THE GIFT OF 
 
 . A4 ?.c,7i ^ :,Z.: rj^/fC 
 
Cornell University Library 
 
 WV19316 
 
 A primer of quaternions, 
 
 3 1924 019 111 008 
 
The original of tliis book is in 
 tine Cornell University Library. 
 
 There are no known copyright restrictions in 
 the United States on the use of the text. 
 
 http://www.archive.org/details/cu31 92401 91 1 1 008 
 
A PRIMER OF QUATERNIONS 
 
.^<mi 
 
A PEIMER 
 
 QUATEENIONS 
 
 ARTHUR S. HATHAWAY 
 
 PROFESSOE OF MATHEMATICS IN THE ROSE POLYTECHNIC 
 INSTITUTE, TERRE HAUTE, IND. 
 
 MACMILLAN AND CO. 
 
 LONDON: MACMILLAN & CO., Ltd. 
 1896 
 
 All rights reserved 
 
COPTEIQHT, 1896, 
 
 By MAOMILLAN AND CO. 
 
 J. S. Gushing & Co. — Berwick & Smitll 
 Norwood Mass. U.S.A. 
 
PREFACE 
 
 The Theory of Quaternions is due to Sir William 
 Rowan Hamilton, Royal Astronomer of Ireland, 
 who presented his first paper on the subject to 
 the Royal Irish Academy in 1843. His Lectures 
 on Quaternions were published in 1853, and his 
 Elements, in 1866, shortly after his death. The 
 Elements of Quaternions by Tait is the accepted 
 text-book for advanced students. 
 
 The following development of the theory is pre- 
 pared for average students with a thorough knowl- 
 edge of the elements of algebra and geometry, 
 and is believed to be a simple and elementary 
 treatment founded directly upon the fundamental 
 ideas of the subject. This theory is applied in the 
 more advanced examples to develop the principal 
 formulas of trigonometry and solid analytical geom- 
 etry, and the general properties and classification of 
 surfaces of second order. 
 
vi PREFACE 
 
 111 the endeavour to bring out the number idea 
 of Quaternions, and at the same time retain the 
 established nomenclature of the analysis, I have 
 found it necessary to abandon the term " vector " 
 for a directed length. I adopt instead Clifford's 
 suggestive name of '■'■step," leaving to ^^ vector" the 
 sole meaning of "right quaternion." This brings 
 out clearly the relations of this number and line, 
 and emphasizes the fact that Quaternions is a nat- 
 ural extension of our fundamental ideas of number, 
 that is subject to ordinary principles of geometric 
 representation, rather than an artificial species of 
 geometrical algebra. 
 
 The physical conceptions and the breadth of idea 
 that the subject of Quaternions will develop are, 
 of themselves, sufficient reward for its study. At 
 the same time, the power, directness, and simplic- 
 ity of its analysis cannot fail to prove useful in 
 all physical and geometrical investigations, to those 
 who have thoroughly grasped its principles. 
 
 On account of the universal use of analytical 
 geometry, many examples have been given to show 
 that Quaternions in its semi-cartesian form is a 
 direct development of that subject. In fact, the 
 present work is the outcome of lectures that I 
 
PREFACE Vll 
 
 have given to my classes for a number of years 
 past as the equivalent of the usual instruction in 
 the analytical geometry of space. The main feat- 
 ures of this primer were therefore developed in 
 the laboratory of the class-room, and I desire to 
 express my thanks to the members of my classes, 
 wherever they may be, for the interest that they 
 have shown, and the readiness with which they 
 have expressed their difficulties, as it has been a 
 constant source of encouragement and assistance in 
 my work. 
 
 I am also otherwise indebted to two of my stu- 
 dents, — to Mr. H. B. Stilz for the accurate con- 
 struction of the diagrams, and to Mr. G. Willius 
 for the plan (upon the cover) of the plagiograph or 
 mechanical quaternion multiplier which was made 
 by him while taking this subject. The theory of 
 this instrument is contained in the step proportions 
 that are given with the diagram.* 
 
 ARTHUR S. HATHAWAY. 
 * See Example 19, Chapter I. 
 
CONTENTS 
 
 
 
 CHAPTER I 
 
 
 
 Steps 
 
 
 
 Definitions and Theorems .... 
 
 ABTICLE8 
 
 . 1-12 
 
 PAGE 
 
 1 
 
 Centre of Gravity 
 
 . 13-16 
 
 9 
 
 Curve Tracing, Tangents .... 
 
 . 17-18 
 
 13 
 
 Parallel Projection 
 
 . 19-20 
 
 19 
 
 Step Proportion 
 
 . 21-22 
 
 21 
 
 Examples 
 
 , 
 
 23 
 
 CHAPTER II 
 
 Rotations. Turns. Arc Steps 
 
 Definitions and Theorems of Rotation . 
 Definitions and Theorems of Turn and Arc Steps 
 Examples 
 
 CHAPTER III 
 
 Quaternions 
 
 Definitions and Theorem 
 
 Examples 
 
 Multiplication 
 
 The Rotator 9 ( ) {"^ 
 
 23-26 28 
 
 27-31 33 
 
 . 41 
 
 32-34 44 
 
 . 47 
 
 35-40 48 
 
 41-42 52 
 
X CONTENTS 
 
 AETICLES PAGE 
 
 Powers and Roots 43 54 
 
 Representation of Vectors, Geometric Theorems . 44-^6 55 
 
 Examples .60 
 
 Addition 47-52 62 
 
 Formulas 53-57 70 
 
 Geometric Theorems 58-60 75 
 
 Examples . , .78 
 
 CHAPTER IV 
 Equations op First Degree 
 
 Scalar Equations, Plane and Straight Line . 
 
 
 61-68 
 
 83 
 
 Examples 
 
 87 
 
 NONIONS 
 
 
 Vector Equations, the Operator <^ . . . . 69-72 
 
 90 
 
 Linear Homogeneous Strains 
 
 
 
 
 73-77 
 
 92 
 
 Finite and Null Strains . 
 
 
 
 
 78-80 
 
 96 
 
 Solution of ^p = 8 . 
 
 
 
 
 81 
 
 100 
 
 Derived Moduli. Latent Roots 
 
 
 
 
 82-83 
 
 101 
 
 Latent Lines and Planes 
 
 
 
 
 84-85 
 
 103 
 
 The Characteristic Equation . 
 
 
 
 
 86 
 
 105 
 
 Conjugate Nonions 
 
 
 
 
 87-89 
 
 105 
 
 Self-conjugate Nonions . 
 
 
 
 
 90-94 
 
 107 
 
 Examples 
 
 
 
 
 . 
 
 109 
 
A PRIMER OF QUATERNIONS 
 
 oXKo 
 
 CHAPTER I 
 Steps 
 
 1. Definition. A step is a given length 
 measured in a given direction. 
 
 E.g., 3 feet east, 3 feet north, 3 feet up, 3 
 feet north-east, 3 feet north-east-up, are steps. 
 
 2. Definition. Two steps are equal when, 
 and only when, they have the same lengths and 
 the same directions. 
 
 E.g., 3 feet east, and 3 feet north, are not 
 equal steps, because they differ in direction, 
 although their lengths are the same ; and 3 
 feet east, 5 feet east, are not equal steps, be- 
 cause their lengths differ, although their direc- 
 tions are the same ; but all steps of 3 feet 
 
2 PBIMEB OF QUATERNIONS 
 
 east are equal steps, whatever the points of 
 departure. 
 
 3. We shall use bold-faced AB to denote 
 the step whose length is AB, and whose direc- 
 tion is from A towards B. 
 
 Two steps AB, CD, are obviously equal when, 
 and only when, ABJDC is a parallelogram. 
 
 4. Definition. If several steps he taken 
 in succession, so that each step begins where the 
 preceding step ends, the step from the beginning 
 of the first to the end of the last step is the 
 sum of those steps. 
 
 E.g., 3 feet east + 3 feet north = 3^2 feet 
 north-east = 3 feet north + 3 feet east. Also 
 AB + BC = AC, whatever points A, B, C, may 
 
STEPS 6 
 
 be. Observe that this equality between steps 
 
 is not a length equality, and therefore does 
 
 N.E 
 
 p' ' ■ ^E 
 not contradict the inequality AB + BC > AC,' 
 just as 5 dollars credit + 2 dollars debit = 3 
 dollars credit does not contradict the inequal- 
 ity 5 dollars + 2 dollars > 3 dollars. 
 
 5. If equal steps he added to equal steps, 
 the sums are equal steps. 
 
 Thus if AB = A'B', and BC=B'C', then AC = 
 A'C, since the triangles ABC, A'B'C must 
 
4 PMIMEB OF QUATEENIONS 
 
 be equal triangles with the corresponding sides 
 in the same direction. 
 
 6. A sum of steps is commutative (i.e., the 
 components of the sum may be added in any 
 order without changing the value of the sum). 
 
 For, in the sum AB + BC + CD + DE + •-, let 
 BC' = CD; then since BGDC is a parallelo- 
 gram, therefore CD = BC, and the sum with BC, 
 CD, interchanged is AB + BC + CD + DE + ■••, 
 which has the same value as before. By such 
 interchanges, the sum can be brought to any 
 order of adding. 
 
 7. A sum of steps is associative (i.e., any 
 number of consecutive terms of the sum may 
 
STEPS 5 
 
 be replaced by their sum without changing 
 the value of the whole sum). 
 
 For, in the sum AB + BC + CD + DE + ---, let 
 BC, CD, be replaced by their sum BD ; then 
 the new sum is AB + BD + DE + •••, whose value 
 is the same as before ; and similarly for other 
 consecutive terms. 
 
 8. The product of a step hy a positive num- 
 her is that step lengthened by the multiplier 
 without change of direction. 
 
 E.g., 2AB = AB + AB, which is AB doubled 
 in length without change of direction ; simi- 
 larly iAB = (step that doubled gives AB) = 
 (AB halved in length without change of direc- 
 tion). In general, mAB = m lengths AB meas- 
 ured in the direction AB: -AB = -th of lena;th 
 
 n n ° 
 
 AB measured in the direction AB ; etc. 
 
 9. The negative of a step is that step re- 
 versed in direction without change of length. 
 
 For the negative of a quantity is that quan- 
 tity which added to it gives zero ; and since 
 
6 PRIMER OF QUATERNIONS 
 
 AB+BA = AA = 0, therefore BA is the negative 
 of AB, or BA=-AB. 
 
 CoE. 1. The product of a step hy a nega- 
 tive number is that step lengthened hy the Clum- 
 ber and reversed in direction. 
 
 For — nAB is the negative of nAB. 
 
 Cor. 2. A step is subtracted by reversing 
 its direction and adding it. 
 
 For the result of subtracting is the result of 
 adding the negative quantity. -E.^., AB — CB = 
 AB+BC = AC. 
 
 10. A sum of steps is multiplied by a given 
 number by multiplying the components of the 
 sum by the number and adding the products. 
 
 Let n-AB = A'B', n-BC=B'C'; then ABC, 
 A'B'C are similar triangles, since the sides 
 
STEPS 7 
 
 about B, B' are proportional, and in the same 
 or opposite directions, according as n is posi- 
 tive or negative; therefore AG, A'C are in 
 the same or opposite directions and in the same 
 ratio ; i.e., nAC = A'C, which is the same as 
 w(AB+BC) = nAB + nBC. 
 
 This result may also be stated in the form : 
 a multiplier is distributive over a sum. 
 
 11. Any step may be resolved into a multiple 
 of a given step parallel to it ; and into a sum of 
 multiples of two given steps in the same plane 
 with it that are not parallel ; and into a sum of 
 multiples of three given steps that are not par- 
 allel to one plane. 
 
 A 
 
 OP=OA+OB+OC= 
 
 a;oA+2/OB + zoc 
 
8 PniMER OF QUATERNIONS 
 
 12. It is obvious that if the sum of two 
 finite steps is zero, then the two steps must be 
 parallel; in fact, if one step is AB, then the 
 other must be equal to BA. Also, if the sum 
 of three finite steps is zero, then the three steps 
 must be parallel to one plane ; in fact, if the 
 first is AB, and the second is BC, then the third 
 must be equal to CA. Hence, if a sum of steps 
 on two lines that are not parallel (or on three 
 lines that are not parallel to one plane) is zero, 
 then the sum of the steps on each line is zero, 
 since, as just shown, the sum of the steps on 
 each line cannot be finite and satisfy the condi- 
 tion that their sum is zero. We thus see that 
 an equation between steps of one plane can be 
 separated into two equations by resolving each 
 step parallel to two intersecting lines of that 
 plane, and that an equation between steps in 
 space can be separated into three equations by 
 resolving each step parallel to three lines of 
 space that are not parallel to one plane. We 
 proceed to give some applications of this and 
 other principles of step analysis in locating a 
 
STEPS 9 
 
 point or a locus of points with respect to given 
 data (Arts. 13-20). 
 
 Centre of Gravity 
 
 13. The point P that satisfies the condition 
 ZAP + mBP = lies upon the line AB and divides 
 AB in the inverse ratio of l:m (i.e., P is the 
 centre of gravity of a mass I at A and a mass 
 m at B). 
 
 The equation gives ZAP = mPB; hence: 
 
 AP, PB are parallel ; P lies on the line AB ; 
 and AP:PB = m:Z = inverse of l:m. 
 
 If l:m is positive, then AP, PB are in the 
 same direction, so that P must lie between A 
 and B ; and if l:m is negative, then P must 
 lie on the line AB produced. If l = m, then 
 P is the middle point of AB ; \il= —m, then 
 there is no finite point P that satisfies the con- 
 dition, but P satisfies it more nearly, the far- 
 ther away it lies upon AB produced-, and this 
 fact is expressed by saying that "P is the 
 point at infinity on the line AB." 
 
10 PBIMER OF QUATERNIONS 
 
 14. By substituting AO + OP for AP and 
 BO + OP for BP in ZAP + mBP = 0, and trans- 
 posing known steps to the second member, we 
 find the point P with respect to any given 
 origin 0, viz., 
 
 (a) (Z + m)OP = ZOA + mOB, lohere P divides 
 AB inversely as l:m. 
 
 CoK. If OC = /OA + mOB, then OC, produced 
 if necessary, cuts AB in the inverse ratio of 
 I : m, and OC is (l + m) times the step from 
 to the point of division. 
 
 For, if P divide AB inversely as I : m, then 
 by (a) and the given equation, we have 
 
 OC = (Z + m)OP. 
 
STEPS 11 
 
 15. The point P that satisfies the condition 
 ZAP + mBP + nCP= lies in the plane of the 
 triangle ABC; AP (produced) cuts BC at a 
 point D that divides BC inversely as m :n, 
 and P divides AD inversely as l:m + n (i.e., 
 P is the center of gravity of a mass I at A, 
 a mass m at B, and a mass n at C). Also 
 the triangles PBC, PCA, PAB, ABC, are 
 proportional to I, m, n, l + m + n. 
 
 The three steps ZAP, mBP, nCP must be 
 parallel to one plane, since their sum is zero, 
 and hence P must lie in the plane of ABC. 
 Since BP=BD + DP, CP = CD + DP, the equa- 
 tion becomes, by making these substitutions, 
 lAP + {m + n)DP+mBD + nCD = 0. This is an 
 
12 PRIMER OF QUATERNION 8 
 
 equation between steps on the two intersecting 
 lines, AD, BC, and hence the resultant step 
 along each line is zero; i.e., mBD + nCD = 
 (or D divides BC inverseljr^s m : n), and 
 
 {a) ZAP + (m + n)DP = 
 
 (or P divides AD inversely as l:m + n). Also, 
 we have, by adding ZPD + ZDP = to (a), 
 
 ZAD + (Z + m + n)DP=0. 
 Hence 
 
 l:l + m + n=PD:kD = PBC:ABC, 
 
 since the triangles PBG, ABC have a com- 
 mon base BC. (We must take the ratio of 
 these triangles as positive or negative accord- 
 ing as the vertices P, A lie on the same or 
 opposite sides of the base BC, since the ratio 
 PD : AD is positive or negative under those 
 circumstances.) Similarly, 
 
 PCA:ABC=m:l + m + n, 
 and PAB : ABC=n:l + m + n. 
 
 Hence, we have, 
 PBC : PCA : PAB : ABC = 1 : m -.n : l + m + n. 
 
STEPS 13 
 
 16. By introducing in ZAP + mBP + wCP= 
 an origin 0, as in Art. 14, we find 
 
 (a) {l + m + n)OP = lOA + 'mOB + nOC, where 
 P divides ABC in the ratio I : m : n. 
 
 Note. As an exercise, extend tMs formula for the 
 center of gravity P, of masses I, m, n, at A, B, C, to 
 four or more masses. 
 
 Curve Tkacing. Tangents. 
 
 17. To draw the locus of a point P that 
 varies according to the law OP=tOk+hfO'Q, 
 where t is a variable number. {E.g., i = num- 
 ber of seconds from a given epoch.) 
 
 Take ^ = — 2, and P is at D', where 
 
 OD'=-2 0A + 2 0B. 
 
 Take i = — 1, and P is at C, where 
 
 OC = - OA + J OB. 
 
 Take i = 0, and P is at 0. Take t= 1, and 
 P is at C, where OC = OA + JOB. Take t = 2, 
 and P is at D, where OD=20A + 20B. It 
 
14 
 
 PRIMER OF QUATERNfONS 
 
 is thus seen that when t varies from -2 to 
 2, then F traces a curve UC'OCD. To draw 
 the curve as accurately as possible, we find 
 
 OA+-(ttJh)OB 
 
 the tangents at the points already found. The 
 method that we employ is perfectly general 
 and applicable to any locus. 
 
STEPS 15 
 
 (a) To find the direction of the tangent to 
 the locus at the point P corresponding to any 
 value of t. 
 
 Let P, Q be two points of the locus that 
 correspond to the values t, t + h of the variable 
 number. We have 
 
 OP =tOA+i fOB, 
 
 OQ = {t + h)OA+i{t + hfOB, 
 
 and therefore 
 
 PQ = OQ - OP = A[OA + (« + J h)OB']. 
 
 Hence (dropping the factor h) we see that 
 OA + {t+ih)OB is always parallel to the 
 chord PQ. Make h approach 0, and then Q 
 approaches P, and the (indefinitely extended) 
 chord PQ approaches coincidence with the 
 tangent at P. Hence making h= 0, in the 
 step that is parallel to the chord, we find that 
 OA + tOB is parallel to the tangent at P. 
 
 Apply this result to the special positions of 
 P already found, and we have : D'A' = OA — 
 2 OB = tangent at D' ; C'S = OA - OB = tangent 
 
16 PRIMER OF QUATERNIONS 
 
 at C; OA = OA + • OB = tangent at 0; 
 SC=OA + OB= tangent at C ; AD = 0A+20B 
 = tangent at D. 
 
 This is the curve described by a heavy par- 
 ticle throv^n from with velocity represented 
 by OA on the same scale in vphich OB repre- 
 sents an acceleration of 32 feet per second per 
 second downwards. For, after t seconds the 
 particle will be displaced a step t ■ OA due to 
 its initial velocity, and a step hf-OB due to 
 the acceleration downwards, so that P is act- 
 ually the step OP = tOk + ifOB from at 
 time t. Similarly, since the velocity of P is 
 increased by a velocity represented by OB in 
 every second of time, therefore P is moving at 
 time t with velocity%.epresented by OA + ^OB, 
 so that this step must be parallel to the tan- 
 gent at P. 
 
 18. To draw the locus of a point P that 
 varies according to the law 
 
 OP = cos {nt -I- e) • OA + sin (nt + e) ■ OB, 
 
STEPS 17 
 
 where OA, OB are steps of equal length and 
 perpendicular to each other, and t is any 
 variable number. 
 
 Witli centre and radius OA draw the circle 
 ABA'B'. Take arc AE = e radians in the 
 direction of the quadrant AB (i. e. an arc of 
 e radii of the circle in length in the direction of 
 AB or AB' according as e is positive or nega- 
 tive). Corresponding to any value of t, lay off 
 arc ^P = nt radians in the direction of the 
 quadrant AB. Then arc AB = nt + e radians. 
 Draw LP perpendicular to OA at L. Then 
 according to the definitions of the trigonometric 
 functions of an angle we have, 
 
 cos {nt + e)= 01/ OP, sin {nt + e) = LP/ OP.* 
 Hence we have for all values of t, 
 
 OL = cos {nt + e) ■ OA, LP = sin {nt + e) • OB, 
 and adding these equations, we find that 
 OP = cos {nt + e) OA + sin {nt + e) OB. 
 
 * Observe the distinctions: OL, a step; OL, a positive or 
 negative length of a directed axis ; OL, a length. 
 
18 
 
 PRIMER OF QUATERNIONS 
 
 Hence, the locus of the required point P is the 
 circle on OA, OB as radii. 
 
 Let t be the number of seconds that have 
 elapsed since epoch. Then, at epoch, t=0, and 
 P is at ^ ; and since in t seconds F has 
 
 moved through an arc UP of nt radians, 
 therefore P moves uniformly round the circle 
 at the rate of n radians per second. Its velocity 
 at time t is therefore represented by n times 
 that radius of the circle which is perpendicular 
 to OP in the direction of its motion, or by 
 
STEPS 19 
 
 OP' = nOQ, where arc PQ=^ -x radians. Hence, 
 since arc -i4.Q=(ni + e + ^] radians, therefore OP' 
 
 = n 
 
 cos fni + e + |j ■ OA + sin (nt + e + |V OB 
 
 The point P' also -moves uniformly in a circle, 
 and this circle is the hodograph of the motion. 
 The velocity in the hodograph (or the accelera- 
 tion of P) is similarly 0P"= n^PO. 
 
 Parallel Projection 
 
 19. If 0? = xOk + yOB, OP'=xOA + t/OB', 
 
 where x, y vary with the arbitrary number t 
 according to any given law so that P, P' describe 
 definite loci {and have definite motions when t 
 denotes time), then the tioo loci {and motions) 
 are parallel projections of each other by rays that 
 are parallel to BB', 
 
 For, by subtracting the two equations we find 
 PP' = ?/BB', so that PP' is always parallel to 
 BB' ; and as P moves in the plane AOB and 
 P' moves in the plane AOB', therefore their 
 
20 
 
 PRIMER OF QUATERNIONS 
 
 loci (and motions) are parallel projections of 
 each other by rays parallel to BB'. The par- 
 allel projection is definite when the two planes 
 coincide, and may be regarded as a projection 
 between two planes AOB, AOB', that make an 
 indefinitely small angle with each other. 
 
 A T 
 
 20. The motion of P that is determined ty 
 
 OP = cos (nt + e) OA + sin [nt + e)OB 
 
 is the parallel projection of uniform circular 
 motion. 
 
 For, draw a step OB' perpendicular to OA 
 and equal to it in length. Then, by Art. 18, 
 the motion of P' determined by 
 
 OP' = cos {nt + e) OA + sin {nt + e) OB' 
 
STEPS 
 
 21 
 
 is a uniform motion in a circle on OA, OB' as 
 radii ; and by Art. 19 this is in parallel perspec- 
 tive with the motion of P. 
 
 Step PROPORXioisr 
 
 21. Definition. Fou?- steps AC, AB, A'C, 
 A'B' are in proportion ivhen the first is to the 
 second in respect to both relative length and 
 relative direction as the third is to the fourth in 
 the same respects. 
 
 This requires, first, that the lengths of the 
 steps are in proportion or 
 
 AC : AB = A'C : A'B' ; 
 .c' 
 
 'B A' 
 
22 PBIMEB OF QUATERNIONS 
 
 and secondly, that AC deviates from AB by the 
 same plane angle in direction and magnitude 
 that A'C deviates from A'B'. 
 
 Hence, first, the triangles ABO, A'B'C are 
 similar, since the angles A, A' are equal and 
 the sides about those angles are proportional ; 
 and secondly, one triangle may be turned in its 
 plane into a position in which its sides lie in the 
 same directions as the corresponding sides of 
 the other triangle. Two such triangles will 
 be called similar and congruent triangles, and 
 corresponding angles will be called congruent 
 angles. 
 
 22. We give the final propositions of Euclid, 
 Book v., as exercises in step proportion. 
 
 (xi.) If four steps are proportionals, they are 
 also proportionals when taken alternately. 
 
 (xii.) If any number of steps are propor- 
 tionals, then as one of the antecedents is to its 
 consequent, so is the sum of the antecedents to the 
 sum of the consequents. 
 
STEPS 23 
 
 (xiii.) If four steps are proportionals, the sum 
 {or difference) of the first and second is to the 
 second as the sum {or difference) of the third and 
 fourth is to the fourth. 
 
 (xiv.) If OA : OB = OP : OQ and OB : 00 = 
 OQ : OR, then OA : 00 = OP : OR. 
 
 (xv.) If OA : OB = 00 : OD and OE : OB = 
 OF : OD, then OA + OE : OB = 00 + OF : OD. 
 
 (xvi.) If OA : OB = OB : OX = 00 : OD = 
 OD : OY, then OA : OX = 00 : OY. 
 
 EXAMPLES 
 
 We shall use i, j, k, as symbols for unit length east, 
 unit length north, and unit length up, respectively. 
 
 •'1. Mark the points whose steps from a given point 
 are i + 2 j, — 3 i — j. Show that the step from the first 
 point to the second is — 4 i — 3 j, and that the length 
 is 5. 
 
 2. Show that the four points whose steps from a 
 given point are 2 i + j, 5 i + 4 j, 4 i + 7 j, i + 4 j are 
 the angular points of a parallelogram. Also determine 
 their centre of gravity, with weights 1, 1, 1, 1; also 
 with weights 1, 2, 3, 4 ; also with weights 1, — 2, 3, — 4. 
 
24 PRIMER OF QUATERNIONS 
 
 3. IfOA = i+2j, OB=4i+3j, OC=2i+3j, 0D= 
 4i+j, find. CD as sums of multiples of CA, CB, and 
 show that CD bisects AS. 
 
 4. If 0P= x\+y], 0P'= x'\ + y'i, then PP'=(a;'-a;)i 
 + (y'- y) j and PP'^ = («' -xf+{y'- y)\ 
 
 5. Show that AB is bisected by OC = OA + OB, and 
 trisected by 0D = 2 0A + 0B, 0E=0A + 2 0B, and 
 divided inversely as 2 : 3 by OF = 20A + 30B. 
 
 6. Show that AA'+BB' = 2MM', where MM are the 
 middle points of AB, A'B', respectively. 
 
 7. Show that 2 AA' + 3 B B' = (2 + 3) CC, where C, 0' 
 are the points that divide AB, A'B', inversely as 2:3. 
 Similarly, when 2, 3 are replaced by /, m. 
 
 8. Show that the point that divides a triangle into 
 three equal triangles is the intersection of the medial 
 lines of the triangle. 
 
 9. Show that the points which divide a triangle 
 into triangles of equal magnitude, one of which is negar 
 tive (the given triangle being positive), are the vertices 
 of the circumscribing triangle with sides parallel to the 
 given triangle. 
 
 10. .If a, b, c are the lengths of the sides BC, GA, 
 
 AB of a triangle, show that - AC ± - AB (drawn from A) 
 
 b c 
 
 are interior and exterior bisectors of the angle A ; and 
 that when produced they cut the opposite side BC in the 
 ratio of the adjacent sides. 
 
STEPS 25 
 
 ABO to any ■] {■ in its plane divide the sides BC, 
 
 CA, AB in ratios whose product is •] _-,; and conversely 
 
 {poTnts'r *^^ \7^ *^^t - ^-^e t^e -^- 
 
 ( meet in a point. 
 ( lie in a line. 
 
 12. Prove by Exs. 10, 11, that the three interior bi- 
 sectors of the angles of a triangle (also an interior and 
 two exterior bisectors) meet in a point; and that the 
 three exterior bisectors (also an exterior and two in- 
 terior bisectors) meet the sides in colinear points. 
 
 13. Determine the locus (and motion) of P, given by 
 OP = OA-|-^OB; also of OP =(1 + 2f)i +(3i5-2)j. 
 
 14. Compare the loci of P determined by the follow- 
 ing pairs of step and length equations : 
 
 AP = 2east, AP = 2; AP = 2 BP, J.P=2 5P; 
 kP+BP = CQ,AP + BP=CD. 
 
 15. Draw, by points and tangents, the locus of P 
 
 determined by each of the following values of OP, in 
 
 which X is any number : 
 
 2 
 x\ + \o?]; a!i-|--j; m + ^s?']; a;i -|- (^ ie' — ar^ -|- 2) j ; 
 
 x\ + — i ; x\ + V4 — v^\ ; a;i -f i V4 — 3?\. 
 
 ar + 4 
 
26 PRIMER OF QUATERNIONS 
 
 16. Take three equal lengths making angles 120° with 
 each other as projections of i, j, k, and construct by points 
 the projection of the locus of P, where OP = 2 (cos a;- i + 
 sina;'j)+a;-k, x varying from to 27r. Show that this 
 curve is one turn of a helix round a vertical cylinder of 
 altitude 2 tt, the base being a horizontal circle of radius 
 2 round as centre. 
 
 17. A circle rolls inside a fixed circle of twice its 
 diameter ; show that any point of the plane of the rolling 
 circle traces a parallel projection of a circle. 
 
 18. A plane carries two pins that slide in two fixed 
 rectangular grooves ; show that any point of the sliding 
 plane traces a parallel projection of a circle. 
 
 19. OAGB is a parallelogram whose sides are rigid 
 and jointed so as to turn round the vertices of the paral- 
 lelogram ; APO, BCQ are rigid similar and congruent 
 triangles. Show that AC : AP = BQ : BC = OQ : OP, and 
 that therefore P, Q trace similar congruent figures when 
 remains stationary (21, 22, xii.). [See cover of book.J 
 
 20. If the plane pencil OA, OB, OG, OD is cut by any 
 straight line in the points P, Q, R, S, show that the 
 cross-ratio (PB : BQ) : (PS : SQ) is constant for all posi- 
 tions of the line. 
 
 [OC = ZOA + mOB = IxOP + myOQ gives 
 PB : BQ = my: Ix}. 
 
 21. Two roads run north, and east, intersecting at 0. 
 A is 60 feet south of 0, walking 3 feet per second north, B 
 
STEPS 27 
 
 is 60 feet west of 0, walking 4 feet per second east. When 
 are A, B nearest together, and what is B's apparent 
 motion as seen by ^ ? 
 
 22. What is B's motion relative to A in Ex. 21 if B 
 is accelerating his walk at the rate of 3 inches per second 
 per second ? 
 
 23. In Ex. 21, let the east road be 20 feet above the 
 level of the north road ; and similarly in Ex. 22. 
 
 24. A massless ring P is attached to several elastic 
 strings that pass respectively through smooth rings at 
 A, B, G, ••• and are attached to fixed points A', B', C", ■•■ 
 such that A' A, B'B, CO, •■■ are the natural lengths of the 
 strings. The first string has a tension I per unit of length 
 that it is stretched (Hooke's law), the second a tension m, 
 the third a tension n, etc. Eind the resultant force on P 
 and its position of equilibrium. 
 
 25. The same as Ex. 24, except that the ring has a 
 mass w. 
 
28 
 
 PRIMER OF QUATERNIONS 
 
 CHAPTER II 
 
 Rotations. Turns. Arc Steps 
 
 23. Definitions of Rotation 
 
 A step is rotated when it is revolved about an 
 axis through, its initial point as a rigid length 
 rigidly attached to the axis. The step describes 
 a conical angle about the axis except when it is 
 perpendicular to the axis. 
 
 If a rotation through a diedral angle of 
 given magnitude and direction in space be 
 applied to the radii of a sphere of unit radius 
 
ROTATIONS. _TVBN8. ARC STEPS 29 
 
 and centre 0, the sphere is rotated as a rigid 
 body about a certain diameter PP' as axis, and 
 a plane through perpendicular to the axis 
 intersects the sphere in the equator of the 
 rotation. 
 
 Either of the two directed arcs of the equator 
 from the initial position A to the final position 
 A' of a point of the rotated sphere that lies on 
 the equator is the arc of the rotation. If these 
 two arcs be bisected at L, M respectively, then 
 the two arcs are 2 AL, 2 AM respectively, and 
 AL, AM are supplementary arcs in opposite 
 directions, each less than a semicircle. When 
 these half-arcs are 0° and 180° respectively, 
 they represent a rotation of the sphere into its 
 original position, whose axis and equator are 
 indeterminate, so that such arcs may be meas- 
 ured on any great circle of the sphere without 
 altering the corresponding rotation. 
 
 24. A rotation is determined hy the position 
 into which it rotates two given non-parallel steps. 
 For let the radii OB, OC rotate into the radii 
 
30 PEIMER OF QUATBBNI0N8 
 
 IS 
 OB', OC. Any axis rotind which OB rotates 
 
 into OB' must be equally inclined to these radii; 
 
 i.e., it is a diameter of the great circle PKL 
 
 that bisects the great arc BB' at right angles. 
 
 E.g., OK, OL, OP, — are such axes. Similarly, 
 the axis that rotates OC into OC must be a 
 diameter of the great circle PiVthat bisects the 
 great arc CC' at right angles. Hence there is 
 but one axis round which OB, 00 rotate into 
 OB', OC; viz., the intersection OP of the planes 
 of these two bisecting great circles : the equator 
 is the great circle whose plane is perpendicular 
 to this axis, and the arcs of the rotation are the 
 intercepts on the equator by the planes through 
 
ROTATIONS. TURNS. ABC STEPS 31 
 
 the axis and either B, JS' or C, C. [When the 
 two bisecting great circles coincide (as when 
 G, C lie on BP, B'P), then their plane bisects 
 the diedral angle BC— O — B'C, whose edge 
 OP is the only axis of rotation.] 
 
 Note. Since BG, B'Q' may be any two positions of 
 a marked arc on the surface of the sphere, we see that 
 any two positions of the sphere with centre fixed deter- 
 mine a definite rotation of the sphere from one position 
 to the other. 
 
 25. A marked arc of a great circle of a 
 rotating sphere makes a constant angle with the 
 equator of the rotation. 
 
 For the plane of the great arc makes a con- 
 stant angle both with the axis and with the 
 equator of the rotation. 
 
 26. If the sphere he given a rotation 
 2 A(,C followed by a rotation 2 CB^^, the resultant 
 rotation of the sphere is 2 AqBo. 
 
 For produce the arcs AqG, B^C to A^, B' 
 respectively, making (JA^ = A^C, B'C=CB^. 
 Then the spherical triangles Afi^G, A^B'C are 
 
32 PEIMEB OF QUATERNIONS 
 
 equal, since the corresponding sides about the 
 equal vertical angles at C are by construc- 
 tion equal. Therefore the sides ^o-^o? -S'JLi are 
 equal in length, and the corresponding angles 
 Aq, A^ and Bo, B' are equal. Therefore, by 
 
 Art. 25, if a marked arc AB of the sphere 
 coincide initially with ^o-^n, the first rotation 
 2 A^C = Af^A-i -wWi bring AB into the position 
 A^B^ on B'A^ produced, and the second rotation 
 2 C-Sq = B'B^ will bring AB into the position 
 A^i, on ^0-^0 produced, where B^A^ = A^^. 
 Hence the resultant rotation of the sphere is 
 2A^B, = AA,. 
 
ROTATIONS. TURNS. ARC STEPS 33 
 
 Note. This tlieoreni enables one to find the result- 
 ant of any number of successive rotations, by replacing 
 any two successive rotations by their resultant, and so 
 on until a single resultant is found. 
 
 27. Definitions of Turn 
 
 A step is turned when it is made to describe 
 a, plane angle round its initial point as centre. 
 
 If a turn through a plane angle of given 
 magnitude and direction in space be applied to 
 the radii of the sphere 0, it turns the great 
 circle that is parallel to the given plane angle 
 as a rigid circle, and does not affect the other 
 radii of the sphere, ^-g-, only horizontal radii 
 
34 PBIMEB OF QUATERNIONS 
 
 can be turned through a horizontal plane angle. 
 The circle that is so turned is the great circle of 
 the turn. 
 
 A directed arc of the great circle of a turn 
 from the initial position A to the final position 
 5 of a point on the great circle, and less than 
 a semi-circumference, is the arc of the turn. 
 When this arc is 0° or 180°, it represents a turn 
 that brings a step back to its original position 
 or that reverses it; and since such turns may 
 take place in any plane with the same results, 
 therefore such arcs may be measured on any 
 great circle of the sphere without altering their 
 corresponding turns. 
 
 The axis of a turn is that radius of the sphere 
 which is perpendicular to its great circle and 
 lies on that side of the great circle from which 
 the arc of the turn appears counter-clockwise. 
 
 28. A turn is determined by the position 
 into which it displaces any given step. 
 
 For, let the radius OA turn into the radius 
 OB. Then, the great circle — AB must be 
 
ROTATIONS. TURNS. ARC STEPS 35 
 
 the great circle of the turn, and AB, the arc of 
 the turn. 
 
 29. Definitions. The resultant of two suc- 
 cessive turns AB, BC is the turn AC. 
 
 B' 
 
 When the arc of the turns are not given with 
 the first ending where the second begins, each 
 arc may be moved as a rigid arc round its great 
 circle until they do so end and begin, without 
 altering their turning value. When the two 
 great circles are not the same, then the common 
 point of the two arcs must be one or the other 
 point of intersection (B, B') of the two great 
 circles. The figure shows that the same result- 
 ant is found from either of these points. 
 
36 pblmeb of quaternions 
 
 Arc Steps 
 
 We may call the great arc AB the arc step 
 from A to B on the surface of the sphere ; and 
 call two arc steps equal when they are arcs of 
 the same great circle of the same length and 
 direction; and call AC the sum of AB, BC or 
 the sum of any arc steps equal to these. The 
 half-arc of a resultant rotation is thus the sum 
 of the half-arcs of its components, and the arc 
 of a resultant turn is the sum of the arcs of the 
 components. The sum of several arcs is found 
 by replacing any two successive arcs of the sum 
 by their sum, and so on, until a single sum is 
 found. An arc of 0° or 180° may be measured 
 on any great circle without altering its value as 
 the representative of a half-rotation, a turn, or 
 an arc step. 
 
 30. The resultant of two successive rotations 
 or turns {i.e., the sum of two arc steps) is com- 
 mutative only when the arcs are cocircular. 
 
 For let the half-arcs of the rotations, or the 
 arcs of the turns, be AB = BA', and C'B = BC; 
 
ROTATIONS. TURNS. ARC STEPS 37 
 
 then the sums AB + BC, C'B + BA' in opposite 
 orders are respectively AC, C'A'; and from the 
 figure those arcs are equal when, and only 
 when, the given arcs are cocircular. 
 
 Cor. 1. An arc of 0° or 180° is commutative 
 with any other arc. 
 
 For it may be taken cocircular with the 
 other arc. 
 
 CoE. 2. The magnitudes of the sums of tioo 
 arcs in opposite orders are equal. 
 
 For ABC, A'BC are equal spherical tri- 
 angles by construction, and therefore AC, C'A' 
 are equal in length. 
 
38 
 
 PRIMER OF QUATERNIONS 
 
 31. A sum of successive arc steps is asso- 
 ciative. 
 
 For, consider first three arcs upon the great 
 circles LQ', Q'R, RL. If the arcs are such as 
 to begin and end successively, the proof is the 
 same as for step addition, e.g., in the sum 
 AQ'+0t + RB = AB, the first two may be 
 replaced by their sum AR, or the second and 
 third by their sum Q'B without altering the 
 whole sum. In the more general case when the 
 three arcs are 
 
 AQ' = s1P', QQ = ER, RB = PT, 
 
ROTATIONS. TUBNS. ABC STEPS 39 
 
 the sum of the first two is AQ= SP, whose sum 
 with the third is ST; and the sum of the 
 second and third is R'B = P'T', whose sum with 
 the first is S'i"; and we must prove that ST, 
 S'T" are equal arcs of the same great circle in 
 the same direction. 
 
 [Observe that in the construction P is deter- 
 mined as the intersection of QA and RB, and 
 P' as the intersection of Q'A and R'B.~\ 
 
 Let the three given arcs be the half-arcs of 
 successive rotations of the sphere 0. Then by 
 Art. 26, the rotation 2AQ = 2SP gives the 
 sphere the same displacement as the first and 
 second rotations, so that 2 ST gives the sphere 
 the same displacement as the three rotations. 
 Similarly, the rotation 2RB=2rT' gives the 
 sphere the same displacement as the second and 
 third rotations, so that 2 S'T" gives the sphere 
 the same displacement as the three rotations. 
 Hence ST, S'T" are arcs of the same great 
 circle, and either equal (and in the same direc- 
 tion) or supplementary (and in opposite direc- 
 
40 PBIMEB OF QUATERNIONS 
 
 tions), since they are half-arcs of the same rota- 
 tion. This is true wherever Q may be. Suppose 
 that Q is slightly displaced towards E; then 
 ST, 8'T' are slightly displaced, and if equal at 
 first, they must remain equal, since a slight 
 change in each of two equal arcs could not 
 change them to supplementary arcs in opposite 
 directions.* Hence by moving Q continuously 
 towards R and finding how the arcs ST, S'T" 
 are related when Q reaches R, we find how they 
 are related for any position of Q, since there is 
 no change in the relation when Q is moved con- 
 tinuously. But when Q is at R, it was shown 
 above that both arcs were equal ; therefore 
 ST, S'T' are always equal. 
 
 So, in general, for a sura of any number of 
 successive arcs, any way of forming the sum by 
 replacing any two successive terms by their 
 sum and so on, must give a half-arc of the result- 
 ant of the rotations through double each of 
 
 * When both arcs are nearly 90°, a slight change in each 
 could change them from equals to supplements in the same 
 direction. 
 
ROTATIONS. TURNS. ARC STEPS 41 
 
 the given arcs. Hence any two such sums are 
 either equal or opposite supplementary arcs of 
 the same great circle ; and since by continuous 
 changes of the component arcs, they may be 
 brought so that each begins where the preceding 
 arc ends, in which position the two sums are 
 equal, therefore they are always equal. 
 
 Cor. 1. An arc of 0° or 180° 'may have any 
 position in a sum. [Art. 30, Cor. 1.] 
 
 CoE. 2. The magnitude of a sum of arcs is 
 not changed hy a cyclic change in the order of its 
 terms. 
 
 For {AB -\-CD+ ■••) + HK and 
 Hk-v{AB+GD+ ■■■) 
 have equal magnitudes. [Art. 30, Cor. 2. J 
 
 EXAMPLES 
 
 1. Show that 2{AB + BG) and 2AB + 2BG are in 
 general unequal. 
 
 2. If (2, 30°) denote a turn of 30° counter-clockwise in 
 the plane of the paper and a doubling, and (3, — 60°) 
 denote a turn of 60° clockwise in the plane of the paper 
 
42 PBIMEB OF QUATERNIONS 
 
 and a trebling, express the resultant of these two com- 
 pound operations (versi-tensors) in the same notation. 
 
 3. Find the resultant of (2, 30°), (3, 60°), (4, - 120°), 
 (1, 180°). 
 
 4. Show that either (2, - 60°) or (2, 120°) taken twice 
 have the resultant (4, - 120°). 
 
 5. Would you consider the resultants of versi-tensors 
 as their sums or their products, and why ? 
 
 6. Let the base QR of a spherical triangle PQB slide 
 as a rigid arc round its fixed great circle, and let the great 
 circles QP, RP, always pass through fixed points A, B 
 respectively. Show that if points S, T lie on the great 
 circles QP, RP so as always to keep PS = QA and 
 PT= RB, then the arc >Sr is an arc of fixed length and 
 direction that slides around a fixed great circle as QR 
 slides round its fixed great circle. [Let P', Q', R', S', T', 
 be given positions of P, Q, R, S, T, and use Art. 31 and 
 figure.] 
 
 7. Show that the locus of the radius OP in Ex. 6 is an 
 oblique circular cone of which OA, OB are two elements, 
 and that the fixed great circles QR, ST are parallel to its 
 circular sections. [Draw a fixed plane parallel to OQR 
 and cutting the radii OA, OB, in the fixed points A', B', 
 and cutting OP in the variable point P', and show that 
 P' describes a circle in this plane through the fixed points 
 A', B' ; similarly, for a fixed plane parallel to OST.^ 
 
ROTATIONS. TURNS. ARC STEPS 43 
 
 Note. — The locus of P on the surface of the sphere 
 is called a spherical conic (the intersection of a sphere 
 about the vertex of a circular cone as centre with the 
 surface of the cone) ; and the great circles QB, ST 
 (parallel to the circular sections of the cone) are the 
 cyclic great circles of the spherical conic. The above 
 properties of a spherical conic and its cyclic great circles 
 become properties of a plane conic and its asymptotes 
 ■when the centre of the sphere is taken at an indefi- 
 nitely great distance. 
 
 8. State and prove Ex. 6 for a plane, and construct 
 the locus of P. 
 
44 PRIMER OF QUATERNIONS 
 
 CHAPTER III 
 Quaternions 
 
 32. Definitions. A quaternion is a number 
 that alters a step in length and direction by a 
 given ratio of extension and a given turn. £J.g., 
 in the notation of Ex. 2, II, (2, 30°), (2, -60°) 
 are quaternions. 
 
 Two quaternions are equal when, and only 
 when, their ratios of extension are equal and 
 their turns are equal. 
 
 A tensor is a quaternion that extends only ; 
 i.e., a tensor is an ordinary positive number. Its 
 turn is 0° in any plane. 
 
 A versor or unit is a quaternion that turns 
 only. E.g., 1, -1 = (1, 180°), (1, 90°), (1, 30°), 
 are versors. 
 
 A scalar is a quaternion whose product lies 
 on the same line or " scale " as the multipli- 
 cand; i.e., a scalar is an ordinary positive or 
 
QUATERNIONS 45 
 
 negative number. Its turn is 0° or 180° in any 
 plane. 
 
 A vector is a quaternion that turns 90°. ^.g., 
 (2, 90°), (1, -90°), are vectors. 
 
 33. Functions of a Quateenion q. The 
 tensor of q, or briefly Tq, is its ratio of exten- 
 sion. JE.g., T2 = 2=T{-2)= T{2, 30°). 
 
 The versor of q ( Uq) is the versor with the 
 same arc of turn as q. E.g., 
 
 U2 = l, U{-2)=-l, U{2, 30°)= (1,30°). 
 
 The arc, angle, axis, great circle, and plane 
 
 of q, are respectively the arc, angular magni- 
 tude, axis, great circle, and plane of its turn. 
 E.g., arc (2, 30°) is a counter-clockwise arc of 
 30° of unit radius in the plane of the paper, 
 and arc (2, —30°) is the same arc oppositely 
 directed; Z (2, 30°) = Z(2, -30°) = 30° = 7r/6 
 radians; axis (2, 30°) is a unit length perpen- 
 dicular to the plane of the paper directed 
 towards the reader, and axis (2, —30°) is the 
 same length oppositely directed ; etc. 
 
46 
 
 PRIMER OF QUATERNIONS 
 
 If qOA= OB, and if i be the foot of the per- 
 pendicular from £ upon the line OA, then OL, 
 LB are called the components of q's product 
 
 B M B 
 
 respectively parallel and perpendicular to the 
 midtiplicand ; also, the projections of OB par- 
 allel and perpendicular to OA. 
 
 The scalar of q (Sq) is the scalar whose prod- 
 uct equals the component of q's product par- 
 allel to the multiplicand ; viz., Sq ■ 0A = OL. 
 
 U.g., /S'(2, 30°) = V3, /S(2, 150°)= - Vs. 
 
 The vector of q (Vq) is the vector whose prod- 
 uct equals the component of q's product perpen- 
 dicular to the multiplicand ; viz., Vq • OA = LB. 
 
 E.g., F(2, 30°) = (1, 90°)= V{2, 150°), 
 7(2, -60°) = (a/3, -90°). 
 
 The reciprocal of q{l/q or q~^) is the quater- 
 nion with reciprocal tensor and reversed turn. 
 i;.g.,{2,SQ°)-' = {i, -30°). 
 
QUATERNIONS 47 
 
 The conjugate of q{Kq) is the quaternion with 
 the same tensor and reversed turn. E.g., 
 
 ^(2, 30°) = (2, -30°). 
 
 34. From the above diagram and the defini- 
 tions of the cosine and sine of an angle, we have 
 
 ,. a OL OL OB OL rr 
 
 /-L\ mir LB OB LB rri 
 
 Note. Arc Vq is a quadrant on the great circle ofq 
 in the direction of arc q. 
 
 EXAMPLES 
 
 1. If equal numbers multiply equal steps, the prod- 
 ucts are equal ; and if they multiply unequal steps, the 
 products are unequal. 
 
 2. If the products of two steps by equal numbers are 
 equal, then the two steps are equal ; and if the products 
 of two equal steps by two numbers are equal, then the 
 numbers are equal. 
 
 3. If several steps be multiplied by equal numbers, 
 then any product is to its multiplicand as any other 
 product is to its multiplicand. 
 
48 PBIMEB OF QUATERNIONS 
 
 4. If two steps be multiplied by reciprocal numbers, 
 then corresponding products and multiplicands are re- 
 ciprocally proportional. 
 
 5. Construct the following products, where OA is a 
 unit step to the right in the plane of the paper, and de- 
 termine the functions of each multiplier that are defined 
 in Art. 33. 
 
 (a) 2 . OA = OL, (4, 60°) • OA = OB, (4, - 60°) • OA 
 = OB ', (2 V3, 90°) • A = M, (2 V3, - 90°) • A = M ', 
 (1, 60°).OA=OBi, (1, -60°).OA = OB/, (1, 90°).OA 
 = OMi, (1, -90°).OA = OMi'. 
 
 (b) The same as (a) with 120° in the place of 60°. 
 
 6- Show that SSq = Sq, SVq = 0, VSq = 0, 
 VVq = Vq, SKq = KSq = Sq, VKq = KVq, USq = ± 1, 
 UTq = 1 = TUq. 
 
 Multiplication 
 
 35. Definition. The product of two or 
 
 more numbers is that number whose extension 
 and turn are the resultants of the successive 
 extensions and turns of the factors (beginning 
 with the right-hand factor). 
 
 E.g., if rOA = OB, ^OB = OC, pOO = OD, then 
 we have pqr • OA=^g'OB=^OC=OD. 
 
QUATERNIONS 49 
 
 36. The product is, however, independent 
 of whether a step OA can be found or not, such 
 that each factor operates upon the product of 
 the preceding factor; i.e., we have by definition, 
 
 (a) T{-pqr) = ■■■ Tp-Tq- Tr. 
 
 (b) arc {—pqr) = arc r + arc q + arcp + -. 
 
 37. The product of a tensor and a versor is 
 a nurnber with that tensor and versor ; and con- 
 versely, a number is the product of its tensor and 
 its versor. 
 
 For if w be a tensor, and q' a versor, then 
 nq' turns by the factor q' and extends by the 
 factor n, and vice versa for q'n ; hence either of 
 the products, nq', q'n, is a quaternion with 
 tensor n and versor q'. Similarly, 
 q=Tq-Uq=Uq- Tq. 
 
 38. Any successive factors of a product may 
 be replaced by their product without altering the 
 value of the whole product ; but in general such 
 factors can be changed in order without altering 
 the value of the product only when those factors 
 are cocircular. 
 
50 PRIMER OF QUATERNIONS 
 
 For replacing successive factors by their 
 product does not alter the tensor of the whole 
 product by Art. 36(a), nor the arc of the prod- 
 uct by Art. 31, 36(&) ; but by Art. 30 the arc 
 of the product is altered if two factors be 
 interchanged except when those factors are 
 cocircular. 
 
 Cor. 1. A scalar factor may have any posi- 
 tion in the product without altering the value of 
 the product. [Art. 31, Cor. l.j 
 
 CoE. 2. TTie angle of a product is not altered 
 by a cyclic change in the order of the factors. 
 [Art. 31, Cor. 2.] 
 
 Cor. 3. The scalar, and the tensor of the vec- 
 tor, of a product are not altered hy a cyclic change 
 in the order of the factors. [Art. 34, a, 6.] 
 
 39. The product of two numbers toith oppo- 
 site turns equals the product of the tensors of the 
 numbers ; and conversely if the product of two 
 numbers is a tensor, then the turns of the factors 
 are opposites. [36 a, &.] . 
 
QUATERNIONS 51 
 
 CoK. 1. The product of two conjugate numbers 
 equals the square of their tensor; and if the 
 product of two numbers with equal tensors is a 
 tensor, then the tioo numbers are conjugates. 
 
 Cor. 2. The conjugate of a product equals the 
 product of the conjugates of the factors in 
 reverse order. 
 
 For (pqr) {Kr ■ Kq- Kp) = {Tpf ■{ Tqf ■ ( ^r)^ 
 since rKr = [Try, may have any place in 
 the product, and may be put first ; and then 
 (qKq) = {Tqf, may be put second, and then 
 (pKp) = {Tpf. [Cor. 1, 38 Cor. 1.] 
 
 Hence, K{pqr) = Kr ■ Kq ■ Kp. [Cor. 1.] 
 
 CoE. 3. The product of two reciprocal num- 
 bers is unity ; and conversely, if the product of 
 two numbers with reciprocal tensors is unity, then 
 the number's are reciprocals. 
 
 CoE. 4. The reciprocal of a product equals 
 the product of the reciprocals of the factors in 
 reverse order. 
 
 For {pqr){r~'^q~^p~^)= 1. 
 
52 PRIMER OF QUATERNIONS 
 
 40. The square of a vector is —\ times the 
 square of its tensor ; and conversely, if the square 
 of a number is a negative scalar, then the number 
 is a vector. [36, a, b.~\ 
 
 CoE. 1. The conjugate of a vector is the nega- 
 tive vector. [39 Cor. 1.] 
 
 CoE. 2. The conjugate of a product of two 
 vectors is the product of the same vectors in 
 reverse order. [Art. 39, Cor. 2.] 
 
 CoE. 3. The conjugate of a product of three 
 vectors is the negative of the product of the same 
 vectors in reverse order. [Art. 39, Cor. 2.] 
 
 The Rotatoe q{ )q~^ 
 
 41. We may consider the ratio of two steps 
 as determining a number, the antecedent being 
 the product and the conseqiaent the multiplicand 
 of the number; viz., OB/OA determines the 
 number r such that rOA = OB. By Art. 21, 
 equal step ratios determine equal numbers. 
 
 If the several pairs of steps that are in a 
 given ratio r be given a rotation whose equato- 
 
qUATEBNIONS 53 
 
 rial arc is 2 arc g*, they are still equal ratios in 
 their new positions and determine a new number 
 / that is called the number r rotated through 
 2 arc q. In other words, the rotation of r pro- 
 duces a number with the same tensor as r, and 
 whose great circle and arc are the rotated great 
 circle and arc of r. 
 
 4[:'2. The number r rotated through 2 arc q is 
 the number qrq"^. 
 
 For, 1st, Tqrq-^ = Tq ■ Tr{Tq)-^ = Tr. 
 
 A 
 
 2d, let A be an intersection of the great circle 
 of r with the great circle of q and construct 
 
 AB = BA' = arc q, AG = arc r, 
 and &B =BC = arc rq-^ ; 
 
 then (fl' = AC" = arc qrq-\ 
 
54 PRIMER OF QUATERNIONS 
 
 But by construction, the spherical triangles 
 ABC, A'BC are equal, and therefore AC and 
 C'A'{= A'C") are arcs of equal length, and the 
 corresponding angles at A, A' are equal. 
 Hence, when arc r { = AC) is rotated through 
 2 arc q{ = AA'), it becomes arc qrq~\ = A' C"). 
 
 Powers and Roots 
 
 43. An integral power, (l" = q- q- q — to n 
 
 factors, is determined by the equations, 
 (a) T-q''=Tq-Tq- Tq - = {TqY. 
 (h) arc q" = arc q + arc q + arc q ■■■ =n arc q ± 
 
 {whole circumferences). 
 1 
 To find q", the number whose nth power is q, 
 
 we have, by replacing q by q" in (a), (b), 
 
 (c) Tq = {T- 2")" or T-q" = {Tqf. 
 
 1 
 {d) Arc q = n arc (f ± whole circumferences, 
 
 i 1 
 
 or, arc q^ = ~ {arc q ± lohole circumferences) = 
 
 - arc q -\ circumferences ± {whole circumfer- 
 
 ences), where m = 0, 1, 2, 3, ••• n — 1, successively. 
 
QUATERNIONS 65 
 
 There are therefore n nth. roots of q whose 
 tensors are all equal and whose arcs lie on the 
 great circle of q. 
 
 When the base is a scalar, its great circle 
 may be any great circle, so that there are an 
 infinite number of quaternion nth. roots of a 
 scalar. On this account, the roots as well as 
 the powers of a scalar are limited to scalars. 
 By ordinary algebra, there are n such nth roots, 
 real and imaginary. There are also imaginary 
 nth roots of q besides the n real roots found 
 above ; i.e., roots of the form a + & V — 1, where 
 a, h are real quaternions. 
 
 Repkesentation op Vbctoes 
 
 44. Bold-face letters will be used as sym- 
 bols of vectors only. In particular, i. j, k will 
 denote unit vectors whose axes are respectively 
 a unit length east, a unit length north, and 
 a unit length up. More generally we shall 
 use the step AB to denote the vector whose 
 axis is a unit length in the direction of AB, 
 
56 PRIMEB OF QUATERNIONS 
 
 and whose tensor is the numerical length of 
 AB ( = AB : unit length). 
 
 This use of a step AB as the symbol of a 
 vector is analogous to the use of AB to repre- 
 sent a tensor {AB : unit length), or of AB to 
 represent a positive or negative scalar, according 
 as it is measured in or against the direction of 
 its axis of measurement. In none of these 
 cases is the concrete quantity an absolute num- 
 ber ; i.e., the value of the number that it repre- 
 sents varies with the assumed unit of length. 
 When desirable, we distinguish between the 
 vector OA and the step OA by enclosing the 
 vector in a parenthesis. 
 
 45. If q{OA) = (OB), then qOA= OB, and 
 conversely. 
 
 The tensor of q in either equation is OB : OA. 
 It is therefore only necessary to show that the 
 arc of q in one equation equals the arc of q 
 in the other equation in order to identify the 
 two numbers that are determined by these two 
 equations as one and the same number. 
 
qUATEBNlONS 
 
 67 
 
 Draw the sphere of unit radius and centre 0, 
 cutting OA, OB in A', B' ; then A'B' is the arc 
 of q in the second equation. Draw the radius 
 OL perpendicular to the plane OA'B' on the 
 counter-clockwise side of A'B', and draw coun- 
 
 ter-clockwise round OA', OB' as axes the quad- 
 rants LM, LN respectively ; then these are the 
 arcs of (OA), (OB) respectively, and since 
 LM+ MN= LN, therefore MN is the arc of q 
 in the first equation. But since LM, LN are 
 quadrants, therefore the plane OMN is perpen- 
 dicular to OL, and must therefore coincide with 
 the plane OA'B', which is by construction also 
 perpendicular to OL. Hence MN lies on the 
 great circle of A'B', and by the construction of 
 
58 PBIMEB OF QUATEBNIONS 
 
 the figure, it must, when advanced 90° on that 
 great circle, coincide with A'B'. Hence the 
 theorem. 
 
 Note. This theorem shows that a number extends 
 and turns vectors into vectors in the same way that it 
 extends and turns steps into steps. Moreover, when the 
 vector is not perpendicular to the axis of the multiplier, 
 there is no resulting vector, since in the case of the 
 corresponding step there is no resulting step. In the 
 case of a vector multiplicand, that is oblique to the axis 
 of q, the product is an actual quaternion that is not a 
 vector, while in the case of the corresponding step multi- 
 plicand the product belongs to that class of products in 
 which the multiplicand does not admit of the operation 
 of the multiplier, as in V2 universities, — 2 countries, etc. 
 
 Cor. 1. The product of two vectors is a 
 vector when, and only ivhen, the factors are 
 perpendicular to each other; the product is 
 perpendicular to both factors; and its length 
 (its tensor) is equal to the area of the rectangle 
 on the lengths of the factors. 
 
 Note. The direction of the product OA • OB = OC 
 is obtained by turning OB about OA as axis through a 
 counter-clockwise right angle ; thus OC lies on that side 
 of the plane OAB from which the right angle AOB 
 appears counter-clockwise. 
 
QUATERNIONS 59 
 
 Cor. 2. The product of two perpendicular 
 vectors changes sign when the factors are inter- 
 changed. (OB ■ OA = OC = - OC.) 
 
 c' 
 
 Cor. 3. The condition that a is perpendicular 
 to y8 is that afi = vector, or Sa^= 0. 
 
 46. If AB, CD are parallel, then AB • CD = 
 CD ■ AB = — AB • CD, a scalar ; and conversely, 
 the product of two vectors is a scalar only when 
 they are parallel. 
 
 Since the axes of the vectors AB, CD are 
 parallel, therefore their product is commutative. 
 When the vectors are in the same direction, 
 then each turns 90° in the same direction, the 
 resultant turn is 180°, and the product is nega- 
 tive ; and when the vectors are in opposite 
 
60 PRIMER OF QUATERNIONS 
 
 direction, their turns are in opposite directions, 
 the resultant turn is 0°, and the product is posi- 
 tive. This is just the opposite of the product of 
 the corresponding scalars AB, CD, which is 
 positive when the scalars are in the same direc- 
 tion (or both of the same sign), and negative 
 when the scalars are in opposite directions ; i.e., 
 ABCD= -ABCI). 
 
 Conversely, the product AB, CD can be a 
 scalar only when the resultant of their two 
 turns of 90° each is a turn of 0° or 180° ; i.e., 
 only when the turns are cocircular, and there- 
 fore their axes parallel. 
 
 CoK. The condition that a is parallel to ^ is 
 a/3 = scalar, or Va^ = 0. 
 
 EXAMPLES 
 
 1. Prove by diagram that (pqY andpV ^^^ in general 
 unequal. 
 
 2. Find the 2d, 3d, 4th, 5th, 6th powers of (2, 90°), 
 (2, - 60°). 
 
 3. Find the square roots and cube roots of (4, 30°), 
 (8, - 120°). 
 
QUATERNIONS 61 
 
 (a) Find the values of [(2, 50°)'^^, [(2, 50°)*]", and 
 (2, 50°)^. 
 
 4. What numbers are represented by 2 feet, 2 feet 
 east, the unit of length being a foot, a yard, an inch ? 
 
 5. Show that P = j2 = k2= ijk =-1; jk=i=-kj; 
 ki=j=— ik; ij = k = — ji. 
 
 6. Let e'*^' denote the versor that turns counter-clock- 
 wise round the axis AB through an arc that is formed 
 by bending the length AB into an arc of unit radius. 
 Show that if facing the west, and holding the paper in a 
 north and south vertical plane, then e', e^', ••• e"', e~^', turn 
 respectively 1, 2, •■• radians counter-clockwise, and 1, 2, 
 ••• radians clockwise in the plane of the paper. Also 
 show that e*^' = ± i, e-"' = - 1, e^"' = 1, where n is any 
 integer. 
 
 7. Show by diagram that Se^' = cos 0, Ve^' = i sin 6, 
 where 6 is any positive or negative number and the unit 
 of angle is a radian. 
 
 8. Show that if OA rotate into B through 2 arc q, 
 then(OB)=q(Ok)q-\ 
 
 9. Show that if a be a vector in the plane of q, then 
 Kq = aqoT^ = a"^ga. 
 
 10. Show that pq rotates into qp, and determine two 
 such rotations. 
 
62 PBIMEB OF QUATERNIONS 
 
 11. Show that SKq = iSq, VKq = - Vq. 
 
 12. Show that Ea.^ = /Sa, Sa^ = Sl3a, Va^ = - F/8a. 
 
 13. Show that ^a^y = -yy8a; Fay87 = Fyj8a; Sa.Py = 
 S^ya = 8yal3 = -Syl3a = -S^ay = - Say/3, (a) Deter- 
 mine the conjugate of a product of n vectors. 
 
 14. Prove by diagram that Kpq = Kg • Kp. 
 
 Addition 
 
 47. Definition. The sum {p + q) is the 
 number determined by the condition that its 
 product is the sum of the products of p and q. 
 
QUATERNIONS 
 
 63 
 
 Thus let OA be any step that is multiplied by 
 both^ and q, and let pOA = OB, qOt<=00, and 
 OB + OC = OD, then {p + q)OA = OD. It is ob- 
 vious that any change in OA alters OB, OC, OD, 
 proportionally, so that the value of the sum 
 p + q{ = OD : OA) is the same for all possible 
 values of OA. 
 
 Similarly, any quaternion, r, may be added to 
 the sum p + q, giving the sum {p + q) + r ; and 
 we may form other sums such as p + {q + r), 
 {q + r)+p, etc. It will be shown later that all 
 such sums of the same numbers are equal, or 
 that quaternion addition is associative and com- 
 mutative. 
 
 48. The sum of a scalar and a vector is a 
 quaternion with that scalar and that vector, and 
 conversely, a quaternion is the sum of its scalar 
 and its vector. 
 
64 
 
 PRIMES OF QUATERNIONS 
 
 For let w be any scalar, and p any vector, 
 and let ioOA = OL, /)0A = OM, then completing 
 the rectangle OLBM, we have {w + p)OA = OB, 
 and the scalar of w + p is w, and its vector is p, 
 since OL, OM are the components of OB par- 
 allel and perpendicular to OA. Similarly, 
 
 q = Sq+ Vq. 
 
 49. The scalar, vector, and conjugate, of 
 any sum equals the like sum of the scalars, 
 vectors, and conjugates of the terms of the sum. 
 \_I.e., S, V, K, are distributive over a sum.'\ 
 
 OM=BE=LN 
 ON=OL+OM 
 MC=ED;LB=NE 
 ND=LB + MC 
 
qUATEUNIONS 65 
 
 For let j90A = 0B, ^OA = OC, 
 
 (^ + ^)OA = OB + OC=OD. 
 
 Then the components of OD parallel and perpen- 
 dicular to OA are, by the figure, the sums of the 
 like components of OB, OC ; i.e., S{p + q) • OA 
 = Sp-OA + Sq-OA, or S{p + q) = Sp + Sq; and 
 V{p + q)-OA=Vp-OA+rq-OA, or V{p + q) 
 = Vp+Vq. 
 
 Also, if OB'UC be the parallelogram that is 
 symmetric to the parallelogram OB DC with 
 reference to OA as axis of symmetry, then 
 KpOk = OB', Kq. 0A = OC, and K{p + q)-OA 
 = 0D', and since OB' + OC'=OD', therefore 
 K(p + q) = Kp + Kq. 
 
 These results extend to any given sum ; e.g., 
 V\_{p + q) + r'] = V{p + q) + Vr = {Vp + Vq) + Vr, 
 etc. 
 
 50. If (OA) + (OB) = (OC), then OA + OB = 
 OC, and conversely. 
 
 For erect a pin OD of unit length perpen- 
 dicular to the plane of the angle AOB on its 
 counter-clockwise side ; and turn A OB round 
 
66 
 
 phimeb of quaternions 
 
 OD as axis through a clockwise right angle as 
 seen from D into the position A' OB'. Then 
 since (OA) is the vector that turns through a 
 counter-clockwise right angle round OA as axis, 
 and extends unit length into length OA = 
 
 length OA', therefore (OA)OD = OA', and simi- 
 larly (OB)OD = OB', and therefore (OC)OD = 
 OA' + 0B'=0C', where OA'C'B' is a parallelo- 
 gram. Hence the step OC of proper length and 
 direction to give the tensor and axis of the 
 vector (OC) must be the diagonal of the par- 
 
QVATEBNIONS 67 
 
 allelogram on OA, OB as sides; and therefore 
 OA + OB = OC. Conversely, if OA + OB = OC, 
 tlien turning the parallelogram OACB into the 
 position OA'C'B', we have, since OA' + OB' = 
 OC, that(OA) + (OB) = (OC). 
 
 Cor. 1. Vectors add in the same way as 
 
 their corresponding steps, and all the laws of 
 
 addition and resolution of steps extend at once 
 to vectors. 
 
 CoK. 2. A sum of quaternions is associative 
 and commutative. 
 
 For since by Cor. 1 a sum of vectors is 
 independent of the way in which its terms are 
 added, and since we know that a sum of scalars 
 {i.e., ordinary numbers) is independent of the 
 way in which its terms are added, therefore 
 by Art. 49 the scalar and the vector of a sum 
 are independent of the way in which the sum 
 is added. Hence the sum is independent of the 
 way in which it is added, since it is equal to the 
 sum of its scalar and its vector. 
 
68 
 
 PRIMES. OF QUATERNIONS 
 
 51. Lemma. If p, q he any quaternions, 
 then (1 +p)q = q +pq- 
 
 For take OB in the intersection of the planes 
 oip, q and draw OA, OC such that g' • 0A = OB, 
 pOB = OC; then (l+p)g- OA = (l+p)OB = 
 OB + 00 = ^OA +pqOA. Hence, 
 {l+p)q = q+pq. 
 
 52. If p, q-, T he any quaternions, then 
 (p + q)r =pr + qr. 
 
 For we have, (1 + qp~'^) p • r — {l + qp'^) • pr, 
 and expanding each member by the preceding 
 lemma, we have, {p + q)r =pr + qr. 
 This result extends to any sum ; e.g., 
 
 {p + q + r + s)t= \_{p + q) + {T + s)]i 
 = {p + q)t + {r + s)t =pt + qt-[-rt + st. 
 
QUATERNIONS 69 
 
 Cor. 1. r{p + q) = rp + rq. 
 
 For let p', q', r' be the conjugates of p, q, r. 
 Then from {p' + q')r' = pY + qy, we have, by 
 taking the conjugates of each member, r{p + q) 
 = rp + rq. [Art. 39, Cor. 2; Art. 49.] 
 
 Cor. 2. A product of sums equals the sum of 
 all partial products that vnay he formed from the 
 given product by multiplying together, in the 
 order in which they stand, a term from each 
 factor of the product. 
 
 E.g., (p + q) {r + s) =pr +ps + qr + qs. 
 
 Note. — This rule should be used even when the 
 factors are commutative, as it prevents all danger of 
 taking out the same partial product twice; e.g., from 
 taking both pr and rp from the above product. To 
 be sure that all the partial products are found, some 
 system of arrangement should be adopted ; also the 
 total number of partial products should be determined. 
 
 E.g., (p + q) ip + q) {p + q) may be arranged 
 according to the degrees of the terms in^, and 
 there are 2x2x2 = 8 terms. This product is 
 then easily seen to be 
 
 y + ip^q +pqp + qp^) + {pq^ + qpq + q^p) + f, 
 
70 PRIMER OF QUATERNIONS 
 
 when p, q are not commutative, and 
 
 p"^ + Sp'^q + 3 pq^ + ^, 
 when p, q are commutative. 
 
 FORMULAS. FOR EXERCISE AND REFERENCE 
 
 53. {a) q=Tq- Uq=Uq-Tq. 
 [1) q=8q + Vq; Kq = Sq - Vq. 
 
 (c) Sq= Tq cos /.q, =r cos 9, say; 
 
 TVq = Tq sm Zq = r sin d. 
 
 (d) Vq= TVq ■ UVq = r sin ^ • e where 
 
 e= UVq. 
 
 ( e ) q = r (cos ^ + e • sin ^) = re*', 
 
 Kq = r (cos ^ — e sin 6*) = re~^'. 
 (/) e''=-e*" = e(*+"''. 
 (^) ^2 = % + ^?); Vq = ^{q-Kq). 
 (h) Tq' =qKq = Kq-q = {Sqf - ( Vqf 
 
 = {Sqf + {TVqf. 
 (i) q-' = Kq/Tq\ 
 
 As a further exercise find the T, U, S, V, K 
 of the T, U, 8, V, K of q, in terms of r, 6, e. 
 
qUATEBNIONS 71 
 
 54. (a) T{ -pqr)= - Tp ■ Tq ■ Tr. 
 
 (b) U{--pqr)= ■■■ Up- Uq- Ur. 
 
 (c) /.{ •■pqr)= Z {r ■••pq)= z {qr--p), etc. 
 {d) S{ ■■■ pqr) = 8{r ■■■ pq) = S{qr ••■ p), etc. 
 (e) TV{ ■■■pqr)=TV{r-pq)= TV{qr -p), 
 
 etc. 
 
 (/) arc ( •■■ pqr) = arc r + arc q + arc p + ••■. 
 
 (g) {•■• pqr)~^ = r~^q~^p~^ •■■. 
 
 (h) K{--pqr) = Kr ■ Kq- Kp ■•■. 
 
 ( i ) 8{xp + yq + zr) = xSp + ySq + zSr, \_x, y, z, 
 scalars] and similarly for V or K in- 
 stead of S. 
 
 55. (a) Ka=-a; Ta' = - a' ; Sa = ; 
 Va = a. 
 
 (b) Eafi = ySa ; Sa^ = S^a ; Va^ = - F/Sa. 
 
 (c) a/3 + ^0L=2 Sa/3, al3~/3a = 2 Va^. 
 
 (d) {a±/3f=a'±2SaP + /3'. 
 
 (e) V{xa + y^) {x'a + y'^) = {xy' - x'y)raft 
 
 xy 
 x'y' 
 
 Va^, say. [x, y, x', y', scalars.] 
 
72 
 
 PRIMER OF QUATERNIONS 
 
 if) V{xa +y^+ zy) {x'a + y'^ + z'y) = 
 
 + 
 
 zx 
 z'x' 
 
 Vya + 
 
 xy 
 
 f / 
 
 xy 
 
 yz 
 y'z' 
 
 ri3y 
 
 Va/3. [x, y, z, x', y', z', scalars. j 
 
 56. (a) Ka^y = - yfia. 
 
 (h) a/Sy - y^a = 2 Sa^y = - 2 Syl3a. 
 
 Hence the scalars of the six products of a, ^, y 
 are equal to one of two negative numbers accord- 
 ing to the cyclic order of the product ; and an 
 interchange in two factors (which changes the 
 cyclic order) changes the sign of the scalar of the 
 product. When two of the three factors are 
 equal, the scalar of their product must therefore 
 be zero, since an interchange of the equal fac- 
 tors changes the sign without changing the value. 
 
 (c) S-ixa + yj3 + zy){x'a + y'/3 + z'y){x"a + y"^ + z"y) 
 
 1 I y"^^ 
 
 +y 
 
 z'x 
 
 + z 
 
 x'y' 
 x'Y 
 
 XSa^y 
 
 X y z 
 x' y' z' 
 x"y"z" 
 
 Sa/3y, say. \_x, y, z, etc., scalars.J 
 
QUATERNIONS 73 
 
 (d) 8al3y = Sa V/3y = 8/3 Vya = Sy Va/S. 
 [Replace fiy by S^y + V/3y, expand by 54 (i), 
 
 and note that S- aS^y=0.'\ 
 
 (e) a^y + yl3a = 2Va^y=2ryl3a. 
 
 Note. — Insert between the two terms of the first 
 member of (e), the null term (a-y/S — ay/3 — -ya/S + ya/8), 
 and it becomes a(/8y + y^) — (ay + ya)^ + y(a|8 + Pa). 
 Hence, using (65 c), we have (/). 
 
 (/) Val3y = aSl3y - /SSya + ySa/S. 
 
 Transpose the first term of the second mem- 
 ber of (/) to the first member, noting that 
 aS^y = V • aSfiy, and ySy — Sfiy = F/Sy, and we 
 have 
 
 (g) Var^y=-l3Sya + ySal3; 
 
 {g') V ■ {V^y)a = fiSya - ySa^. 
 
 (h) r-iral3)ryB=-ySal3S + SSa^y [(g), (d)] 
 = aSfiy8-fiSayS. [(/), (cZ)] 
 
 (i) SSafiy = aS^yh + /SSyaS + ySafiS. [{h)'] 
 
 Replace a, ft, y, by V/3y, Vya, Vy^, noting 
 that V ■ {Vya ■ Vafi) = — aSa^y, etc., and that 
 
74 
 
 PEIMER OF QUATEENIONS 
 
 S{Vfiy ■ Vya ■ Ya^) = - {Sa/Syf, and we have 
 0) SSafiy = V^ySaS + VyaSfiS + Va/SSyS. 
 
 Note. — 0, (f) may be obtained directly by putting 
 8 = xa + yl3 + zy or x V^y + y Fya + z Vaft, and finding 
 X, y, z, by multiplying in the first case by Py, ya, aft, and 
 in the second case by a, (3, y, and taking the scalars of 
 the several products. 
 
 57. (a) P =f = k^ = ijk = - 1; jk = i = - kj; 
 ki = j = — ik, ij = k = — ji. 
 
 (&) p = — '\S'\p —jSip — kSkp. [56 (i) or (j) or 
 directly as in note.] 
 
 Let p = x\ + yj + zk, p = x'\ + y'] + z'k, etc. 
 [x, y, z, etc. scalars. J 
 
 Then, prove by direct multiplication, 
 
 (c) -p' = x^ + y'' + z''= Tp\ 
 
 {d) — Spp' = xx' + yy' + zz' = — Sp'p. 
 
 (e) Vpp' = 
 
 yz 
 y'z' 
 
 i + 
 
 zx 
 z'x' 
 
 
 xy z 
 
 if) -8pp'p" = 
 
 x'y'z' 
 
 
 
 x"i 
 
 l"z" 
 
 j + 
 
 xy 
 x'y' 
 
 k = - Vp'p. 
 
 = -SpVp'p". 
 
quaternions 76 
 
 Geometric Theorems 
 
 58. The angle of a/3 equals the supplement 
 of the angle 6 ietiveen a, ^. 
 
 For, since a/3 • /3~^ = a, therefore a/3 turns 
 through the angle from ^~'^ to a, which is the 
 supplement of the angle 6 from a to y8. 
 
 Cor. Safi = - Ta/3 cos 6, TVafi = Ta^ sin 6. 
 \^Sq = Tq cos Z q, etc.] 
 
 59. The scalar of a^ equals the product of a 
 and the projection of ^ upon it ; the vector of 
 a/3 equals the product of a and the projection of 
 /8 perpendicular to it, and Va^ is a vector per- 
 
76 
 
 PRIMER OF QUATERNIONS 
 
 pendicular to a, /3 on their counter-clockwise side 
 whose length equals the area of the parallelogram 
 on a, /3 as sides. 
 
 Let /3i, ^2 be the components of /3 parallel 
 and perpendicular to a, then /S = /8i + ySg and 
 tt/S = aySi + aySa = scalar + vector. Hence 
 
 Sa^ = a|8i, as stated ; and Fa/8 = a^^) 
 
 which is /3<i turned a counter-clockwise right 
 angle round a and lengthened by Ta. Hence 
 Va/3 is perpendicular to a, /S on their counter- 
 clockwise side (towards the reader in the figure), 
 and its length is Ta ■ T^^ = area parallelogram 
 on a, /3 as sides.* 
 
 * The parallelogram on a, /? may be considered as bounded 
 by the path of a point that receives the displacement a, then 
 the displacement |8, then the displacement — a, then the dis- 
 placement —p. This area is therefore bounded counter-clock- 
 
QUATERNIONS 77 
 
 Cor. 1. The projections of yS parallel and 
 perpendicular to a equal a'^Safi and a'^Vafi. 
 
 Cor. 2. The scalar measure of the projection 
 of /8 upon a is — Sa.fi/Ta, and the tensor 
 measure of the projection of /3 perpendicular to 
 a is TVafi/Ta. [Also from 58, Cor.] 
 
 Cor. 3. If he the angle between a, fi, then 
 cos6= - SajS/ Tafi, sin = TVa.fi/ Tafi. [Also 
 from 58, Cor.] 
 
 60. The volume of a parallelopiped on 
 a, fi, y as edges is — Safiy {the volume being 
 positive or negative according as a lies on the 
 counter-clockwise or clockwise side of fi, y). 
 
 wise round Va^ as axis ; and Fa/S may therefore be called the 
 vector measure of the area of this directed parallelogram or of 
 any parallel plane area of the same magnitude and direction of 
 boundary. 
 
78 PRIMER OF QUATERNIONS 
 
 For let tti be the projection of a upon V^y ; 
 then taking the face ^, y of the parallelepiped 
 as the base, we have by Art. 59 that TV^y is 
 the area of the base ; also Joti is the altitude. 
 Hence numerical volume 
 
 = Tai ■TVPy=Ta^ V/Sy [Art. 46 .] 
 
 = T /Sa F^y = T Sal3y. [59, 56, d.] 
 
 The upper or lower sign must be taken accord- 
 ing as ttj, F/Sy are in the same or opposite 
 directions. This numerical result must be mul- 
 tiplied by — 1 when a lies on the clockwise side 
 of fi,y; i.e., when aj, V/3y are opposites (since 
 7/3y lies on the counter-clockwise side of /3, y). 
 Hence — Sa^y is the required algebraic volume. 
 
 Cor. The condition that a, yS, y are coplanar 
 vectors is that Sa^y = (or a^y = a vector). 
 
 EXAMPLES 
 
 1. Expand {p + q + rf, (a + ^ + yf, (p + q)ip- q), 
 
 (p -q){p + q), (p + q)K{p + q)- show that r(p + qf 
 
 = Tp^ + 2SpEq+ Tq\ 
 
 2. Solve 5^ + 4 kg — 8 = for q. [g must be cocircular 
 with k. Hence g = — 2 k ± 2 are the real solutions.] 
 
QUATERNIONS 79 
 
 3. rind the tensor, versor, scalar, vector, and angle 
 of each of the numbers : 2, — 3, 3 i, 2 + 3 i, i + j, 3 i + 4 j, 
 
 6e^', (2i + 3j + 6k)l 
 
 4. Show that the three quaternion cube roots of — 1, 
 ■with horizontal great circle, are — 1, ^ ± ^ V3 k. 
 
 5. Sho-w geometrically that e'^ + e~^' = 2 cos 6, c*' — 
 e-»' = 2esine. 
 
 6. The numbers e" and 
 
 are equal. Verify this approximately by geometric 
 construction when Ta = 1, and when Ta = 2. [For the 
 series, construct OA, AB = aOA, BC = -^-aAB, CD = ^aBC, 
 DE = ^aCD, etc.] 
 
 7. In the plane triangle ABO, whose sides opposite 
 A, B, C, are a, b, c, show by squaring BC= AC— AB, 
 that a^ = b^ + c' — 2bc cos A ; also from 
 
 FBCCA = FCAAB = FA BBC 
 
 show that a : 6 : c = sin ^ : sin B : sin C. 
 
 8. From e^' = cos ^ + e sin 6, e^' ■ e^^ = e(«+»''=, show that 
 
 cos (e + e')= cos 9 cos e' — sin ^ sin 6', 
 sin (0 + 6')= sin 9 cos 6' + cos 6 sin 6'. 
 
 9. Show that (cos 9 + e sin^)" = cos n6+ € sin n9. 
 
80 PRIMER OF QUATERNIONS 
 
 10. Show that Spq = SpSq + S{Vp-Vq), and hence 
 that cos Zpq 
 
 = cos Zp ■ cos Z q + sirxZp ■ sinZq -cosZCVp-Vq). 
 
 11. If - arc q = BA, arc|) = -4(7, show that the last 
 equation of Ex. 10 is the property 
 
 cos a = cos b cos c + sin 6 sin c cos A 
 
 of the spherical triangle ABQ. [Draw B'A = arc Vq, 
 AC = arc Vp.] 
 
 12. If a, p, y, a', j8', y' are vectors from to the vertices 
 A, B, C, A', B', C" of two spherical triangles on the unit 
 sphere 0, where a' = ?7F/3y, P' = UVya, y' = UV{aj3); 
 then a = U'F/3'y') P = UVy'a.', y = f7Fa'/8', and the two 
 triangles are polar triangles. 
 
 13. In Ex. 12 show that cos a = — SPy, sin a = TVPy, 
 etc. ; cos J. = S{UVya. ■ UVa^) = /S/3'y' = - cos a', etc. 
 Hence ZA, Za' are supplements, etc. 
 
 14. Show that the equation of Ex. 11 follows from the 
 
 identity, - SPy = 8(J3y ■ ya) = Sj3y ■ Sya + S{ Vya ■ Va^). 
 
 15. From ViVya-VaP) = — aSa^y, and the similar 
 equations found by advancing the cyclic order a, p, y, 
 show that we have in the spherical triangle ABC, 
 
 sin a : sin 6 : sin c = sin A : sin B : sin G. 
 
 16. Show that if a, j8, y are coplanar unit vectors, then 
 a^y = — aji~^ • y = (y turned through the angle from /8 to 
 a and reversed) = (;8 rotated 180° about the exterior bi- 
 sector of the angle between a, y) = (a — y)j8(a — y)~^ 
 
QUATERNIONS 81 
 
 17. Show that (FABCD)-' ^SACFABCD is the shortest 
 vector from the line AB to the line CD. [Project AC 
 upon the common perpendicular to AB, OD.^ 
 
 18. If a, j8, y be the vector edges about a vertex of an 
 equilateral pyramid (-whose edges are unit lengths), 
 then fi — y, y — a, a — p, are the remaining vector edges. 
 Hence show that Sfiy = Sya. = SaP = —\, and FaF/Jy 
 = (- P)Sya. + ySaP = \(fi - y). Also show that : 
 
 (a) The face angles are 60°, the area of a face is 
 \ V3, and its altitude is \ VS. 
 
 (&) Opposite edges are perpendicular, and their shortest 
 distance is ^ V2. 
 
 (c) The angle between a face and an edge is cos"'^ VS. 
 
 (d) The angle between two adjacent faces is sin-'-| V2. 
 
 (e) The volume and altitude of the pyramid are 
 
 19. The cosines of the angles that a vector makes with 
 i, j, k, are called its direction cosines. Find the lengths 
 and direction cosines of 
 
 2 i — 3 j + 6 k, i + 2 j - 2 k, a;i + 2/J + zk. 
 
 20. Show that the sum of the squares of the direction 
 cosines of a line equals 1. 
 
 21. If (I, m, n), (I'f m', n') are the direction cosines of 
 two lines, show that Zi + mj + wk, Z'i + m'j + n'k, are unit 
 vectors in the directions of the lines, and that if 6 be the 
 angle between the lines, then cos = 11' + mm' + nn' ; also 
 
82 PlilMER OF QUATEENIONS 
 
 that sin^^ = 
 
 mn 
 
 nl 
 n'V 
 
 + 
 
 Im 
 
 I'm' 
 
 , and that the three terms 
 
 of the second member, respectively divided by sin^0, are 
 the squares of the direction cosines of a line that is per- 
 pendicular to the given lines. [Art. 57.] 
 
 22. If be a given origin, then the vector OP = p = 
 x\ -\-yj+ zk say, is called the vector of P with reference to 
 the given origin. If OX, Y, OZ be axes in the directions 
 of i, j, k, the scalar values of the projections of OP upon 
 these axes, i.e., (x, y, z), are called the coordinates of P 
 with reference to the given axes. Let the coordinates of 
 the vertices of the pyramid ABCD be, respectively, 
 (8, 2, 7), (10, 6, 3), (1, 6, 3), (9, 10, 11). Draw this pyramid 
 with reference to a perspective of i, j, k, showing coordi- 
 nates and vectors. Also : 
 
 (a) Find the vectors and coordinates of the middle 
 points of the edges. [OM = ^(OA -|- OB), etc.] 
 
 (6) Find the lengths and direction cosines of the 
 edges. [- AB2 = ^S^etc.] 
 
 (c) Find vectors that bisect the face angles. [fJAC 
 iJJAD bisects ZC-4Z).] 
 
 (d) Find altitudes of the faces and the vectors of their 
 feet. [If L be the foot of the perpendicular from B on 
 AG, then AL= AB->^ABAG, etc.] 
 
 (e) Find the areas of the faces. 
 
 (/) Find the volume and altitudes of the pyramid. 
 {g) Find the angles between opposite edges, and their 
 (shortest) distance apart. [Ex. 17.] 
 
 Qi) Find the angle between two adjacent faces. 
 
EQUATIONS OF FIRST DEGBME 83 
 
 CHAPTER IV 
 Equations of First Degree 
 
 61. The general equation of first degree in 
 an unknown vector p is of tlie form, 
 
 («) qipri + q2pr2 + -=q, 
 
 where q, q^, r^, q^, r^,--- are known numbers. 
 
 This equation may be resolved into two 
 equations by taking the scalar and the vector 
 of each member ; and we shall consider these 
 equations separately. 
 
 62. Taking the scalar of (a), Art. 61, the 
 term Sq^pr^ becomes, by a cyclic change in the 
 factors, S ■ rj qip, and this becomes [by dropping 
 the vector {Sr-^q-^p, since its scalar is zero] 
 S{Vr-^qi- p); and similarly for the other terms. 
 Hence if we put Vriq-^+ Vr^q^^ — = S, and 
 Sq = d, the general scalar equation of first 
 degree in p becomes, 
 
 (a) SSp = d or SS{p-dS-') = 0. 
 
84 PRIMER OF QUATERNIONS 
 
 One solution of this equation is obviously 
 pzrzdS"^. This is not the only solution, since 
 by Art. 45, Cor. 3, the second factor may be 
 any vector that is perpendicular to S. Hence 
 the general solution is p = dS~^ + V(TS, where cr 
 is an arbitrary vector. 
 
 63. Hence, draw OD = S, take iVon the line 
 OD so that ON = <iS"^, and draw any vector 
 N P = VaS that is perpendicular to the line OD ; 
 then p = OP is a solution of the equation 
 SSp = d. 
 
 The locus of P is therefore a plane perpen- 
 dicular to OD at the point iV; and this plane 
 is called the locus of the equation S8p = d, ivith 
 respect to the origin 0. [The locus is the 
 
EQUATIONS OF FIRST DEGREE 85 
 
 assemblage of all points that satisfy the equa- 
 tion.] 
 
 64. The vector perpendicular distance from 
 the plane SSp — d=0 to the point P' {whose vec- 
 tor is p) is S^\SSp' — d), and the corresponding 
 scalar distance 'measured upon 8 is 
 
 -{S8p'-d)/T8. 
 For the perpendicular distance of P' is the 
 projection of \^F={p-dS-% upon 0D = 8. 
 
 65. The locus of the simultaneous equations 
 Sap = a, S^p = h is a straight line, viz., the inter- 
 section of the tivo plane loci of these equations 
 taken separately. 
 
 For in order that p = OP may satisfy both 
 equations, P must lie in both planes, and its 
 locus is therefore the intersection, of those 
 planes. 
 
 66. The equation VSp = S', or 
 
 Vh{p-8-'S') = 0, 
 is a consistent equation only when S' is perpen- 
 dicular to 8, since VSp is always perpendicular 
 
86 
 
 PRIMER OF QUATERNIONS 
 
 to S. When 8' is perpendicular to S, then 8~^S' 
 is a vector (Art. 45, Cor. 1), and the general 
 solution of this equation is p = S~'S' + xS, where 
 X is an arbitrary scalar (Art. 46, Cor.). Hence 
 draw 0N = 8"^S', and NP = xS (any vector par- 
 allel to 8), and then p = OP is a solution of the 
 
 given equation. The locus of P is therefore the 
 straight line through iV" parallel to 8, and ON is 
 the perpendicular from the origin upon the line. 
 The equations of Art. 65 take this form by 
 multiplying the first by /3, the second by a, and 
 subtracting, remembering that 
 
 r{rafi-p) = aSfip-l3Sap. 
 
 67. The vector perpendicular distance from 
 the line VSp - 8'= to the point P' is 8'\rBp-S'), 
 where p'=OP'. 
 
EQUATIONS OF FIRST DEGREE 87 
 
 For the required perpendicular distance of 
 P' is the projection of NP', =(/3' — S"^S'), per- 
 pendicular to 8. 
 
 68. The point of intersection of the three 
 planes Sap = a, Sfip = b, Syp = c is 
 
 p = {a Vfiy + hVya + c Va^ySafiy. 
 [Art. 56, (i).J 
 
 EXAMPLES 
 
 1. Find the equation of the locus of a point that moves 
 so that its numerical distances from two fixed points are 
 equal. 
 
 2. A point moves so that its scalar distances from two 
 fixed planes are equal ; show that its locus is a plane 
 bisector of the diedral angle of the given planes. 
 
 3. A point moves so that the sum or difference of its 
 scalar distances from two fixed planes is constant ; show 
 that its locus is a plane parallel to the interior or exterior 
 bisector of the diedral angle of the given planes. 
 
 4. A point moves so that the ratio of its scalar dis- 
 tances from two fixed planes is constant ; show that its 
 locus is a plane. 
 
 5. A point moves so that its numerical distances from 
 two intersecting lines are equal; find its locus. [Take 
 the point of intersection as origin.] 
 
88 PRIMER OF QUATERNIONS 
 
 6. A point moves so tliat its numerical distances from 
 three fixed points are equal ; find its locus. 
 
 (a) The same with coplanar lines instead of points. 
 [Four straight lines perpendicular to the plane of the 
 lines.] 
 
 7. Find the vector of the centre of the sphere whose 
 surface passes through four given points. 
 
 8. A point moves so that its tangential distances from 
 two given spheres are numerically equal ; find its locus. 
 
 9. On the chord OQ of a given sphere a point P is 
 taken so that OP • OQ = — d'; when Q moves round the 
 sphere find the locus of P. [A plane perpendicular to 
 the diameter OD.^ 
 
 10. The locus of the point P whose coordinates (x, y, z) 
 satisfy Ix + my + nz + d = is a plane perpendicular to 
 the vector Zi + mj + wk, at a distance from the origin of 
 — d/y/(P + m^ + w^), measured in the direction of this 
 vector, (a) Show that the equation of this plane may be 
 put in the form aj cos a + y cos ;8 + z cos y —p = 0, where 
 p is the perpendicular distance from to the plane and 
 the cosines are the direction cosines of this perpendicular. 
 
 11. Find the perpendicular distance of P' = (x', y', z'), 
 from the plane of Ex. 10, 
 
 [m' cos a + 2/' cos j8 + z'cosy—p']. 
 
 12. The locus of the point P whose coordinates (x, y, z) 
 satisfy (x — a)/l=(y — b)/m=(z — c)/n is a line paral- 
 
EQUATIONS OF FIBST BEGEEE 89 
 
 lei to the vector l\ + mj + vk through the point (a, h, c). 
 If P satisfy the first two of these three equations, its 
 locus is a plane through the line, perpendicular to the 
 plane of XOY. [If t be the common value of the three 
 ratios, then p = ai + &j + ck + t(V\ + mj + wk) .J 
 
 13. Find the perpendicular distance of P' = (a:', y', z') 
 from the line of Ex. 12. 
 
 In the following examples A, B, C, D, P are points 
 whose coordinates are (8, 2, 7), (10, 6, 3), (1, 6, 3), (9, 
 10, 11), («, y, z). 
 
 14. The equation of the plane through A perpendicular 
 to Oi) is /SODAP = 0, or 9 a; + 10 2/ + 11 2 = 169. 
 
 15. The equation of the plane through AB parallel to 
 CI> is aS-APFABCD = 0, or2a;-22/-« = 5. 
 
 16. The equation of the plane ABC is ^S'- APFABAC 
 = or 2/ + 2 = 9. 
 
 17. Find the perpendicular distance of D from the 
 planes in Exs. 14, 15, 16. 
 
 18. The equation of the plane through AB that con- 
 tains the common perpendicular to AB, CD is 
 
 /S- APF(ABFABCD)=0, or 2a; + i/ + 22 = 32. 
 
 19. The equation of the line through A parallel to OD 
 is FODAP = or AP = «OD, or (a; - 8)/ 9 =(2/ -2)/ 10 
 = (2 -7)/ 11. 
 
90 PlilMEE OF QUATERNIONS 
 
 20. The equation of the line AB is, FABAP = 0, or 
 AP = iAB or (x - 8)/2 ={y - 2)/4 = {z - 7)/- 4. 
 
 21. The equation of the common perpendicular to 
 AB, CD is the equation of Ex. 18 and a; + 2 2/ - 2 2 = 7. 
 
 22. Find the distance of D from the lines in Exs. 19, 
 20, 21. 
 
 23. Eind OD in the form 10 A + mOB + nOC, and find 
 the ratios in which OD cuts the triangle ABC. 
 
 NONIONS 
 
 69. The vector equation of first degree is 
 (a) Vq^pr^ + Vq^pr^ + ... = Vq. 
 
 To solve this equation we resolve it along 
 i, j, k, by multiplying it by these vectors and 
 taking the scalars of the products. We thus 
 find three scalar equations of first degree from 
 which p may be immediately found as in 
 Art. 68. Hence (a) has in general one, and only 
 one, solution which corresponds to the intersec- 
 tion of three given planes. [See further Art. 81.] 
 
 70. The first member of Art. 69 (a) is a 
 linear, homogeneous vector function of p ; i.e., it 
 
EQUATIONS OF FIRST DEGREE 91 
 
 is of first degree in p, every term is of the same 
 degree in p, and it is a vector. 
 We may denote the operator 
 
 by a single letter, <f), so that (ftp, (f)a;-- denote 
 the vectors that result from putting p, a, — in 
 the places occupied by the parenthesis. 
 
 71. The operator ^ is distributive over a sum 
 and commutative ivith scalars ; i.e., 
 
 <f){xp +ycr) = X(^p + y<l>(T. 
 This is immediately verified by putting 
 xp + ycr in the places occupied by the paren- 
 theses of (j) and expanding the several terms. 
 
 72. We have p = xcL + y/3 + zy, where a, ft, y 
 are given non-coplanar vectors, and x, y, z, are 
 scalars, each of first degree in p, as shown in 
 56 (i) with p in the place of 8 ; hence, 
 
 {a) (ftp = X(f>a + y(j)ft + z<f)y. 
 
 The complete operation of <f> is therefore 
 determined when the three vectors <^a, <f>ft, (f)y 
 are known. Since each of these vectors involves 
 
92 PRIMER OF QUATERNIONS 
 
 three scalar constants (e.g., the multiples of the 
 given non-coplanar vectors a, yS, y, that express 
 it), therefore the value of <^ depends upon nine 
 scalar constants. The operator (j) may therefore 
 be called a nonion. Scalars and rotators are 
 particular forms of nonions. 
 
 Note. — It is readily shown that nonions have the 
 same laws of addition and multiplication among them- 
 selves as qtiaternions. Products are not in general com- 
 mutative. A product 
 
 (<^ - 9i) (<t> - 92) (i> - 93), 
 where gi, g,, g^ are scalars, is commutative, since <^ is 
 commutative with scalars by Art. 71. Hence this prod- 
 uct multiplies out as if 4> were a scalar, and is 
 
 4'^ - (91 + 92 + 93)<t>'' + (929s + 9i9i + 9i92)<l> - 9i9-£s- 
 
 Linear Homogeneous Strain 
 
 73. An elastic solid is subjected to the strain 
 ((> with respect to an origin 0, when all its parti- 
 cles, A, B, C, etc., are displaced to positions A', 
 B', C, etc., that are determined by OA' = <^0A, 
 OB' = (/jOB, OC = <^0C, etc. In general, any 
 particle P whose vector is OP = p occupies after 
 
EQUATIONS OF FIRST DEGREE 
 
 93 
 
 the strain the position P', whose vector is 
 OP' = ^p. The particle at is not moved, 
 since its vector after strain is ^00 = ^0 = 0. 
 
 (a) We have, also, (^AP = A'P', etc. 
 For, A'P' = OP' - OA' = <^0P - <^0A 
 = <^(0P - OA) = ^kP, etc. 
 
 74. A straight line of particles parallel to a. 
 is homogeneously stretched and turned hy the 
 strain <f> into a straight line of particles parallel 
 to <f)a, and the ratio of extension and turning 
 is (f)a/a. 
 
 For let AP be a line parallel to a, and let 
 
94 PRIMER OF QUATERNIONS 
 
 A, P strain into A', P'. Then, since kP — xa, 
 therefore, by 73a, k'P' = x^a, and the ratio of 
 extension and turning is A'P'/AP = ^a/a. 
 
 Note. — This property that parallel lengths of the 
 substance strain into parallel lengths and are stretched 
 proportionally, is the physical definition of linear homo- 
 geneous strain. 
 
 75. A plane of particles parallel to a, ^ is 
 homogeneously spread and turned by the strain ^ 
 into a plane of particles parallel to <^a, ^/3, and 
 the ratio of extension and turning is V(j>a(j)j3/Va^. 
 
 For let APQ be a plane parallel to a, /3, and 
 let A, P, Q strain into A', P', Q'. Then, since 
 AP = xa + yfi, AQ = x'a + y'fi, therefore 
 
 A'P' = x4>a + 2/(^/3, A'Q' = jc'c^a + y'(f,l3. 
 
 By Arts. 59, 55, (e), the directed area of the 
 triangle APQ is iV ■ AP ■ AQ = i{xy' - x'y) Va/S, 
 and the directed area of the triangle AP'Q' is 
 the same multiple of V(f>a^l3- Hence the ratio 
 of the extension and turning of directed area is 
 
EQUATIONS OF FIRST DEGREE 95 
 
 76. A volume of particles is homogeneously 
 dilated hy the strain <f> in the ratio 
 
 where a, /3, y are any given non-coplanar vectors. 
 For let the pyramid APQR strain into the 
 pyramid A'P'Q'R'. Then since 
 
 k? = xa + y^ + zy, AQ = x'a + y'^ + z'y, 
 
 AR = x"a + 2/"y8 + z"y, therefore A'P', A'Q', A'R' 
 have these values with ^a., <f)j3, <^y instead of 
 a, P, y. The volume of the pyramid APQR 
 relative to the order AP, AQ, AR of its edges 
 is, by Arts. 60, 56, (c), 
 
 X y z 
 -iSAPAQAR= - * xy'z' Sa^y, 
 
 x'Y'z" 
 
 while that of the strained pyramid is the same 
 multiple of S(j}a<f)/3(fiy. Hence the ratio of 
 dilation of volume is S<f)(x<l)fi<^y/ SajBy. 
 
 77. The ratio of dilation of ^ is called its 
 modulus. 
 
 (a) It is obvious from the signification of the 
 
96 PRIMER OF QUATERNIONS 
 
 modulus that the modulus of a product of 
 nonions equals the product of the moduli of the 
 factors ; e.g., mod (f>\jj = mod ^ • modxp. 
 
 When mod (j) is positive, the parts of the 
 volume are in the same order before and after 
 strain. When mod (j) is negative, the order of 
 the parts is reversed by the strain ; i.e., if AP 
 lie on the counter-clockwise side of the plane 
 AQR, then A'F' lies on the clockwise side of 
 A'Q'R', so that the particles along AP have 
 been strained through the particles of the 
 plane AQR. Such a strain is obviously not a 
 physical possibility. 
 
 Finite and Null Strains 
 
 78. If an elastic solid which fills all space 
 he subjected to a strain (j), the strained solid fills 
 all space if mod cj) he finite, and it fills only an 
 indefinite plane or line through the origin or 
 reduces to the origin if mod 4> he zero. 
 
 For if S(j)a(p/3(j)y be finite, then (f>a, ^^, <^y are 
 non-coplanar vectors, so that 
 
 <l)p{ = x<f)a + y^^ + zj)y) 
 
EQUATIONS OF FIBST DEGREE 97 
 
 may be made any vector by properly cboosing 
 p{ = xa + y^ + zy). But if S(f)a(f>^(f)y = 0, then 
 <f)a, (f)fi, (j)y are coplanar vectors or colinear 
 vectors or each zero, so that <f)p will be a vector 
 in a given plane or line through or the vector 
 of 0, whatever value be given to p. 
 
 When mod <j> is zero, is called a null 
 nonion ; and it is called singly or doubly or 
 triply null, according as it strains a solid into a 
 plane or a line or a point. If ^a = 0, then a is 
 called a null direction of <^. 
 
 79. Null strains, and only null strains, can 
 have null directions; a singly null strain has only 
 one null direction; a douhly nidi strain has a 
 plane of nidi directions only ; a triply null strain 
 has all directions null. 
 
 For when mod ^ = 0, then ^a, ^/8, ^y are 
 coplanar or colinear vectors, and we have a 
 relation l(fia + m<f)/3 + n(f)y=0 ; i.e., la + m/3 + ny 
 is a null direction of 0. Conversely, if <^ have 
 a null direction, take one of the three non- 
 coplanar vectors a, jS, y, in that diiection, say a, 
 
98 PRIMER OF QUATERNIONS 
 
 and we have S(jia(j)/3(fyY = 0, since <^a = 0, and 
 therefore mod (^ = 0. 
 
 Also, if (f> have only one null direction, a, 
 then <j)fi, <f)y, are not parallel, since (j)fi = l<j)y 
 makes ^ — ly a, second null direction. Since 
 ^ = xa + ?/y8 + zy, therefore (f)p = ycj)^ + Z(j)y, which 
 is any vector in the plane through parallel to 
 (f>fi, <f>y ; hence (f> is singly null. 
 
 But if <f) have two null directions, a, ^, then 
 <f)p = z<f)y, which is any vector in the line 
 through parallel to ^y, and therefore <j) is 
 doubly null. Also, since <j){xa + y/3) = 0, there- 
 fore any direction in the plane of a, /8 is a null 
 direction of <f). 
 
 If (^ have three non-coplanar null directions 
 a, /3, y, then <f>p = for all values of p ; i.e., a 
 triply null nonion is identically zero. 
 
 80. A singly null nonion strains each line 
 in its null direction into a definite point of its 
 plane ; and a doiibly null nonion strains each 
 plane that is parallel to its null plane into a 
 definite point of its line. 
 
EQUATIONS OF FIB ST DEGREE 99 
 
 For when ^ is singly null, say <^a=0, then 
 X(f)^ + y(})y is the vector of any point in the 
 plane of <f>, and all particles that strain into 
 this point have the vectors p = xa + y/3 + zy, 
 where x is arbitrary, since ^a = ; i.e., they are 
 particles of a line parallel to a. So, if (^ is 
 doubly null, say <^a =0, ^/S = 0, then any point 
 of the line of ^ is z^y, and the particles that 
 strain into this point have the vectors 
 
 p = x<x-\-y^ + Zy, 
 
 in which x, y are arbitrary ; i.e., they are par- 
 ticles of a plane parallel to a, y8. 
 
 Note. — It follows similarly that the strain <^ alters 
 the dimensions of a line, plane, or volume by as many 
 dimensions as the substance strained contains inde- 
 pendent null directions of <^, and no more. Hence, a 
 product t^i/' has the null directions of the first factor, and 
 the null directions of the second factor that lie in the 
 figure into which the first factor strains, and so on ; the 
 order of nullity of a product cannot exceed the sum of 
 the orders of its factors, and may be less ; etc. 
 
100 PBIMEB OF QUATERNIONS 
 
 Solution of <j)p = S 
 
 81. The solutions of ^p = S are, by defini- 
 tion of the strain <p, the vectors of the particles 
 that strain into the position whose vector is 8. 
 Hence : 
 
 1. When (f) is finite, there is one, and only 
 one, solution. 
 
 2. When cf) is singly null, and 8 does not lie 
 in the plane of (f>, there is no finite solution. 
 Divide the equation by Tp, and make Tp infinite, 
 and we find (j)Up = 0; i.e., the vector of the 
 point at infinity in the null direction of ^ is a 
 solution. 
 
 3. When (f> is singly null, and 8 lies in the 
 plane of ^, there are an infinite number of solu- 
 tions, viz., the vectors of the particles of a line 
 that is parallel to the null direction of <^. 
 
 4. When ^ is doubly null, and 8 does not lie 
 in the line of <^, there is no finite solution. As 
 in (2) the vectors of the points of the line at 
 infinity in the null plane of ^ are solutions. 
 
EQUATIONS OF FIRST DUGBEE 101 
 
 5. When <f) is doubly null, and 8 lies in the 
 line of <f), there are an infinite number of solu- 
 tions, viz., the vectors of the particles of a 
 plane that is parallel to the null plane of ^. 
 
 These results correspond to the intersections 
 of three planes, viz. : 
 
 1. The three plane's meet in a point. 
 
 2. The three planes parallel to a line. 
 
 3. The three planes meet in a common line. 
 
 4. The three planes parallel. 
 
 5. The three planes coincide. 
 
 Derived Moduli op <f> 
 
 82. The ratio in which the nonion <l) + g 
 dilates volume is, 
 
 mod {<}> + g) = S{<iia + ga){^^ + g^){<t>y + gy)/Safiy. 
 
 This is independent of the values of the non- 
 coplanar vectors a, /3, y in terms of which it 
 is expressed. If ^ is a scalar, this modulus is 
 an ordinary cubic in g, whose coefficients will 
 therefore depend only upon ^. The constant 
 
102 PRIMER OF QUATERNIONS 
 
 term is mod <^, and the coefficients of g, g^, are 
 called modi i>> modg cf), so that, 
 (a) mod{(f) + g) = g^ + g^ mod^cj) + g modii^ + mod<j). 
 [modi ^ = Sia^P^y + fi^y^a + y<^a^^)/SaPy ; 
 modjc^ = S{^y^a + ya^^ + ay8^7)/yS'a;8y]. 
 
 83. The roots g^, g^, g^, of the cubic 
 mod (^ — ^) = 
 are called the latent roots of (f). We have from 
 82 (a) with — (/ in the place of g, and the theory 
 of equations, 
 
 mod(j> = g^g^gs, mod,(^ = g^g^ + g^g^ + g^g^, 
 mod^<f> = g^ + g, + g^. 
 
 (a) The latent roots of cjt — g^ are those of (f> 
 diminished by g^. 
 
 For the roots of mod {^ — gi — g) = 0, are 
 
 9 = ^,92- 9i^ 9b - 9i- ^■9-^ 9 = 92-9i gives 
 mod l<j>-gi- {g, - ^,)] = mod (</. - g,) = 0. 
 
 (&) ITie order of nullity of (jy cannot exceed the 
 number of its zero latent roots. 
 
 For if (j) has one null direction a, then <^a = 
 makes mod <^ = 0, so that at least one of the 
 
EQUATIONS OF FIliST DEGREE 103 
 
 latent roots is zero, say g^ ; and if <^ has a 
 second null direction ^8, then <^a =0, <f>^= 0, 
 makes modi^=0 or g,j,g.i = 0, so that another 
 latent root is zero, etc. 
 
 (c) The order of nullity of (ft — g^ cannot exceed 
 the number of latent roots of <^ that equal g^. 
 
 [(«), (&)] 
 
 Latent Lines and Planes op <j) 
 
 84. Those lines and planes that remain 
 unaltered in geometrical position by the strain 
 <j) are called latent lines and planes of <j). 
 
 (a) The latent directions of <f) are the null 
 directions of (ji-gu ^-(72; <l>-gs, cmd g^, g^, g^ 
 are the corresponding ratios of extension in 
 those directions. 
 
 For if p is any latent direction, and g is the 
 ratio of extension in that direction, then we 
 have <^p = gp or (^ — g)p = 0, Hence (j) — g is a, 
 null nonion, or mod {(f) — g) = 0, so that ^ is a 
 latent root of <f) ; also p is a null direction of 
 
 ^-9- 
 
104 PBIMER OF qUATEHNIONS 
 
 Note. — Since a cubic with real coefficients has at 
 least one real root, therefore a real nonion has at least 
 one latent direction. Also if two roots are imaginary, 
 they are conjugate imaginaries, and the corresponding 
 latent directions must also be conjugate imaginaries. 
 
 85. If a, p, y he the latent directions corre- 
 sponding to gi, g^, g^, then (/3, y), {y, a), (a, |8) 
 determine latent planes of (^ in which the ratios 
 of spreading are g,gs, g-^g,, g^g^. E.g., 
 
 Y^^i>y = Y{g,^ ■g,y)= g,g, Vfiy. 
 
 Hence, in the general case when the latent 
 roots are all unequal, the latent vectors a, /8, y 
 must form a non-coplanar system, since any 
 two of the latent lines or planes determined by 
 them have unequal ratios of extension, and 
 cannot, therefore, coincide. 
 
 (a) The plane of <f) — g^ is the latent plane cor- 
 responding to g2, gs- 
 
 For {<l>-g,)p = y{g2-gi)/3 + z{gs-gi)y, 
 (= plane of /S, y). [The plane and null line of 
 (j) — gi uiay be called corresponding latents of (f>.] 
 
EQUATIONS OF FIBST BEGEFE 105 
 
 The Characteristic Equation of <j> 
 
 86. We have also, 
 
 (a) {(f>-g,){(j)-g^){<^~gg)=0. For the first 
 member has the three non-coplanar null direc- 
 tions a, fi, y. [See 80 note, 72 note.] 
 
 Conjugate Nonions 
 
 87. Two nonions c^, (ft are conjugate when 
 (a) Sp(f>T = S(r(f)'p for all values of p, a-. 
 When (f) is known, this determines ^' without 
 
 ambiguity. Thus, put o- = i, j, k, in turn, and 
 we have by Art. 57 (b), 
 
 cfi'p = — '\Sp(f)'\ — ']Sp(f>'] — kSp(f)k. 
 
 Conversely, this function satisfies (a), for we 
 have ScT<p'p = Sp(f){ — '\Sicr — \S] r — k^kcr) = Sp(f)(T. 
 
 88. From this definition of conjugate strains 
 we have 
 
 (a) {a4> + hxjj)' = a4' + hxjs' ; {<j>xpy = f f . 
 (&) {Vqi )qy = Vqi )p, [aS^{ )]' = /3Sa{ ). 
 
106 PRIMER OF QUATERNIONS 
 
 E.g., Scr{(f>}pyp = Sp^\^(r = Sp(f>{xlicr) 
 
 = S^crcji'p = 8(T\\i<^'p, 
 
 and therefore ((^i//)' = »/»'(^'. [If Scr{a-p) = {) 
 for all values of a-, then a — /S = 0, since no 
 vector is perpendicular to every vector cr. 
 Hence, comparing the first and last member of 
 the above equation, we have (<^i/»)'/3 = ^'4>'p-] 
 
 89. Two conjugate strains have the same 
 latent roots and moduli, and a latent plane of 
 one is perpendicular to the corresponding latent 
 line of the other. 
 
 For since ((^ — g'l) a = 0, therefore 
 
 = Sp{<f)-g,)a= Sa{4,' - gi)p, 
 
 and therefore (f)' — g^ is a null nonion whose 
 plane is perpendicular to a. Hence g^ is a 
 latent root of (/>', and the latent plane of (j/ 
 corresponding to (f/ — g^ is perpendicular to 
 the latent line of ^ corresponding to <^ — g^. 
 [Art. 85, (a).] 
 
equations of first degree 107 
 
 Self-Conjugate Nonions 
 
 90. A nonion ^ is self -conjugate when <ji' = 4> 
 or when Spcjicr = Scrcjip for all values of p, cr. 
 In consequence of this relation a self -con jugate 
 strain has only six scalar constants, three of 
 the nine being equal to three others, viz., 
 
 S\(j>} = S\(l>\, S\(l>k = Sk<l>\, S]c{>k = Sk<l>]. 
 
 91. A self-conjugate strain has by Art. 88 
 three mutually perpendicular latent directions, 
 and conversely, if (f> have three mutually per- 
 pendicular latent directions, i, j, k, correspond- 
 ing to latent roots a, h, c, then 
 
 ^/D = — a'\S'\p — b'jSip — ckSkp, 
 which is self-conjugate. [68 &.] 
 
 92. A real self -conjugate strain has real 
 latent roots. 
 
 For let a' = a + /SV^l, /S' = a - ^V^i be 
 latent directions corresponding to conjugate 
 imaginary roots a, 6 of a real nonion ^ ; then, 
 if (^ is self-conjugate, we have 
 
 Sa4>^' = Sfi'<^a' = hSa/3' = aSa/3', 
 
108 PRIMER OF QUATERNIONS 
 
 or, since a, h are unequal, therefore Safi' = ; 
 but this is impossible, since Sa'fi' = a? + yS^, a 
 negative quantity. Therefore is not self-con- 
 jugate if it has imaginary latent roots. 
 
 93. A nonion ^ is negatively self -conjugate 
 when (f)' = — (j), or when Scrcjyp = — Spffxr. Such 
 a nonion has therefore only three scalar con- 
 stants, since S'\(f)\ = — S\(f)\ shows that S\(]}\ = 0, 
 and similarly, /S'j^j = 0, 8k(f>k = 0, while the 
 other six constants occur in negative pairs 
 
 S\(j)i = — S\^\, etc. 
 
 (a) The identity Sp<^p = gives (by putting 
 /) = jci + 2/j + sk where x, y, z are arbitrary) all 
 the above relations between the constants of 
 ^, and is therefore the sufficient condition that 
 ^ is negatively self-conjugate. It shows that ^p 
 is perpendicular to p or that <l>p = Vep, where e 
 must be independent of p since (f)p is linear in p. 
 
 94. -Any nonion tj) may be resolved into a 
 sum of a conjugate and a negatively self-conjugate 
 nonion in only one way. 
 
EQUATIONS OF FIRST DEGREE 109 
 
 For if ^ = ^ + xp, where ^ =^, xjj' = - xjt, 
 then <fi' = <f> — xfj, and adding and subtracting, we 
 have tl)=i{(l> + ^')\p= i{(f,- ^'), and 
 
 (a) (l>p=i{<j> + (j)')p + J(<^ - 4>')p='^p + Vep. 
 To find e in terms of the constants of (j), we 
 have p= — '\S\p — ]S\p — kSkp, and therefore 
 <^p= ~ ^\8\p — etc. 
 <^> = - \Sp<i>\ - etc. [88 &.] 
 Hence i(^ - 4')p = i{\Sp(f>'\ - (f>'\8\p) + etc. 
 
 = iV-{V\(j>\)p + etc.= Vep, 
 and therefore 
 
 (a) e = J F(i<^i + j^j + k^k). 
 
 EXAMPLES 
 
 1. rind the equation of a sphere whose centre is 
 ^(OA = a) and radius a. 
 
 2. Show that the square of the vector tangent from 
 the sphere of Ex. 1 to P' is (p' - af + a\ 
 
 3. Find the locus of the point P such that PP' is cut 
 in opposite ratios by the sphere of Ex. 1 ; show that it is 
 the plane of contact of the tangent cone from P' to the 
 sphere and is perpendicular to AP'- 
 
 4. Let P' be any point on the sphere A of Ex. 1, and 
 take P on OP' so that OP • OP' + c^ = ; find the locus 
 
110 PRIMER OF QUATERyiONS 
 
 of P. [P, P' are called inverse points with respect to 0, 
 and the locus of P is the inverse of the given sphere A. 
 It is a sphere with centre A' on OA, or a plane perpen- 
 dicular to 0^1 if the given sphere A pass through 0.] 
 
 5. Show that the inverse of a plane is a sphere 
 through 0. 
 
 6. Show that the general scalar equation of second 
 degree is Sp<t>p + 2SSp+ d = 0, where <^ is a self-conju- 
 gate nonion. 
 
 7. Show that Sptfip = is the equation of a cone with 
 vertex at 0. 
 
 8. Show that the line p=a + x^ cuts the quadric 
 surface of Ex. 6 in two points ; apply the theory of equa- 
 tions to determine the condition that this line is a tan- 
 gent to the surface, or an element of the surface, or that 
 it meets the surface in one finite point and one point at 
 infinity, or that the point whose vector is a lies midway 
 between the points of intersection. 
 
 9. Show that the solution of c^p -|- 8 = is the vector 
 of a centre of symmetry of the quadric surface of Ex. 6. 
 Hence classify quadric surfaces as central, non-central, 
 axial, non-aadal, centro-planar. 
 
 10. Show that the locus of the middle points of chords 
 parallel to /3 is a diametric plane perpendicular to <;f>;8. 
 
 11. Show that an axial quadric is a cylinder with ele- 
 ments parallel to the null direction of its nonion <^. 
 
EQUATIONS OF FIRST DEGHEE 111 
 
 12. Show that a non-axial quadric is a cylinder with 
 elements parallel to the mill plane of its nonion tj) and 
 perpendicular to its vector 8. 
 
 13. Show that a centro-planar quadric consists of two 
 planes parallel to the null plane of its nonion <^. 
 
 14. Show that the equation of a central quadric 
 referred to its centre as origin is Spfjtp + 1 = 0. Show 
 that the latent lines and planes of (^ are axes and planes 
 of symmetry of the quadric ; also that <f>p is perpendicu- 
 lar to the tangent plane at the point whose vector is p. 
 (a) Show that the axes and planes of symmetry of the gen- 
 eral quadric are parallel to the latent lines and planes of <^. 
 
 15. Show that if \f/^ = <f>, then the equation of the 
 central quadric is (i/'p)^ + 1 = 0; and that therefore the 
 quadric surface when straiaed by tj/ becomes a spherical 
 surface of unit radius. 
 
 16. Show that if g, a are corresponding latent root and 
 direction of <^, then g", a are the same for <^". Find the 
 latent lines and planes, the latent roots and moduli of 
 the following nonions and their powers : 
 
 (a) (aaSliyp + bfiSyap + CySal3p)/SaPy. 
 
 (6) [aaS/Syp +(aP + ba)Syap +(cy<Sa/3p] / Safiy. 
 
 (c) [aaS^yp + (aft + ba.)Syap + (ay + C/3)<Sa/8|o] / Safiy. 
 
 (d) Vcp,qpq-\ 
 
 17. Show that the latent roots of ep—fVap^ (/>0, 
 Za = Ty8 = 1) are e+f, e +fSaP, e—f, corresponding to 
 
112 PBIMME OF QUATEHNIONS 
 
 latent directions a + ^, Va/S, « — /? ; and that this is 
 therefore a general form for self -conjugate nonions. 
 Determine the latent directions and roots in the limiting 
 case when a. = j3, or — j8 or/= 0. 
 
 18. Show that the nonion of Ex. 16 takes the form 
 hp —f(aS^p + pSap), where b is the mean latent root. 
 
 19. Substitute the nonion of Ex. 18 for <^ in Ex. 6 and 
 show that the quadric surface is cut in circles by planes 
 perpendicular to a or ^. When is the surface one of 
 revolution ? 
 
 20. If the conjugate of a nonion is its reciprocal, and 
 the modulus is positive, then the nonion is a rotation; 
 and conversely every rotation satisfies this condition. 
 [If B, B~^ are conjugate nonions, then p^ = 8pIi~^Ep 
 = SBpBp = (Bpy- ; i.e., TBp /p = 1. Also 8pa- = SpB-'Ba- 
 = SBpBa- and therefore the angle between p,a- = Z 
 between Bp, Bu. Therefore B strains a sphere with 
 centre into another sphere with centre in which the 
 angles between corresponding radii are equal and their 
 order in space is the same, since mod B is positive. 
 Hence the strain is a rotation.] 
 
 (a) Show that (B<l>B-y = B<j>"B-^. 
 
 21. Show that <^'</i is self-conjugate, and that its latent 
 roots are positive, and that therefore there are four real 
 values of i/^ that satisfy i//^ = <^'<^, mod \j/ = mod </>. [Let 
 <p'4>\ = a'\ ; then a = — ^i<^'<^i =(r</>i)-.] 
 
EQUATIONS OF FIRST BEGBEF 113 
 
 22. If ^ = Rt^i, where i/' is the self-conjugate strain 
 V^^, then i? is a rotation. So <^ = x-Bj where x = BijjR-' 
 
 23. Show that <^' • F<^;8<^y = V^y ■ mod <^. [66/] 
 
 24. Show that the strain <^p = p — aaS^p, where a, (3, 
 are perpendicular unit vectors, consists of S. shearing of 
 all planes perpendicular to yS, the amount and direction 
 of sliding of each plane being aa per unit distance of the 
 plane from 0. 
 
 25. Determine i(/ and R of Ex. 22 for the strain of Ex. 
 24, and find the latent directions and roots of \p. 
 
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