ellipsoid. For if p be a radius of the sphere, or the vector into which it is changed by the strain, we have and Tp = C, from which we obtain r-'(r^-'(7=-C', or, finally, S . (Kjr'a- = - C. X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 189 This is the equation of a central surface of the second degree ; and, therefore, of course, from the nature of the problem, an ellipsoid. II. To find the vectors -whose direction is unchanged by the strain. Here p must be parallel to p or ^P = gp- This gives <^'p=g'p, &c., so that by {k) we hare f/ - 7)i,sr° + m,g -VI- 0. This must have one real root, and may have three. Suppose g^ to be a root, then ^P - 9,P = 0> and therefore, whatever be X, iSXp — <7,»S'Xp = 0, or S.p{i>'X-g,\) = 0. Thus it appears that the operator 4>' — ffi <"its off from any vector X the part -which is parallel to the required value of p, and there- fore that -we have p\\3ir.{4>'-g,)\('-g,)fji \\{m4r'-g^{m,-' + m^tji -m ^ The same result may more easily be obtained thus. The expression (<^' - vi^'^" + m,(^ _ m) /) = 0, being true for all vectors whatever, may be written 190 QUATERNIONS. [CHAP. and it is obvious that each of these factors deprives p of the por- tion correspoading to it : i. e. <^ — p", applied to p cuts ofl" the part parallel to the root of so that the operator (<^ - g^) ((j> - g^ when applied to a vector leaves only that part of it which is parallel to o- where III. Thus it appears that there is always one vector, and that there may be three vectors, whose direction is unchaoged by the strain. When there are three they are perpendicular to each other, if the strain he pure. For, in this case, the roots of are real. Let them be such that then ^i^a'Sp.p, = >Sct>Pip, = Spi'f'Pa (because, by hypotliesis, the strain is pure) = 9'^PiPii for iP, = 9>P, and p(^(T) i. e. the lengths of vectors, and their inclinations to one another, are unaltered. In this case, therefore, the strain can be nothing but a rotation. It is easy to see that the second of these equa- tions includes the first; so that if, for variety, we take as represented in equations (a), and write p = xa. + yli + zy, we have, for all values of the six scalars x, y, z, i, 17, Z, tte follow* jng identity ; ] W' 192 QUATERNIONS. [CHAP. - {xi + yr, + «o = -S" ■ K + yl^' + «7') (i<^' + v^+ ^yO + {xr, + y$) Sa'^ + {yi, + zr,) Sfi'y' + {zi + x^) Sy'a'. This necessitates Sa'P' = Sfi'y' = Sy'a' = OJ ^ ^' i. e. the vectors a, /3', / form, like a, (3, y, a rectangular unit system. And it is evident that any and every such system satis- fies the given conditions. VII. It may be interesting to form, for this particular case, the equation giving the values of g. We have S.( + g)a( + g)p{^ + g)y '~ S.ajiy S.(a' + ga){p'+gP)(Y+gy) S. a/3y = l-gS{afi'y' + a:py'+a'l3'y) - g^S (apY + a^'y + a')8y) + g". Recollecting that a, j3, y ; a', jS', y are systems of rectangular unit vectors, we find that this may be written M^=l-{g + g')S{o.a' + fi^ ^yy')+g'' = {g+\)[g'-g{l + S{aa'+ p^' +yy')\+\\ Hence the roots of are in this case ; first and always, which refers to the axis about which the rotation takes place : secondly, the roots of g'-g{\ + S{aa' + /3y3' + yy')} + 1 = 0. Now the roots of this equation are imaginary so long as the coefficient of the first power of g lies between the limits ± 2. Also the values of the several quantities Saa, Sp^', Syy can never exceed the limits ± 1. When the system a, p, y coincidjep X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 193 ■with a', /3', y', the value of each of the scalars is — 1, and the coefficient of the first power of ^ is + 2. "When two of them are equal to + 1 and the third to — 1 we have the coefficient of the first power of g = — 2. These are the only two cases in which the three values of g are all real. In the first, all three values o{ g are equal to — 1, i. e. p = p for all values of p, and there is no rotation whatever. In the second case there is a rotation through two right angles about the axis of the — 1 value of g. VIII. It is an exceedingly remarkable fact that, however a body may be homogeneously strained, there is always at least one vector whose direction remains unchanged. The proof is simply based on the fact that the strain-function depends on a cubic equa- tion (with real coefficients) which must have at least one real root. IX. As an illustration of what precedes (though one which must be approached cautiously), suppose a body to be strained so that three vectors, a", /3", y" (not coplanar, and not necessarily at right angles to one another), preserve their parallelism, be- coming e,a", e^jS", e^y". Then we have ^pS. a"l3"y" = e^a"S. j8"y"p + e,P"'S- y"a"p + e^y"S. a"y3"p. By the formulae (m, s) we have '' S.a"P"y" s {a"^'f<^<^" + y"4"^"n -c, II nil It *i*2*a ' lb . a p y O "13" " — «2''3 + *3*l+ ^A) o . a p y so that we have by (k) (<^-e,)(<^-e,)(.^- 63)9 = 0. Though the values of g are here all real, we must not rashly adopt the conclusions of (iv.), for we must remember that a", /J", /' do not, like a, P, y, necessarily form a rectangular system. T. Q. 13 194 QUATERNI0N3. [CHAP, In this case we have 4,'pS . a" /3" y" = e, V^"y"Sa"p + eJy"c!'S^"p + eJa"P'Sy"p. So that, by (A), 2c = Y. {eyV^Y + e^p'rya" + e^y"Va"n This vanishes, or the strain is pure, if either 1 . Sa"^' = Sfi"y" = Sy"a" = 0, i.e. if a", fi", y" are rectangular, in -which case e^, e,, e, may have any values ; or 2. 6^ = 6,, = 63, in which case 4,'pS. a''P"y" = e, {Fy8"y"^a"p + F/a";S/3"p + Fa"^"Vp} = e,pS..a"P"y" by (69. 14), 80 that 'p = eip = p for every vector : a general uniform dilatation unaccompanied by change of direction. 3. e, = Sj, and a" and P" both perpendicular to y". From what precedes it is evident that for the complete study of a strain we must endeavour to distinguish in each case between the pure strain and the merely rotational part. If a strain be capable of being decomposed into 1st a pure strain, 2nd a rotation, it is obvious that the vectors which in the altered state of the body become the axes of the strain-ellipsoid (i.) must have been originally at right angles to one another. The equation of the strain-ellipsoid is Spct>-'p = -c\ and in this it is obvious that ~''p is the normal to the ellipsoid at p, X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 195 the bar above being used to shew that the non-conjugate term has been omitted, No-w (k') wi<^~' = m, - m,^ + ', ■whence m~' = — ' (nil - ^ai> + 4'') -»*, + <#>, or m^ffr' = {m' — mm^ — {miin^ -m)^ + mi'. Now, by (J) p = 'p = {^+n){^p+Vep) = ^"p +V-.e^p + ^Fep+ V. tVep, whence ~'p = (m^ — mm^) p — (miin^ — m)p+ mi (c^^'p + V. tVep), which must be = m'hp, if p is an axis of the strain- ellipsoid. We have to shew that, if p, and p^ are two of the three vectors which satisfy this equation, we have not only Spip^ = 0, but also S . (^~'pi<^~'pj = 0. By the help of the expressions above this is easily effected. But the result is much more easily obtained as an immediate con- sequence of a somewhat different mode of treating the question, one which we will now give : — If q be any quaternion, the operator q { ) q~^ turns the vector, quaternion, or body operated on round an axis perpendicular to the plame of (^ artd through an angle equal to double that of q. The proof of this extremely important proposition is very simple; but we refer the reader to Hamilton's Lectures, § 282, 13—2 196 QUATERNIONS. [CHAP. Elements, § 179 (1), or Tait, § 353. It is obvious that the tensor of q may be taken to be unity, i.e. q may be considered as a mere versor, because the value of its tensor does not affect that of the operator. A very simplg but important example of this proposition is given by supposing q and r to be both vectors, a and /8 let us say. Then is the result of turning ft conical ly through 2 right angles about u, i. e. if a be the normal to a reflecting surface and /8 the incident ray, - aPoT' is the reflected ray. Now let the strain be eflfected by (1) a pure strain ct (self- conjugate of course) followed by the rotation g ( ) g"'. We have, for all values of p, 4>p = q{^p)q" (v). whence ^'p = 5 (q'^pq). We may of course put, as in Chap, vi, CT-p = e^aSap + e^^Sfip + e^ySyp, where a, /?, y form a rectangular system. Hence tjip = e^qaq'^Sap + e^q^q-^S^p + e^qyq'^Syp. Here the axes are parallel to qaq-\ qj3q-\ qyq~\ and we have S . qaq-'qPq-' = S . qa^q'' = Sa^ = 0, &c. So far the matter is nearly self-evident, but we now come to the important question of the separation of the pure strain from the rotation. By the formulae above we see that 'ft>p = ''!yq~^(j>pq = -^q-^{qwpq-')q so that we have in symbols, for the determination of ot, the equation X.] VECTOR EQUATIONS OF THE EIRST DEGREE. 197 To solve this equation we employ expressions like (Jc). <^'<^ being a known function, let us call it w, and form its equation as 0)' — OT^w' + mjoi — m = 0. Also suppose that the corresponding equation in 5 is ■where g, g„ g^ are unknown scalars. By the help of the given relation ot^ = to, we may modify this last equation as follows : '5ia-g^ia + g^-g = Q, _ g + g (0 whence w = — ^-^~ : g, + u> i. e. ra- is given definitely in terms of the known function m, as soon as the quantities g are found. But our given equation — 2 C7 = 0) may now be written or o,' - {gj- - 2g^ a>' + {g^' - ^gg^) a. - ^' = 0. As this is an equation between co and constants it must be equivalent to that already given : so that, comparing coefficients, we have 91-^992 = ^1, g' =m; from which, by elimination of g and g,, we have (^?)'^--'.' The solution of the problem is therefore reduced to that of this biquadratic equation ; for, when g^^ is found, g^ is given linearly in terms of it. It is to be observed that in the operations above we have not 198 QUATERNIONS. [CHAP. been particular as to the arrangement of factors. This is due to the fact that any functions of the same operator are commutative in their application. Having thus found the pure part of the strain we have at once the rotation, for {v) gives _ or, as it may more expressively be written, If instead of (v) we write p = w{rpr~') (v), we assume that the rotation takes place first, and is succeeded by the pure strain. This form gives p = is equivalent to the pure strain ^'. This leads us, as an example, to find the condition that a given strain is rotational only, i. e. that a quaternion q can be found such that = q( )q-\ Here we have ^' = g'"' ( ) q, or '= + (ji", or m' = nil ~ "iz"^ + ^^ whose conjugate is mcji = wij — wij"^' ' + ",} X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 199 and the eliminatioa of <^' between these two equations gives 1. e. (m'm, — mm^m^ + in') — (nv' — mm' + 2m^m,, — m) ' -m^' = (m'm, — vim^m^ + m') — (m^ — mm^' + 2m ^m^ — (7)»»l, - 2m, - »!./) ' — 27n^(f>' by using the expression for tjy* from the cubic in ff>. Now this last expression can be nothing else than the cubic in c^ itself, else would have two different sets of constants in the form (k), which is absurd, as these constants, from the mode in which they are determined, can have but single values. Thus we have, by comparing coefEcients, m/ = 2oti + m/ — mmj — wi, m^m^ = iii' — mm^ + 2m^^ — m ' mm^ = m'm^ — Tnm^m,^ + m^ The first gives i)\ = mm„ by the help of which the second and third each become m^-m = 0. The value m, = IS to be rejected, as otherwise we should have been working with non-existent terms, and m as the ratio of the volumes of two tetra- hedra is positive, so that finally m= 1, m, = m,, and the cubic for a rotational strain is, therefore, or (<^-l){<^''+(l-m,)<^+l} = 0. ■where mIj is left undetermined. 200 QUATERNIONS, [CHAP. By comparison with the result of (vii.) we see that in the notation there employed m, = -S{aa' + P^ + yy'). The student will perhaps here require to be reminded that in the section juat referred to we employed the positive sign in operators such as +g- In the one case the coefficients in the cubic are all positive, in the other they are alternately posi- tive and negative. The example we have given is a particularly valuable one, as it gives a glimpse of the exteot to which the separation of symbols can be safely carried in dealing with these questions. Dep. a simple shear is a homogeneous strain in which all planes parallel to a fixed plane are displaced in the same direction parallel to that plane, and therefore through spaces proportional to their distances from that plane. Let a be normal to the plane, )8 the direction of displacement, the former being considered as an unit-vector, and the tensor of the latter being the displacement of points at unit distance from the plane. We obviously have, by the definition, Sal3=0. Now if p be the vector of any point, drawn from an origin in the fixed plane, the distance of the point from the plane ia — Sap, Hence, if o- be the vector of the point after the shear, 17 = (j>p = p — ^Sap. This gives <}>'p = p-aSPp, which may be written as = p-Tp.aS. UPp, so that the conjugate of a simple shear is another simple shear equal to the former. But the direction of displacement in each shear is perpendicular to the unaltered planes in the other. X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 201 The equation for ^ is easily found (by calculating m, m^, ni^ from (7)1, s,)) to be ^'-3.^' + 3^- 1=0. Putting <^'^ = i/^, we easily find (with b = TjS) i^-{3+b')il,'+{3 + b')xl,-l = 0. Solving by the process lately described, we find If 6 = 2, this gives g^ = 1, and the fartlier equation ^.' + ?i'-13^,-21 = 0, of which ^j = — 3 is a root, so that and ff^=1^2j2. We leave to the student the selection (by trial) of the proper root, and the formation of the complete expressions for the pure and rotational parts of the strain in this simple and yet very interesting case. As a simple example of the case in which two of the roots of the ciibic are unreal, take the vector function when the strain is equivalent to a rotation 9 about the unit vector a ; the others of the rectangular system being )S, y. Here we have, obviously, ^a = a, l3 = /3 cos ^ + y sin ^, (^y = y cos 6 - )8 sin 6, whence at once --l) {-cos6- J^ sinO) {cl>-cos6 + ^^ siue) = 0. Now — (^ - 1) p = (1 — cos 6) (aSap + p) — sin 6 Vap, — (^ - COS ^ — J— 1 sin 6) p = (l— cos 6) aSap + sin 6 (p J— 1 — Vap), — ( — I) a = 0, which is, of course, true. Again (<^ - 1) (<^ - cos e-^-ITsin e) p = - sin (9 (1 -e"*^) (J-la+ 1) Vap, which we leave to the student to verify. The imaginary directions which correspond to the unreal roots are thus, in this case, parallel to the Bivectors (a±yrr)Fap. Here, however, we reach notions which, though by no means difficult, cannot well be called elementary. A very curious case, whose special interest however is rather mathematical than physical, is presented by the assumptions a' = P + y, ^ = V + a, for then p = (Ji + y) Sap + (y + a) Sfip .+ (a + P) Syp = {a+ P +y) S {a + fi + y) p - (aSap + jSS^p + ySyp) = 38SSp + p, where 8 is a known unit vector. This function is obviously self- conjugate. Its cubic is ^»-3<;!) + 2 = = (^-l)'(<^ + 2), X.J VECTOE EQUATIONS OF THE FIRST DEGREE. 203 which might easily have been seen from the facts that 1st, <^8 = -28, 2nd, <^a = a, if SaS = 0. The case is but slightly altered when the signs of o', ;S', y are changed. Then p = — 3SSBp — p, and the cubic is ^'_3<^_2 = (<^+l)'(<^-2) = 0. These are mere particular cases of extension parallel to the single axis S. The general expression for such extension is obviously ^p = p— ehSSp, and we have for its cubic (.^-l)'{.^-(l+e)} = 0. We will conclude our treatment of strains by solving the following problem : Find the conditions which must be satisfied by a simple sliear which is capable of reducing a given strain to a pure strain. Let = (j> + fiS . a^, = 'ap, so that the requisite conditions are contained in the sole equation 2£ = Ffa|8. This gives (1) ^.^< = 0, (2) S'a€=0=Sa€. But (3) Sa^ = (by the conditions of a shear), 60 that xa = V. ^e. Hence we may assume any vector perpendicular to e for fi, and a is immediately determined. When two of the roots of the cubic in <^ are imaginary let us suppose the three roota to be Let jS and y be such that Then it is obvious that, by changing throughout the sign of the imaginary quantity, we have .These two equations, when expanded, unite in giving by equating the real and imaginary parts the values To find the values of a, /?, y we must, as before, operate on any vector by two of the factors of the cubic. As an example, take the very simple case = 0, that is ^((^ + ev/^)(<^-e \/^) = 0. As operand take p = ix + jy + hz, then a II Y { + e J - i) (<^ - e J^l) p \\eV . (4, + e 'J'-i){ki/-jz- ps/^) Wi-Jy-Icz+p) X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 205 Again y3-7^/^|l<^(<^ + e^/3^)p \\e^{ky-jz + J -\p) II -jy -hz+ sj-\{ky -jz) WJy + hz-J -\{jz-hy). With a change of sign in the imaginary part, this will represent P + ysl^l, so that p=jy + kz, y=jz- ky, Thus, as the student will easily find by trial, j3 and y form with a a rectangular system. But for all that the system of principal vectors of , viz. a, /3±y\/^l does not satisfy the conditions of rectangularity. In fact we see by the above values of y3 and y that S.(^ + yJ^){p-yJ~l)^P' + f^-2{y' + :^). It may be well to call the student's attention at this point to the fact that the tensors of these imaginary vectors vanish, for T= (/? ± y n/^) = - /S- (/? ± y ^/ri) (^ ± y V^) = y= - ;8' = 0. This gives a simple example of the new and very curious modifications which our results undergo when we pass to Bivectors; or, more generally, to Biquaternions. As a pendant to the last problem we may investigate the relation of two vector-functions whose successive application pro- duces rotation merely. Here <^ = i/fx"' is such that by (w) i.e. x'-y=xr'. _ or X'X = "AV = ■«^'» 206 QUATEENT0N3. [CHAP. since each, of these functions is evidently self-coujugate. This shews that the pure parts of the strains i/r and ;!^ are the same, which is the sole condition. One solution is, obviously, x'=x-'\ v-r\ i. e. each of the two is itself a rotation ; and a new proof that any number of successive rotations can be compounded into a single one may easily be given from this. But we may also suppose either of i/f, ;(, suppose the latter, to be self-conjugate, so that X' = X = X) or '/''«/' = ?, which leads to previous results. Examples to Chapter X. 1. If a, ;8, y be a rectangular unit system S.Va<^aVp<^^Vy<^y^-m8.P4,'-'aS.p(^-^')a, iind therefore vanishes if ^ be self-conjugate. State in words the theorem expressed by its vanishing. 2. With the same supposition find the values of 2 F. Va^a . rpl3 and' of ^S. Va^. Also of 2 . aSatfia. 3. When are two simple shears commutative ! 4. Expand ^j — in powers of , and reduce the result to three terms by the cubic in <^. 6. Shew that 4,' V. <#.p<#.V = ^'/pfpfP J. p^p " • PW9 P — mVptpp. C. Why cannot we expand ^' in terms of ^°, <^, (^"l X.] VECTOK EQUATIONS OF THE FIRST DEGREE. 207 7. Express Vpt^p in terms of p, 4>p, <^'P) ^^'^ ^o™ *^^ result find the conditions that <^p shall be parallel to p. 8. Given the coefficients of the cubic in , find those of the cubics in tji', <^', &c. <^". 9. Prove (f>V. a<^'a — mV . a^'~'a= 0, (<^ + ni,)F.a^'a = ra^"a. 10. li m= A, h, c shew that i/", = may be written as A h, c a. B, c' a', b', C or e" (^ ■*■■") TO = 0. 11. Interpret the invariants m^ and m^ in connection with Homogeneous Strain. 12. The cubics in ^ij/ and i//<^ are the same. 13. Find the unknown strains and ^ from the equations ^ + X = ^> 14. Shew that the value of V {axa + ^^xP + i'jX'i) i^ ^^^ same, whatever rectangular unit system is denoted by a, /3, y. 15. Find a system of simple ishears whose successive applica- tion results in a pure strain. 16. Shew that, if V V _ v-m From either of them we obtain the equation ! 208 QUATERNIONS. [CHAP. X. 17. Shew that in general any self-conjugate linear and vector function may be expressed in terms of two given ones, the ex- pression involving terms of the second order. Shew also that we may write ^ + z = a (nT + xy + b (■or + x) ( + y)', where a, b, c, x, y, z are scalars,- and ■sr, =pa + qp} then FD=pa + qP-ntat and RS=^RP + PS^RQ+ QS gives {l+m)a + x (pa + J)8 - ma) = (1 + m) jS + ^ {pix + q^ - m/3), 1 +»» whence "^ = III and as = -^ {pa + ql3) = —^ AD. 14—2 212 ' QUATERNIONS. Chap. III. Ex. 5. Let A BCD be the quadrilateral ; DA, DB, DC, a, P, y respectively. Now ^(y-a) + (y-a)/3 = y(/3-a) + (/3-a)y + a(y-^) + (y-;3)a. Taking scalars, and applying 22. 3, there results, Sp{y-a) = Sy{fi-a) + S<.{y-^), •B'hich is the proposition. Ex. 6. If u, p, y be the rectors OA, OB, OC corresponding to the edges a,h,c; we have V(CA.CB)=V(a-y){p-y) = V{aP + Py + ya) = abk + hci + caj, the negative square of which is the proposition given. Ex.7. If Sa (13 -y) = and SI3{a-y) = 0, then, by sub- traction, will Sy (a - ;8) = 0. Ex. 8. If a'=(p- yf, /3» = (y - a)', y" = (a - /3)'; then will *(^-i)(^°-f)=». *«■*«•. for these are the same equations in another form ; and they prove that the corresponding vectors are at right angles to one another. Ex. 9. If OA, OB, OC, OD are a, ^, y, 8 ; triangle Z>4.B : DAC :: tetrahedron ODAB : ODAQ :: Sa^S : SayS :: triangle OAB : OAC, because the angles which 8 makes with the planes OAB, OAC are equal. APPENDIX. 213 Chap. IV. Ex. 1. Let be the middle point of the common perpendi- cular to the two given lines ; a, —a, the vectora from to those lines, unit vectors along which are (3, y; p the vector to a point /" in a line QR which joins the given lines ; P beiug such that JiP = mPQ ; therefore p + a — yy = m {a + x^ — p). Now since a is perpendicular to both )3 and y, the equation gives (1 + m) Sap = (m — 1) a' ; a plane. Ex. 2. Retaining what is necessary of the notation of the last example, let OS=S. If PJi perpendicular on y meet yS in Q, we have — a + yy + EP = p, which gives yy' = Syp ; PQ = 2a + xP — yy, which gives yy^ = xS^y ; and SP'' = e'PQ' gives {p-By = e'{a + xli-py Syp Sidy' which being of the second degree in p shews that the locus is a surface of the second order. See Chap. VI. Ex. 3. The equation of the plane is Syp = a, which, being substituted in the equation of the surface, gives what is obviously the equation of a circle. Ex. 4. "With the notation of Ex. 1, let 8, 8' be the perpen- diculars on the lines, then p+S = a + xP gives V/3S = - V^{p- a), and the condition given may be writtea P/3S = eTV8'; ='[''^sjry'^-p) 214, QUATERNIONS. Now (22. 9) V'P{p-a)=-p'{p-a)' + S'p{p-a), Tvlience p'-2 Sap + a" + S^^p = e' (p' + 2Sap + a' + S'yp), a surface of the second order. Ex. 6. Sp (^ + y) = c, a plane perpendicular to the line which bisects the angle which parallels to the given lines drawn through make with one another. Ex. 7. a, /3 the vectors to the given points A, B, Syp = a, JSSp = b the equations of the planes, y, 8 being unit vectors. xy, 2/S the vector perpendiculars from A on the planes, then X = Say — a,y = SaZ — b, .: x + y = Sa{y + h)-{a + b) (1). Hence by the question Sa(y+B)=S^(y + B) or S{J3-a){y + S) = (2). Now equation (1) will give the sum of the perpendiculars on the planes from auy other point in the line AB by simply writing a + 3 (^ - a) in place of a ; and from equation (2) this will pro- duce no change. Ex. 8. If P' be the vector to C, equation (2) of the last example gives S(P-a){y + i) = 0, SiP'-a)(y+&) = 0. Now the sum of the perpendiculars from any other point in the plane will be found from equation (1) by writing a + a(j8-a) + «'(/3'-a) in place of a. Hence the proposition. Ex. 10. Tait's Quaternions, Art. 213, -Appendix. 2l£ (!)• Ex. 11. Let a, p, y, 8 be the vectors OA, OB, OC, OD ; then (34. 5, Cor.) 8 = ;S' . aj3y . ( Fa^ + V/By + Fya)"' _ ahc (bci + caj+ abk) ~ {abf + {bcf + (cdY No-w triangle ABB : triangle ABO :: tetrahedron OABB : tetrahedron OABO :: A^.a^S : S.a^y :: S.abijh : S.abcijh :: [aby : {abf + (be)' + (ca)' :: (triangle ^Oi?)'' : (triangle ^i?C)'. (Chap. III., Additional Ex. 6.) Ex. 12. This is merely the equation. p = ai + t' with t eliminated by taking the product of Vap, Vjip. (See 55. 3.) Chap. Y. Ex. 3. Let a, a' be the radii of the circles ; a, p the vectors from the centre of one of them to that of the other, and to the point ■whose locus is required ; then Tp^T{p-a) ^ a a' Ex. 7. This is the polar reciprocal of Ex, 3, Art. 40. Ex. &. Let yl be the origin, AB^P, AC = y, the vector to the centre a : then -Y{AB.BO . CA) = V.p{y-li)y = 2pSay — 2ySaP from the circle ; .: S.ar(AB.£C._CA) = 0. 216 QUATEENIONS. Ex. 9. Tait, Art. 222. Ex. 10. Tait, Art. 221. Ex. 11. Tait. Art. 223. Ex. 12. Tait, Art. 232. Chap. VJ. Ex. 1. Let 8 be the vector to the given point, ir the vector to the point of bisection of a chord, /3 a vector parallel to, the chord, all measured from the centre ; then h = ir + xp, SwS = S^Tr (48); from which by making we get >^PP = T 'SS'^Sj an ellipse whose centre is at the point of bisection of the line which joins the given point with the centre of the given ellipse. Ex. 2. Let 26 be the shortest distance between the given lines ; $ their angle of inclination ; 2a the line of constant length ; then as in Ex. 2, Chap. IV., -4a'' = (2a+a;;3-^/y)^ 2p = x^ + yiy;, the former gives a;=' + y''-2a^cos^ = 4(a»-i'). (1), the latter , , . , ip==(x + y)(^+y)+{x-y)(J3-y), which, since P+y, P -y are vectors bisecting the angles "between the lines and therefore at right angles to one another, is an equa- tion of the form of that in Art. 55. 2; whilst equation (1) satisfies the condition {x + yf+m{x-yY = c, which is requisite for an ellipse. . .'. APPENDIX. 217 Ex. 3. Let a be a vector semi-diameter, parallel to a cliord through ; S the vector to : then p = 8 + cca gives aS'8^8 + 2xS^a. + x'Satfta ■-■= 1 , which, since Saa=l, shews that the product of the two values of x is constant ; hence the rectangle by the segments of the chord varies as a", which is the proposition. Ex. 4. "With the usual notation, let CE, CE' be semi- diameters parallel to DP, D'P, and let their vectors be m(a — P), n(a + P) ; then since P, J), E, E' are points in the ellipse, .: 2ot'=]. Similarly 2m^ = 1, ni = ii, and DP : D'P :: r(a-/3) : T(a. + P) :: Tm{a-P) : Tn{a + P) :: CE : CE'. Cor. Since m==^, CE : DP :: I : J2. Ex. 5. Put na, np in place of a, p in equation (1), Art. 43. Ex. 6, 7. "With everything as in Ex. 4, CE, CE' being now semi-diameters in the direction of diagonals of the parallelogram, SCE4>CE' =^S{a-p)i>{a + l3) = 0j hence CE, CE' ai-e conjugate. Ex. 8. S{a + P)(a + fi) = 2 gives an ellipse, whose equation is Sp'p = 1, whei-e <{>'= ^ ; hence the diametere of the locus are to those of the given ellipse :: v/2 : 1, 218 QUATERNIONS. Ex. 9. If y be a unit vector to which the lines are parallel, p, p points in which the lines cut the ellipse, p=ai + my, p' = bj + ny, and iSp4>P = 1 gives Similarly 2bSjp' = an Siy + bniSjy + mnSy^y — 0, by equations (1) ; .'. p, p' ai'e conjugate. Cor. The same demonstration applies when the diameters from whose extremities parallels are drawn, are any conjugate diameters whatever, i, j being parallel to those diameters. Ex. 10. Let OF, CP" be any two semi-diameters, their vec- tors being a, a ; PQ the semi-ordinate to GP'; CQ = na' ; then S{PQ.a:)=0 gives >S' (a - na) a = 0, .■..«. = Sa(f>a'. Now the area of the triangle QCP is proportional to V{CP.CQ), i. e. to n Vaa or to Sa(f>a.' . Vaa', which, being symmetrical in a, a, proves the proposition. Ex. 11, If the tangent at P' meet GP produced in T, CT = ma; then, since P'F is perpendicular to <^a', SiCT-a')cj>a' = 0. 1 oap = c, y a vector along PQR ; then ^'yy and since CQ = a + y8 + a?y, S{CQy) = 0; hence PH is conjugate to CQ, and therefore bisected at Q. Ex. 13. This is simply a combination of 49. 2 and 49. 1. Chap. VII. Ex. 3. The equation of the circle is / aV 9 . ■which by 52. 1 gives 5 (a'- Sap)' — a' Sap = y^ a*, ■•■ Sap=j, which (52. ] 1) is the proposition. Ex. 5. If be the centre of the circle, Q a point at which it meets the tangent at .4 ; then, with the notation of 55. 1, QO' = {aa +\ (p - aa) - z^Y = J (P " ««)'. 220 QUATERNIONS. .-. z'li'-zSI3p + aSap = 0, i.e. z' — zy+2 =0, which gives two equal values of »; hence the proposition. Ex. 6. With any point as origin, let )8, y be the vectors to the two given points, ir the vector to the focus of one of the parabolas. Write aa in place of a in equation (1), Art. 52, a being a unit vector ; then -(l3-7ry = {a+Sa{P-7r)Y (1) -{y-^f = {a+Sa(y-^)Y, whence, by subtraction, li'-f-2S7r{^-y)= -Sa{P-y){2a + Sa{l3-y)-2SaTr}, which gives a by a simple equation in tt; and then equation (1) becomes a quadratic in tt. Ex. 8. If two tangents meet at T, it is easy, as in Ex. 5, Art. 55, with the notation available for the focus, to find e-p_yy'„ , y + y' o „„ yy_ _ ,y + y and S {ST . ST') = will follow at once, from the fact that Ex. 9. Let P be the point of contact, FQ the chord, TUF the line parallel to the axis cutting the curve in Hj; E the origin ; EpJ^o. + tp, £T=-^^a, , I U' whence ^"^J~l'^ ^"~'2 ' APPENDIX. 221 .-. PF : FQ :: t : t' ■■ 2 ■ 2 :: TE : EF. Ex. 10. This is evident from equation (1) Art. 52. Ex. 11. With the notation of Art. 52, let SQ = xFS = —xp, Ay = — x'AP = a;' ( - - p j , ,•. X (a— 2/D) = a+ 8, x'{a:'-2Sap) = a\ But p,—xp being vectors to the parabola, equation (1), Art. 52, gives x^a^-Sapy = {a? + xSa.pY, .'. x{a.' - Sap) = a^ + xSap, X (a' - 2Sp = 1 , Sir(f>p => 1, with the condition ■ff=x<^p, give 1 IT* —J (SW^"V =1, — = 1 respectively, therefore STir= 1 is the equation of the plane of contact, and if sk be the point in which this plane cuts the axis of z, zSk^ir = 1, i. e. zSTr(f>k= 1, gives «. Now ^i is a multiple of k, and since Sirk is constant, z is constant. Ex. 7. The equations of the ellipsoids Spp = 1, S(p-a.)4>(p-a) = l, give Sptjia = const, as the plane of contact. Ex. 8. If^a be the vector to the point in the line OA; the equation of its .polar plane is Spacfip = 1 ; and the square of the reciprocal of the perpendicular from the centre on this plane is —p^ (<^a)^ Hence the conclusion by Ex. 8, Art. 64. Ex. 9. Let p be the vector to F; a, )S, y vector radii parallel to the chords; then p +xa, p+yji, p + »y. APPENDIX, 1 225; will bei the vectors to A, B, ; and since P, A, B, G are points in the ellipsoid Spp=l, 2Spy (Ta)' " Spct>a "^ {T^}' ' Sp^) ' which is satisfied by w — p = mPf where ; "U(W ^ (W ^ (^1 " ^ ' and therefore Ex. 4 above gives 1 i 1 Chap. IX. Ex. 2 and 3. Employ formula 11. Ex. 5. Since a'fiy=al3y.ypa, formulae 4 and 6 give the required result. Ex. 6. Apply formula 10 to Ex. 5. Ex. 8. (aiSy)' = aiSy . a^y = af3y {S . a^y + V . OjSy) = aPy{S.al3y+ V-yfia) = al3y{yPa + 2S.al3y) = a'py+2al3yS.apy. T. q, 13 226^. quAternioiJS. Ex. 9. Formula 10 gives the vector of the product of three vectors a, p, y, under the form a — /3' + y where a = aS^y, &0. Hence the, Required scalax may be written S.{a'-P' + y'){a' + P'-y')(-a' + ^' + y')i and as the scalar part of this product is that which involves all of the three vectorg a', jS', y' we have exactly as in the demonstra- tion of formula 5, S(ra^yri3yaryal3) ■ ~^- °-',-^',y' ■ a',/3',-y' 10. The scalar part, by formula 16, is reduced to SaSS^y - SaySjSS - SahS^y + SajSSyS + SayS^y - Sa^SyS, which is identically 0. The vector part, by formula 12, is aS . yS/S - j3S . ySa + aS . Sl3y -yS .B/3a + aS . fiyB- is . Pya, which, by formula 13, reduces to 2aS.PyS. 12. If, for brevity, we denote S . a;8y, F. a/?y respectively by J5±S..45C = 0; and since tlie scalar of the product of tliis veotor by the vector perpendicular to tlie phme in whieJi A, B, C, D lie gives tlie riglit- hand side of Ex. 13, we obtain a. BCD - p ,CDA + y . DAB -Z . ABC =0. CAMBKllKJK : FBIXTED BY C, J . ClAT, M-i. AT THE TSl^KKSITT PRfSS. ..nell University libraries APR 6 ISO. ^'.AT'-^WIATICS LIBRA:^.V jneil University Libraries MATHEMATICS UBRAR'