Production Note Cornell University Library produced this volume to replace the irreparably deteriorated original. It was scanned using Xerox software and equipment at 600 dots per inch resolution and compressed prior to storage using CCITT Group 4 compression. The digital data were used to create Cornell's replacement volume on paper that meets the ANSI Standard Z39.48-1984. The production of this volume was supported in part by the Commission on Preservation and Access and the Xerox Corporation. 1990.(Üorncll Httiwwiig J ibtatü BOUGHT WITH THE INCOME FROM THE SAGE ENDOWMENT FUND THE GIFT OF Heiwpu W. Sage 1891 <2 3/r/9^ ^ukUBuxm. OUTLINES OF QUATERNIONSTHE OUTLINES OF QUATERNIONS LIEUT.-COLONEL H. W. L. HIME (late) royal artillery LONDON LONGMANS, GREEN, AND CO. AND NEW YORK : 16 EAST 16“ STREET 1894 i, iM All rights reservedErrata Page 38, line 10 from top, for ij =ji, read ij ^ji ., 52, ., 13 „ „ „ T, „ fm „ 56, „ 11 „ bottom, for Wpt =\JVp?<>, read, UVp±‘= Wp±ce ., „ „ 10 „ „ „ tpl=Lpe<>, „ Lp±l=Lpic9 ,, „ „ 1 „ p'6, ., Piee 98, „ 2 „ „ „ Vi, „ Vil „ 190, 1, col. 2, for ternions, read, QuaternionsCONTENTS PART I SUBTRACTION AND ADDITION OF VECTORS CHAPTER I FIRST PRINCIPLES OF VECTORS Section 1 The Nature of a Vector ABT. PAG* 1°. Definition of a vector..............................1 2°. A vector implies an operation of transference. . . 1 3°. Actual and null vectors.............................2 4°. A vector implicitly involves three numbers . . . . 2 5°. Distinction between a vector and a radius vector . 2 6°. Opposite, coinitial, successive, coplanar, and diplanar vectors 2 Section 2 Equality and Inequality of Vectors 7°. Definition of equal vectors.........................2 8°. Example of equal vectors............................3 9°. All equal vectors are denoted by the same symbol . . . 3 Section 3 Subtraction and Addition of Two Vectors 10°. Definition of the difference of any two vectors ... 3 11°. „ „ „ sum of two successive vectors . . 4 12°. „ „ „ „ „ any two vectors ... .4vi OUTLINES OF QUATERNIONS ABT. FA GB 13°. a = —( — a); — ( + a) = + ( — a) = — a . . . . 4 14°. Directions can be assigned to the sides of a plane triangle such that the sum of two sides shall be equal to the third 4 16°. Sum and difference of two coinitial sides of a parallelogram 5 Section 4 Addition amd Subtraction of Yectort in general. 16°. Definition of the sum of any number of vectors. . 5 17°. The subtraction and addition of vectors are associative and commutative operations................................5 Section 5 Coefficients cf Vectors 18°. aro ± ya = (x ± y)a ; x(ya) = (xy)a = xya .... 6 19°. The quotient of two parallel vectors is a number . . . 6 20°. Parallel vectors have the same ratio as their lengths . . 7 21°. ®(o ± /3) = xa ± ar0....................................7 Section 6 Scalars, Unit· Vectors, and Tensors 22°. Definition of a scalar..................................8 23°. „ „ „ unit-vector and a tensor..................8 24°. Distinction between a scalar and a tensor .... 8 CHAPTER II ON points and vectobs in a given plane Section 1 Linear Equations connecting Two Vectors 25°. If pa + qfr = o, and p and q have real and actual values. then a || 0......................... . . . 10 26°. If pa + q0 = o, and o and 0 are oblique ; thenj? = o, £ = o 10 Section 2 Linear Equations connecting Three Vectors 27°. If la + m0 + ny = o, o, 0, y are coplanar .11 28°. If la + m0 + ny « o, and a, 0, y terminate in a straight line ; then Z + m + n = o..............................12CONTENTS vii ABT. PAG· 29°. Anharmonic and harmonic section of vectors .13 30°. (a) Investigation of the equation la + m/3 + ny ■» o, when l + m + n ? o.....................................14 (b) The equation of the six segments..................15 (c) On the sign of geometric figures..................15 31°. Investigation of the equation a + 0 + 7 = o 16 CHAPTER III ILLUSTRATIONS IN COPLANAR VECTORS 32°. Illustrations.........................................19 CHAPTER IV ON POINTS AND VECTORS IN SPACE Section 1 The Mean Point 33°. Definitions of the simple and complex mean points . . 25 34°. The position of the mean point depends absolutely upon the configuration of the system.....................25 35°. The sum of the vectors drawn from the mean point to all the points of the system is zero....................26 36°. The projection of the mean point is the mean point of the projected system............................. . 26 Section 2 Linear Equations connecting Four Noncoplanar Vectors 37°. Vector-diagonal of a parallelopiped...................27 38°. If ha + l& + my + nb = o, and a, 0, 7, S terminate in a plane; then h + l + m + n = o.....................28 33°. Equation of a plane in terms of its intercepts on the Cartesian axes of coordinates.......................297111 OUTLINES OF QUATERNIONS PAET II DIVISION AND MULTIPLICATION OF TWO VECTORS CHAPTER I PIEST PRINCIPLES OF QUATERNIONS Section l Definitions ART. PAGB 1°. (a) Definition of a quaternion..............30 (*> „a"1 . . 30 (c) £ = /3a-' ; j8a = . 30 a (d) If i = -, a a ”1 qa = j8 ; if q' — &a, q'a “1 = $ . . . 30 (e) Definition of the angle of a quaternion . . 31 (/) „ „ plane of a quaternion . . 31 GO „ „ coplanar quaternions . . 31 W „ „ diplanar quaternions . . 31 5 7 5 (0 - ± : - a a * \ &C a . 31 (j) Miscellaneous definitions...........................81 Section 2 The Nature of a Quaternion 2°. (a) Nature of the symbol-...............................31 a (&) „ „ „ symbol j8a..............................82 3°. The operation of tension............................,33 4°. „ „ „ version...................................33 (a) The nature of a versor.............................33 (b) Positive and negative rotation.....................34 (c) Symbolic expression of a versor .34 (d) A versor implicitly involves three numbers . . . 36 6°. A quaternion implicitly involves four numbers .36CONTENTS IX CHAPTER II THE PROPERTIES OF A SYSTEM OF THREE MUTUALLY RECTANGULAR UN IT-YECTORS, i, j, k AST. FAQ· 6°. Definition of the function of a unit-vector, in the first power, as a versor...............................................37 7°. ij = k; ij fji;(- h) = - jk.................................37 8°. i (J ± h) - ij ± ih.........................................38 9®. iH _ = k..........................................38 j 1 10°. ijh = i* = p = & = - 1; ^ %*; . . 39 11°. Real geometric signification of V — 1 . . . .40 12°. i = - i.......................................................41 % 13°. i — H-2®. Graphic representation of a vector and its a Ta reciprocal................................................41 14°. »i = ii.......................................................42 % i CHAPTER m THE VARIOUS FORMS OF A QUATERNION Section 1 A Quaternion as the Product of a Tensor and a Versor 43 44 15°. q = TqJJq = 16°. U (Vq) = Uq . Section 2 A Quaternion as the Sum of a Scalar and a Vector 17°. q = Sq + Vq...............................................44 18°. Recapitulation of the formulae of 17°.....................48 19°. On the symbol, cos 0 + c sin 0............................49 20°. The sum of a scalar and a vector is a quaternion . . . 49 21°. Vector multiplication is not commutative . . .50 22°. Sa0 = ¿(a/3 + 0a); Va0 = |(a0 - 0a).......................50 23°. If Vq - o, the constituent vectors are parallel .50 24°. The square of a vector is a negative scalar . . 51X OUTLINES OF QUATERNIONS ART. PAG! 25°. If S^ = o, the constituent vectors are at right angles . . 51 26°. Vaj8 represents the vector-area of the parallelogram whose coinitial sides are a and b.............................51 Section 3 A Quaternion as the Power of a Vector *¿7°. Definition of the function of a unit-vector, in any power, as a versor............................................. 62 28°. The versors of the four quaternions of 17° as the powers of unit-vectors..............................................52 29°. f-l =1; « — I £ _ «*; - £ (- t)<........................53 €* 30°. The algebraic laws of indices apply to versors and quater- nions ............................................................54 31°. Geometric interpretation of Moivre’s theorem . . .55 32°. ^ = ±1; ,« + 1 = ± n..........................................56 33°. The power of a vector is a quaternion. The amplitude of a quaternion................................................56 34°. Every quaternion may be represented by the power of a vector...................................................57 Section 4 Quadrinomial Form of a Quaternion 35°. q = w + xi + yj + zk...................................57 36°. (a) Second proof that vector multiplication is not commu- tative ................................................58 (V) Vector multiplication is distributive (, . • Product of two right quaternions at right angles to each other is a third right quaternion at right angles to both. 85 85 Section 4 Circular Vector-Arcs ^8°. Definition of equal vector-arcs.............................86 ^9°. Addition of vector-arcs is equivalent to multiplication of quaternions.............................................87 70°. Addition of diplanar vector-arcs is non-commutative . . 87 71°. Geometric meaning of the inequality, qq' q'q . . . 88 72°. Addition of coplanar vector-arcs is commutative . . . 88 78°. Geometric meaning of the equation, Kq'q = KqKq'. . . 89 74°. Composition of two right rotations..........................89 75°. Conical rotation, and the operator q ( )£-* . . . 90 CHAPTER IX FOBÜULÆ Various Formulae........................93CONTENTS Xlll CHAPTER X INTERPRETATION OP QUATERNION EXPRESSIONS ART. 93° Geometric meaning of Sa&y = o................. 94°. „ „ „ SajS7, when Safry J o 95°. If Saj37<0, Sa07 = — ................. 96°. Geometric meaning of Vafiy, when Saj37 = o . 97° 1 109°' | Pinions °I various loci................... CHAPTER XI DIFFERENTIATION OF VECTORS AND QUATERNIONS Section 1 General Principles 110°. Definition of the differential of any variable 111°. Differential of the square of a scalar . 112°. „ „ an area........................... Section 2 Differential of a Vector 113°. General formula for the differential of a vector 114°. Differential of p- — 2 Section 3 Differential of a Quaternion 115°. General formula for the differential of a quaternion 116°. Differential of Q = q2.......................... Section 4 Miscellaneous Examples 117°-129°. qr = qdr + dq . r, «fee. «fee. PAG* 99 99 100 101 102 . Ill . 113 . 114 . 115 . 116 . 117 . 117 . . 118XIV OUTLINES OF QUATERNIONS CHAPTER XII EQUATIONS Section 1 Scalar Equations art. page 130°. A vector cannot be determined by two scalar equations . 124 131°. „ „ can always be determined by three scalar equations 124 132°. „ „ cannot be eliminated by fewer than four scalar equations .......... 126 Section 2 Linear Vector Equations 133®. Definition of the general form of a linear vector equation . 126 134°. Conjugate and self-conjugate functions..................127 135°. Properties of

«', p = yr-'F, y = sT1/ . . . . (1) Substituting these values of a, /?, y, successively, in the given equation, we get lx~la! -f m/3 4- wy = ol la + my~ip{ 4- wy = o j-...............(2) la + mfi + nz~li = oj But a', ft y; a, ¡3', y ; a, /?, y’, are coinitial vectors terminating, respectively, in the straight lines BC, CA, AB. Therefore, 28°, lx~] 4- m 4- w = o | l 4- my~l 4- n = o L. I 4- w 4- n%~1 = o JLINEAR EQUATIONS CONNECTING THREE VECTORS 15 Hence, get -l ; y , „ ; z = . . (3) m + n n 4- L l -f- m Substituting these values of sc, y, 3, in equation (1), we < ~la . ff _ - raff. „/ _ - ny a\ ’ n + l’ ^ l + m ' m -)- n Substituting the same values of sc, y, 3, in the equations of (2), , _m/3 -f ny t ot___ny -{■ la t ,___la -f- m/3 ' _ n + l ’ 7 a' = · ; F = m 4- n l -r m ■ (5) Equations (4) and (5) give the sought values of the vectors of the points A', B', 0'. Comparing the equations of (5) with the equations of (3), 28°, it is evident that BC/ l^. CA' m. AB' n fiv C A ” m; AB n ’ B' C l * ' ' ‘ W (b). If we multiply together the three ratios, equations (6), we get the equation of the Six Segments, AB' BC' CA' _ . B'C C'A A B (7) as the condition of concurrence of the lines A A', BB', CC. We have, also (6), Therefore, l : m = BC' : C'A = OBC :OCA,) m : n = CA' : A'B = OCA : OAB, -. nil = AB' : B'C = OAB : OBC. I : m : n = OBC : OCA : OAB • · (8) where OBC, &c., are the areas of the respective triangles. (c). In such equations as those of (8), attention must be paid to the signs of figures, plane and solid. Any plane figure is positive or negative according as the rotation of a particle round its periphery, as seen from a given aspect of the plane, is right-handed or left-handed. Thus, for any triangle ABC, we have ABC = BCA = CAB = - BAC = - ACB = - CBA ; and as for a line we have AB + BA = o ;16 LINEAR EQUATIONS CONNECTING THREE VECTORS so, for an area, we have ABC + CBA = o. The case of solids is strictly analogous ; the sign of the tetrahedron, fig. 6, being positive or negative according as the rotation of a particle round one of its faces is right-handed or left-handed, as seen from the opposite apex. Thus, AiA2A3A4 = — A^.^A.tAg = — A2A1A3A4 = A2A3A1A4 = — A2A3A^Au &c., &c. In this case we have AiA2A3A4 + A2A3A4Aj — o. 31°. If l = m = n, the equation la + m[3 + ny = o becomes a + /J + y = 0 ; and the three vectors can obviously be made the sides of a triangle ABC, fig. 5, by 14°. In this case, BC = a, CA = ft, AB = y. Let A A', BB', be drawn, cutting BC and CA respectively in given ratios; and through their cross D (instead of O, fig. 5) draw a line from C cutting AB in C'. It is required to determine the ratios AD : DA'; BD : DB' · CD : DC'; and AC' : C B. Let the given ratios be BA' : A'C = 1 — m : m ; CB' : B'A = n : 1 - n; so that we have A'C = ma, and CB' = n/3. (1) To determine AD : DA' and BD : DB'. Since BA' : A'C = 1 — m : m, by 28° AA' = my — (1 — m) ft „ = — ma — ¡3. Similarly, BB' = a + nfi. Let AD = pAAf; BD = gBB'. Then, CA + AD = CD = CB + BD; or, ¡3 — pma — pfi = — a + qa + ; (1 - p - qn) P = (— 1 + q + pm) a.LINEAR EQUATIONS CONNECTING THREE VECTORS 17 But, since a and ¡3 are not parallel, the coefficients of both must be zero, 26°. Therefore, 1 — n 1 — m p = -----· q —----------. 1 — mn 1 — mn \ —p — n(l-m). 2 _ m(l - n) 1 — mn 3 1 — mn Therefore, AD _ p ________ \ — n % BD ____ q _____ 1 —- ra ,,, DA' 1 -p~n(l-m)'3 ‘ ‘ U (2) To determine CD : DC' and AC' : C'B. Let CC' = xGD; AC' = yAB = yy = y (- a - p). Then, since CC' = CA + AC', aCD = ft — ya — yp. But, since AD : DA' = (1 — n) : n (1 — m) — __ n (1 — m) ft - (1 — n) 1 — mn Therefore, xn (l —m) ft — xm (1 — n) a = (1 — mn) (p — yp — ya\ {xn (1 — m) — (1 — mn) + y (1 — mn)} p = {xm (1 — n) — y (1 mn)} a. Equating the coefficients to zero, 1 — mn m (\ — n) OJ — ___________· qj —— _V_____/ · m — 2mn + n ’ m — 2mn + w J x _ 1 _ (1 -*»)(!- w) ! _ w (1 - m) m — 2mw + n 3 m — 2mn + n Therefore, CD __ 1 __ m (1 — n) + n (1 — m) # DC' x — 1 (1 — m) (1 — n) 3 AC' __ y _______m (1 — n) C'B 1 — y n (1 — m) Had the ratios been given in the form BA' . A'C = mi i m2 ] CB' i B'A = nj ; n2 \ c18 LINEAR EQUATIONS CONNECTING THREE VECTORS we should have had AD DA' CD DC' (m} 4- m2) m BD _ m, (n] 4- n9) /«v mlnl ’ DB' ra2w2 _ m]n] 4- m9n2 # AC' m2w2 /a\ 9 C'B mlnl Knowing AB, AC, and the ratio BA' : A'C, we obtain A A' by (3) of 28°. BB' and CC' are similarly obtained. Again, knowing AC, AC', and the ratio CD : DC', we obtain AD, and consequently DA'. BD, DB' and CD, DC' are similarly obtained.19 CHAPTER III ILLUSTRATIONS IN COPLANAR VECTORS 32°. 1. The Mean Point of a triangle, M. In the following illustrations, the successive sides of the triangle ABC—BC, CA, AB—will be represented by the vectors a, (3, y respectively, so that a + P + y = o. The tensors of these vectors, or the lengths of the sides of the triangle, will be represented by a, b, c. Let A' and B' be the middle points of BC and CA, and let AA' and BB' cross in M. Produce CM to meet AB in C'. Then, making m = n = ^ in (1) of 31°, we get AM 2 . BM 9 MA' ' MB' and (2) of 31°, CM _ 9 . AC' MC' " * C'B Therefore, the medians of a triangle meet in a point which trisects them. This point is called the Mean Point. AM = f AA' = § ...........(U 2. The Incentre, I, and the (b) Excentre, Eb. I is the cross of AA' and BB', the bisectors of the angles CAB and ABC respectively. Produce Cl to meet AB in C'. Then, proceeding by the method of 31°, we get AI _b + c ' BI _c + a. Cl _a + b. AC'_b m IA' a ' IB'' b ’ 1C' c 9 C'B a ‘ ‘ ' The last equation shows that the internal angle-bisectors are concurrent. c 220 ILLUSTRATIONS IN COPLANAR VECTORS By (3) of 28°, BB' = ay CC' = b + c * c + a These three equations are of the form p = x (US ± Ue) . . Jy/t e.g., AA' = ^ (Uy-V/3), aft — ha a + 6 (2) (3) the negative sign showing that A A' is the bisector of the supplement of the angle between ¡3 and y. Were the direction of ft reversed, we would have ÂÂJ = -~rc( U/? + Uy). Equation (3), consequently, is the general expression for an angle-bisector, the tensors of the lines containing the angle being arbitrary. By (3) of 28° and (1) of this article AI by — c/3 a + b + c BI = c(k — ay a + b + c CI== afi — ha a + b + c •(4) From (2) and (4), IB' = -h - . ca-~ ay-.......(5) c + aa-f-6 + c Eb is the cross of the external angle-bisectors at C and A respectively. Let CE produced meet AB' in C" ; AE meet BC in A" ; and let the external angle-bisector at B meet CA in B". Then, a — b C^A = a — 6 a — b (7) Hence, Therefore, 29°, (BC'AC") = - 1 ; or, the sides of a triangle are cut harmonically by the internal and external bisectors of the opposite angles. We also have C"A' + A"C' __ e + a ,gv C"B' b + c ’ B'C' b - c ‘ * * # U Therefore, the points A', B', C" and C', B', A" are respectively collinear.ILLUSTRATIONS IN COPLANAR VECTORS 21 Again, A"C' _c-a, C b-c’ . . (9) or, the crosses of the external angle-bisectors with the opposite sides are collinear. Let BE be drawn, and we have BE = ca — ay c + a — b ca c + a — b (Ua- Uy). • (10) Therefore, BE bisects the angle ABC ; BB' and BE coincide ; and a line drawn from any corner of a triangle to the excentre of the opposite side passes through the incentre. By subtraction of BB' from BE, we have B'E = b ca — ay c + a c + a — b% Therefore, (4) and (5), (BIB'E) = - 1 ; or, the internal angle-bisector is cut harmonically by centres of the in- and excircles. 3. The Orthocentre, P. By the methods explained in 31°, ——- y tan B — /3 tan C tan A + tan B -f- tan C* 4. The Circumcentre, Q. By 31°, —— (tan C -J- tan A) y — (tan A 4- tan B) ft ^ 2 (tan A + tan B + tan C) the 5. The Midcentre, N. Let M1? M2, M3 be the middle points of the sides BC, CA, AB of the triangle ABC. Then the circumcentre, N, of the triangle M1M3M2 is the mid centre of the triangle ABC. AN = AMj + MjN ; which becomes, by 1 and 4, ___y — /3 (tan A + tan B) ¡3 — (tan C + tan A) y ” ’’’ 2 4 (tan A + tan B + tan C) (tan A + 2 tan B + tan C) y—(tan A + tan B 4- 2 tan C)/? ” 4 2 tans 1 j (tan C + tan A) y — (tan A + tan B) ft ” 2 ( 2 2 tans y tan B — tan C) 2 tans i „ =| (AQ + AP), 4 and 3.22 ILLUSTRATIONS IN COPLANAR VECTORS Hence, AQ + AP — 2 ANT = o ; but 1 + 1 — 2 = o. Therefore, 28°, the three coinitial vectors AP, AN, AQ, terminate in a straight line which is bisected in N) or, the orthocentre, the midcentre, and the circumcentre are collinear, and N bisects PQ. The mean point, M, also lies upon the line PQ. For « — ITn (tan ^ + tan A) y — (tan A ■+■ tan B) /3 2 AQ + AP = ^Sn^ y tan B — ft tan C t ____ _____ ________________S tans 5 2 AQ + AP = y — ƒ? = 3 AM, 1 ; hence, 2 AQ + AP — 3 AM = o ; but 2 + 1 — 3 = o ; therefore, the three coinitial vectors AQ, AP, AM, terminate in a straight line which is trisected in M. Therefore, the orthocentre, the midcentre, the mean point, and the circumcentre are collinear ; and the line upon which they lie is bisected in N, and trisected in M. Hence, (PNMQ) = - 1. 6. Let I' be the point in which a line drawn from A through I, the incentre of ABC, cuts the line PQ. Then PT' = 2 cos A. IQ If A -= 60°, PI' = I'Q ; and I' is the midcentre of the triangle. 7. To find the vector-expressions for the isogonal and the isotomic of a given line. Let 8 be a vector drawn from the corner B of a triangle ABC, cutting CA in D so that ^ ; and let ABD=<£. Thpn IP _ AD __ csin) cr_ sin ap Let 8', the isogonal of 8, cut CA in D'. Then, AD' c sin (B — <£)_____cV andILLUSTRATIONS IN COPLANAR VECTORS 23 Therefore, 8 = ^^; S' = C-^ p + r ^ If 8" be the isotomic of S, -a2 a?py c*r ay (1) g// — m-py............/2\ p + r Let p : r = cn : a\ Then, 3 — c a ~~ g Y . 3/ _ ft «■ * ““ cn + an 3 n ” c*-* »-2a — cn~2y 4- au Let p : r = cn~2 : an_2c Then, 8" « = dr“wa — c" (3) c"”2 + a""2 Therefore, S'w = S"n_2 ,* or, the isotomic of S„_2 is the isogonal of Sn ; a theorem which suggests a simple geometric construction for the successive centres of gravity of weights placed at the corners of the triangle proportional to the 0, 1, 2 . . . nth powers of the opposite sides. 8. Suppose it be desired to obtain symmetric expressions for the vectors of the mean point, incentre, &c., &c., instead of the unsymmetric expressions obtained in the preceding illus- trations. Let O be any point in the plane, and let OA = a, OB = /?, OC = y. Then the vector of the orthocentre, 3, may be written: Tp V tan B — /?' tan C # ^= 2 tans , where a', ¡3', y' are the vectors along the sides of ABO. But, a =■ y — ¡3 ) f3' = a — y ; y' = j3 — a. Therefore, OP = a + AP = a + ~-a)-n-| ¿s ~ y) tan C __ a tan A + ¡3 tan B + y tan C ” 2 tans Similarly, for the circumcentre, —- (tan B 4- tan C)a+(tan C+tan A)/?+(tan A + tan B)y 0Q= 2 2 tans and for the midcentre, --- (2 tans + tan A)a + (2 tans + tan B)j8 + (2 tans + tan C)y ON =—------------—“— ---------------------------* · (1) (2) 4 2 tans • · (3)24 ILLUSTRATIONS IN COPLANAR VECTORS If Gn be the centre of gravity of weights placed at the corners of the triangle proportional to the n%]x power of the opposite sides, ™ _ ana + bnp + cny n an + bn + cn (4) If n = o, OG0 = a — the vector of the mean point. o (5) If n = 1, OG, — ®aJr ^ the vector of the incentre ... (6) 1 a -f 6 + c If n == 2, OGo= cy, the vector of the symmedian point.. (7) 1 a1 -f b2 + c225 CHAPTER IY ON POINTS AND VECTORS IN SPACE Section 1 On the Mean Point 33°. Definition.—If the sum, 2a, of m coinitial vectors, coplanar or non-coplanar, OA] = aj, OA2 == a2 OAm = am, divided by their number, m, ¿/¿e resulting vector, /a = OM = 2 (QA) m is the Simple Mean of those m vectors, and its term, M, is the Mean Point of the system of points, A1? A2 . . . Am. If we are given such a system of points, a1? a2 . . . an, and also a system of scalars, pu p2 . · . pn ; the vector y = OC = P\ «I + P‘1 a, some vector in the plane OAB, fig. 7, and p 4- my = q£, some vector in the plane containing and y, i.e., some plane different from OAB, OBC, and OCA. Hence the expression, ha 4- 4- my, represents some fourth vector, say — nS} whose coefficient, n, is J zero. Fig. 7. unless all three coefficients vanish. For28 LINEAR EQUATIONS CONNECTING POUR VECTORS or, In symbols, ha + ip -f my -f n8 · 8 — —— « + —— ¡3 + - n n = o; - m If we take OA'= —- a, OB' = —- /?, OC'= ■—— y, and n * w ^ n n complete the parallelopiped OA,,B,/C", we determine a point D, such that, OD = OC"+ C77D = OC"+ OC'= OA'+ OB'+ OC' — A , — £o . — n = ---- a f·----/5 +----- n n n > = 8. Hence, since a, ¡3, y may be any actual vectors, and since A, £, m may have any values whatever, the sum of the three coinitial edges of a parallelopiped is the internal coinitial diagonal. 38°. If Aa + Ip + my -f w8 = o, what must be the re- lation between the scalars in order that the point D may be situated in the fourth given plane ABC ; or, in other words, what is the condition of coplanarity of the four points, B, C, D ? If D lies in the plane ABC, the vectors DA, DB, DC are coplanar, and are consequently, 27°, connected together by an equation of the form jt?DA + #DB + rDC = o ; or, p (a — 8) + q (/3 — 8) + r (y — 8) = o ; or, pa + qfi + ry — (^ + ^ + r)8 = o. But Aa + 1/3 + my + nh = o. Hence, eliminating 8 from the last two equations, {A (p + q + r) + np} a + {l (p + q + r) + nq) ¡3 + {m (p + q + r) + nr) y = o. Now, if the coefficients have an actual value, a, (3, y are coplanar. But, by hypothesis, a, ¡3, y are not coplanar. Therefore the coefficients have not an actual value, and must be equated to zero. Therefore, h =--------L— ; l — ----------— ; m — —----------; p + q + r p + q + r p + q + TLINEAR EQUATIONS CONNECTING FOUR VECTORS 29 and liArl+m+n— — n f—£--------j--i-----}- — ---)+w = o; xp + q + r p + q + r p + q + rj the required condition of coplanarity of the four points, 39°. The equation just deduced may be written, — n — n — n But, by construction, OA7 = OA7 . OA ' l = OB' . m ^OC' -n OB ’ — n 00 * Therefore, OA7 OB7 00' , OA + OB + ÔC ’ the equation of a plane in terms of the intercepts it makes on the Cartesian axes of coordinates OA, OB, OC.30 PART II DIVISION AND MULTIPLICATION OF TWO VECTORS CHAPTER I FIRST PRINCIPLES OF QUATERNIONS Section 1 Definitions 1°. (a). A Quaternion is an operator which turns any one vector into another (Clifford). (b). The Reciprocal of any vector, a, which is written, as in Algebra, - or a-1, is another vector whose unit-vector is the a opposite of the unit-vector, and whose tensor is the reciprocal of the tensor of the vector a. - Ua <“>· ! = = (d). If a and ¡3 be any two vectors, and if a then, whatever be the nature of q, Ta (1) If qa = = ¡3 a ?' = /2a = 4T; g'a-1 =0a. a-* = fa''= ¡3 ... (2) a 1 thenDEFINITIONS 31 The two vectors will in all cases be supposed to be co- initial, and to be inclined to one another at a Euclidian angle, between zero and tt, unless the contrary be stated. ¡3 is the Multiplier and a the Multiplicand of the product /?a. The multiplier is always written to the left, the multi- plicand to the right; and the symbol /3a is to be read—1 a multiplied by (3/ or, shortly, ‘ (3 into a.’ It follows from (1) and (2) that the quotient and product of two vectors are quaternions. (e) . The Angle of a quaternion, in the form of a quotient, is the angle contained by its constituent vectors. The Angle of a quaternion, in the form of a product, is the supplement of the angle contained by its constituent vectors. (f) . The Plane of a quaternion is the plane containing its constituent vectors. {g). Coplanar Quaternions are those whose planes are coin- cident or parallel. (h) . Diplanar Quaternions are those whose planes are not parallel. (i) . If a, /3, y, &c., be any three vectors, _4_ a 8 ± a 8 . a 8. 8 a 8 ~ P P J ¡3/3 a3 a y y , If qa = ¡3 and q'a = (3, then q' = q. 8' __ 8 ” . / — i = y, „ 3' = 8. y y 5) 8' = 8, „ y' = y. „ 8y = » y' = y> ,, 8' = 8. » }> » 8' = s, „ y' = y. Section 2 The Nature of a Quaternion 2°. (a)- Let OA = a, OB = ¡3, fig. 8, be any two vec- tors in the plane AOB, inclined g to one another at an angle 0. By definition, 1° (d), if ^ = q\ then \ n a j q'a = Qa = [3 ; or qor is such a * A X & a Fig. 8. factor that when it operates upon the divisor, a, it trans- forms it into the dividend, ¡3. Now, since a differs from32 THE NATURE OF A QUATERNION ft, not only in lengthy but in direction,, it is clear that two independent operations, of a totally different nature, are necessary in order to transform a into ft. The one is an operation of tension, the other an operation of torsion, or version ; and the order in which the two operations take place is immaterial. We may make a rotate round the point O until its direction coincides with that of ¡3, and then alter its length until it is equal to that of ¡3; or we may alter its length until it is equal to that of ¡3, and then make it rotate round O until its direction coincides with that of ¡3. Now, a may acquire the direction of j8 either by a rotation round O in the plane AOB, or by conical rotation round a third coinitial vector bisecting the angle 6. To avoid ambiquity it is defined that the rotation from a to ¡3 takes place in the plane of the two vectors, AOB. Further, a may rotate in the plane AOB into the direction of ¡3 through either the angle 0 or the amplitude, 2ir — 6. For the same reason it is defined that rotation from a to ¡3, in the plane OAB, takes place through the angle 0, which, 1° (d), lies between zero and tt. (b). Let OA = a, OB = ft, fig. 9, still represent any two vectors in the plane AOB, inclined to one another at an angle 6 ; and let OA' be the reciprocal of OA, 1° (b), or Then, since Ua~1 is the opposite of Ua, the angle BOA'= 7r — 0. By definition, 1° («d), if ft; or q", or ft a, is such a factor that when it operates upon the reciprocal of the multi- plicand, a-1, it transforms it into the multiplier, ft. As in the previous case, two operations are necessary in order to effect this transformation—one of tension and one of version, the order of which is immaterial. As before, also, it is defined that rotation from a-1 to ft takes place in the plane of the vectors, BOA', and through the angle between them, ir — 0, which lies between zero and tt when 6 lies rr and zero. The vector a-1, then, may be transformed into ft by altering its length from OA' to OA"= OB, and then making the altered vector rotate in the plane BOA', through the angle (7r — 6) into the direction of ft. Such is the nature of the symbols qf or and qn or fta.THE NATURE OF A QUATERNION 83 Both, as factors, imply two operations—one of tension and one of version—which are heterogeneous and absolutely inde- pendent. No mere change of length can in any way affect the direction of a vector; no amount of rotation can alter its length. 3°. The operation of Tension is purely metric, and we need only one number to carry it out—namely, the number (whole or fractional) by which we must multiply the length of one line in order to make it equal to the length of another. Given this number, we can make the length of the one line equal to that of the other, without knowing the absolute length of either of them. 4°· The operation of Version is of a more complex nature, and will be found to involve a knowledge of three numbers. The first point to be explained is the means of giving rotation to a vector. (a). Let _____ OA = Ua, OB = U/?, fig. 10, be any two unit- vectors in the plane of the paper, inclined to each other at any angle 6 ; and let OA' = - Ua, OB'= - Ufl, be the opposites, or recipro- cals, of Ua and U/3, 1° (b). Let OX be a unit-vector perpendicular to the plane of the paper, drawn from the origin, O, towards the reader as he reads the book ; and let OX' be the opposite unit-vector of OX, drawn from the reader through the leaf of the book. Conceive OA and OA' to be two very fine wires so connected at O with twro other very fine wires, OX and OX', that by twisting either OX or OX' about its longest axis, a motion of rotation is communicated at will to either OA or OA'. Motion of rotation would thus be communicated to OA or OA', as the case might be, in exactly the same way as if OA or OA' were the minute-hand, and OX or OX' the key of a clock which it was necessary to set ; the key being applied in the case of OX to the face of the clock, and to the back of the clock in the case of OX'. Thus, if we conceive the unit-vectors to be gifted with the powers of the wires, by means of OX or OX', we can make Ua or — Ua, or any Du THE NATURE OF A QUATERNION coinitial vector in the plane of the paper, rotate into the direction of Uj3, or any other direction in that plane. Generally, rotation is given to any vector lying in any plane by operating on it with a coinitial unit-vector perpen- dicular to that plane. When employed to give rotation to other vectors in planes perpendicular to themselves, unit-vectors are called Yersors (vertere, to turn). Yersors can only operate upon—that is, give rotation to— vectors perpendicular to themselves. (6). When twisting the wire OX about its longer axis, the reader is supposed to be in the position he occupies while reading the book—with his eye at X, looking towards O. When twisting OX' he is supposed to have moved to a position beyond X', facing his former position, with his eye at X', looking towards 0. These two positions bear exactly the same relation to one another as the two positions one successively occupies when locking a door on the outside and on the inside. And as one sees different sides of the door when locking it on the outside and on the inside, so one sees different sides, or aspects, of the plane A'BA, when twisting OX' and when twisting OX. Furthermore, a right-handed twist given to OX at X appears to be a left-handed twist when seen from X' ; just as locking the door on the inside by a right-handed turn of the key would appear to be locking it by a left-handed turn of the key to anyone viewing the operation through a glass door from the outside. In order, then, to estimate the direction of the twist, we must imagine ourselves to be in the position of the person giving the twist. Right-handed rotation—the rotation of the hands of a clock when looked at to take the time—will be considered as positive ; left-handed, or anti-clockwise rotation as negative, in this book. (c). Ua may be made to rotate through the angle 6 into the direction of U/3 by giving either a negative twist to OX, or a positive twist to OX'. To avoid ambiguity, it is defined that OX', which turns I7a into the direction of U/3 by positive rotation, is the versor by which this operation is to be carried out. ___ Similarly, OX, which turns — Ua into the direction of U/3 by positive rotation, through the angle tt — 0, is defined to be the versor by which this operation is to be carried out.THE NATURE OF A QUATERNION 35 Generally, rotation is communicated to any vector by that versor which turns it by positive rotation into the required direction. We must now place a limitation to the turning powers of versors. Every versor possesses the power of turning any vector perpendicular to itself, by positive rotation, through a certain angle; but it cannot produce^rotation through any other angle, greater or less. Thus, if OX possesses the power of turning OB through a definite angle 0, it can turn any other vectors, OP, OQ, &c., in the plane AOB, through the angle 0 ; but it can turn them through the angle 0 only. In consequence of this limitation, we must modify our terminology. There are two different, although coincident, versors along OX : (a) the versor which turns U/3 through the angle 0 into the direction of Ua; (6) the versor which turns — Ua through the angle tt — 6 into the direction of U/3. There are also two different, although coincident, versors along OX': (c) the versor that turns Ua through the angle 0 into the direction of U/3; (d) the versor that turns — U/3 through the angle tt — 0 into the direction of Ua. For the moment we will designate these four different versors by the following symbols : (a) by_GX# (b) by OX,-,; (c) by OX'„; (d) by OX',-* Since Ua and I!¡3 are of equal length, when U ¡3 is turned by OX9 into the direction of Ua, it becomes equivalent to, or is transformed into, Ua. In symbols, (1) (2) oxfl. up = Ua. Therefore, 1° (d), OX9 = ^ · · · · · Similarly, OX* _0.Ua~1 = U/3. Therefore, 1° (d), bx„_9 = tJ^ = U/3Ua . . ........... OX'„-9 = UaU/3 . . . . In like manner, (3) (4)36 THE NATURE OF A QUATERNION It follows that the quotient or product of any two unit- vectors is a third coinitial unit-vector perpendicular to their plane. (d). Since an operation of version implies rotation in a certain plane, towards a certain hand, through a certain angle, at least three numbers are required for the complete determina- tion of a versor ; one to represent the magnitude of the angle of rotation, and two to fix the direction of the plane of rota- tion, or of the versor itself. It will be observed that, as in the case of a vector, two numbers are required to determine its direction, and a third to define the amount of motion of translation communicated to a particle in that direction ; so in the case of a versor, two numbers are required to determine its direction, and a third to define the amount of motion of rotation communicated to a vector at right angles to that direction. 5°. Since the operation of Tension depends upon one number, while the operation of Version depends upon three numbers, it is clear that for the complete determination of a quaternion a set of four numbers is required. Hence the name quaternion ( — i, — j, — k, denote six of the geometric square roots of negative unity, or, i = j = k =\/ — 1 = — i = — j = — k . . . (8) In the present Calculus, therefore, the imaginary of Algebra, n/ — 1 admits of a geometrically real interpretation as an indeterminate right versor. It is indeterminate, because its direction is indeterminate. In other words, the equation, i2 -f l=o, has indefinitely many roots, all of which are geometric reals. We have now reached the point which Sir W. H. Hamilton pronounced (‘ Life, &c.,' III., 90) to be ‘ the difficulty in the theory of the geometrical interpretation of Quaternions ’ ; namely, the two meanings of i, as a vector, and as a versor. In the equation, ij = k, j and k are vectors, while i is a versor. In fact we may regard k as the vector generated, when the versor i operates upon the vector j. Further, i is a perfectly definite versor, operating in the plane of j and k, to which it is perpendicular, so that rotation round it from^* to k is positive. In the equation, i = \/ - 1, by analytically determining a value of i independent of j and k, we have abstracted from the conception of i the idea of a plane in which, and conse- quently of a hand towards which, it operates. Equation (8) is a corollary of definition (c), 6°. It asserts that all right versors are equivalent in respect to angle; and it asserts no more. (6). It maybe said that equation (8) violates the definition of equal vectors. This is not so. The definition asserts that unit-vectors with given directions are only equal, as vectors, when those directions are similar. Equation (8) asserts that all unit-vectors in the first power are equal, as versors, in respect to angle.THREE MUTUALLY RECTANGULAR VECTORS 41 (c). Again, it may be said that since i = jk, and also i — k =j = >/ — 1; we must have, >/ — 1 = >/ — 1. \/ — 1 = — 1, which is absurd. We cannot substitute s/ — 1 for i and k in the equation, i because i and k are vectors, and the symbol, n/ — 1, represents them only in their character of indeterminate right versors. It would be a contradiction to write, i = V — 1 . k; for in this case k and i being given, the versor is not the indeterminate s/ —1, but the definite versor j. 12°. By 9° (3), (4), k . ? ; t = — »- J K i k 1 i But £ = 1 : .: therefore, — i = - = i~1 : or, the reci- k j ^ % procal of a unit-vector is its opposite. Hence, i~lj = - ij = - k. In words, j may be turned into the direction of — k, either by operating upon it with — i and turning it through a positive quadrant, or by operating upon it with i and turning it through a negative quadrant; definition (c), 6°. But 4° (c), — i (not i~l) is the versor of the quaternion——, , . J because it operates positively. 13°. By 12° we obviously have, a i = i_ _L = i, tj v a Ta Ua TaK ^ — Ua Ta ’ To exhibit graphically a vector and its reciprocal: through any point C in the diameter A'A of a circle, fig. 13, draw ED perpendicular to A'A ; draw any chord FG through C; and let CE, or CD, be the unit of length. Then, CF . CG = CD. CE = CD2 = CE2=1; and consequently, CF = — CG Therefore, since T . CF = I . CG and IJ^CF = - U . CG; if CF = p, then CG = p~ Fig. 142 THREE MUTUALLY RECTANGULAR VECTORS 14°. Since ii~l = i (— i) = ikj = kji, and also, i~li = (— i) i kji ; we have ii~1 = i~ 1i ; or, a unit-vector and its reciprocal are commutative, this only holds good for unit-vectors of the same name, ii “1 = i ” ; but hi “1 i But43 CHAPTER III THE VARIOUS FORMS OF A QUATERNION Section 1 A Quaternion as the Product of a Tensor and a Versor 15°. A quaternion may be thrown into the form of the product of a tensor and a versor. Let qx = fig. 10. Then, resolving a and ¡3 into their tensors and unit-vectors, we have, TaTTa Ta ITa Ta ■ 01 = T/3U/3 T/3 JJ/3 T/3 = ^ OX*, 4° (c), (1). Ta ITa ¡jr| is called the tensor, and ^ 0r OX*, the versor, of the quaternion In symbols, Ta T».=T“iI=i|;lTi. = ul=wl=ox.; fi T/3 /* U/3 Similarly, if q2 = /3a, we have • (1) = T/3Ta . OX*.* 4° (c), (2). T/3Ta is called the tensor, and UjSUa, or OX^-*, the versor, of the quaternion /3a. In symbols, Tq2 = Tj3a = T/3Ta ; TJq2 = TJ/3a = OXT_, m q2 =/3a = T£a . U£a = Tq2Vq2 \ · W In general, q = TgTJ#.....................(3)44 A QUATERNION AS PRODUCT OF TENSOR AND VERSOR In this equation, Tq is the quotient or •product of the two tensors, according as the quaternion is the quotient or product of the two vectors. In both cases 17<7 is a unit-vector drawn jrom the assumed origin, O, perpendicular to the plane of the quaternion, such that rotation round it, from the divisor to the dividend, or from the multiplier to the multiplicand, is positive. As a quaternion is the product of a tensor and a versor, so, conversely, every product of a tensor and a versor is a qua- ternion. For if (Tq . Ug) operate upon any vector at right angles to itself, it will obviously alter both its length and direction, that is, transform it into another vector, 1° (a). 16°. The versor of a versor is the versor itself : U (U/3) = 17/?; 17 (Uq) = 17?. It may be observed that the value of a composite vector expression is not altered by altering the order of the numerical quantities it may contain : ftbyfdh = bftfyhd = bdfftyl. Section 2 A Quaternion as the sum of a Scalar and a Vector 17°. We are now in a position to investigate, an expression for a quaternion explicitly involving its angle. (a). Let OA(I7«) and OB (17/?), fig. 14, be the unit-vectors of any two given vectors, a, ft, inclined to one another at an angle 0, different from zero, and tt ; and draw OC, of unit length, perpendicular to OB in the plane AOB. Let fall per- pendiculars from A on OB, OC, cutting them in A', A" respectively ; and draw from O, towards us, the unit-vector OX, perpendicular to the plane AOB. Then, 4° (e), (1), Ua OA OA' A'A OA' OA" OXi “ U/3 - OB - OB + OB _ OB + OB ‘ OA' Now, is a scalar—the ratio of the lengths of OA' andA QUATERNION AS SUM OF A SCALAR AND A VECTOR 45 OB. The length of OA' = OA cos 0 = cos 0, since OA = 1 : the length of OB is unity. Therefore, cos 0. OA"= TOA". UOA". OB But TOA" = OA sin 0 = sin 0, and UOA" - OC. ___ ___ OA" OC Therefore, OA" = sin 0 . OC, and = sin 0 . But OC, OB are rectangular unit-vectors. Their quotient, therefore, is a versor in the first power in the direction of OX, which we will call c. In symbols, OC , OA" . , OB =£’and OB = esmft Therefore, TJa OA' OA" OX# — U/?— OB OB ^ 0 "h € sm 0 . . (1) or, U^ = cos 0 + c sin 0 . (2) Equation (2) gives us an expression for the versor of the quaternion, -j*, in terms of its angle, which will enable us for the future to discard the temporary symbol OX*. If 0 == o, we have = 1, or TJfi = Ua. If $ =ir, = - 1, or U/3 = - Ua. If 0 = = e which is merely the equation, with different symbols. Ta Multiplying equation (2) by we get, 15°,46 A QUATERNION AS SUM OF A SCALAR AND A VECTOR Equation (3) gives us an expression for the quater- Tct nion ^ as the product of its tensor, and its versor, (cos 0 + € sin 0), which turns /? into the direction of a by positive rotation through the angle 0. Equation (4) gives us the expression for the quaternion Ta as the sum of a scalar and a vector quantity. — cos 0 is called Ta a the Scalar ; — sin 0 . c, the Vector, of the quaternion -. In symbols, n fl Ta /i tt a Ta · f\ /k\ s/3=t/?cos*; · · · (0) = S“ + V^ P P P . (6) Ta T0 sin 6, (5), is called the Tensor of the Vector; e, the Versor of the Vector, of the quaternion In symbols, TY“ = J“sin*;UY“=£ .... (7) For shortness’ sake the versor of the vector is often called the Axis of a quaternion. It is a unit-vector coincident with the versor of the quaternion, Uq ; but the two are generally unequal versors, because although they produce rotation in the same direction they operate in general through different angles. UVg, or Ax . q, is always a quadrantal versor : Uq is a quadrantal versor only when L q = TT 2’ The term, cos 0 (2), is called the Scalar of the Versor; sin 0 . e, the Vector of the Versor; sin 6, the Tensor of the Vector of the Versor of the quaternion a P' In symbols, SU!| = cos0; YU= e sin 0; TVU~ = sin 0 . . (8)A QUATERNION AS SUM OE A SCALAR AND A VECTOR 47 ^7 A (b). Let OA, OB, fig. 15, be the same unit-vectors, at the same angle, as in (a). Produce BO until OB' = OB, and draw OC, of unit length, perpendicular to OB in the plane AOB. Let fall per- pendiculars from A on OB, OC, cutting them in A', A" respec- tively ; and draw from 0, from us, the unit-vector OX', per- pendicular to the plane AOB. Then, 4° (c), (4), OX'_0 = UaU^ = Z^ = OA OB' Fig. 15. OA' OA^' OB' + OB' ’ OP cos 0 4- sin 0 ; OB (9) • (10) or, TJa¡3 == — cos 0 — € sin 0 . . Multiplying across by TaT/3, a/3 = TaT/3 (— cos 0 — € sin 0) ) „ = — TaT/3 cos 0 — TaT/3 sin 0 . c J In this case ( — cos 0 — € sin 0), or — (cos 0 + c sin 0), is the value of the versor OX'ff _ #, which turns ~ 1 through the angle (7r — 0) positively, into the direction of a. Equations (10) are expressed in terms of 0, the angle of the quaternion To express them in terms of (7r — 0), the P angle of a/3, we have merely to write, a/3 = TaT/3 {cos (7r — 0) — € sin (jr — 0)} .... (11) Sa/3 = - TaT/3 cos 0; Va/3 = — TaT/3 sin 0 . c . . (12) a/3 = Sa/3 4- Ya/3..................................(13) TVa/3= TaT/3 sin 0 ; UVa/3 = - €.......................(14) SUa/3 = — cos 0; VTJa/3 = — c sin 0; TVTJa/3 = sin 0 . . (15) (c). It may be similarly shown, by slight modifications of fig. 14 and 15, that ^ (cos 0 — € sin 0) . ..·.... (16) a la T/3 , T p . a = Ti cos 6 ~ fT sin 6 ’ (17) /3a = T/3Ta (— cos 0 + € sin 0)..........(18) ,, = — T/3Ta COS 0 + T/3Ta sin 0 . e . . . (19)48 A QUATERNION AS SUM OF A SCALAR AND A VECTOR 18°· Recapitulation of the formula of 17°. a q = t q = Φ T q = —. q Ύβ Tq = ΤαΤβ. JJq = cos 0 + € sin Θ. Q Τα a S? = Ίβ cos Θ. U# == — cos 0 — € sin 0. Sq = — ΤαΤβ cos 0. Vq = Sin θ ' e- Yq = — ΤαΤβ sin 0 . e. TVq = ^ Sin θ· TY# = ΤαΤβ sin 0. U Yq = €. SJJq = cos 0. YU# = € sin 0. TYU# = sin 0. U Yq = - €. SU# = — cos 0. YU# = — € sin 0. TYU# = sin 0. ? = -· a q = βα. T? = ^. H Τα JJq = cos 0 — € sin 0. T q — ΤβΤα. U# = — cos 0 4- c sin 0. S? = cos Θ. * Τα S# = — ΤβΤα COS 0. Vq = - ^ sin Θ . €. Y# = ΤβΤα sin 0 . €. TV? = ^ sin Θ. Τα TYq = ΤβΤα sin 0. UY# = - C. SU# = cos 0. YU# = — € sin 0. TYU# = sin 0. UY# = €. SU# = — cos 0. YU# = e sin 0. TYU# = sin 0. -β = ^ (cos θ + £ sin 0)......................(A) αβ = ΤαΤβ (— COS 0 — € sin 0) ) „ = ΤαΤβ {cos (7r — 0) — € sin (π — 0)} ) * ' ' ' 'A QUATERNION AS SUM OF A SCALAR AND A VECTOR 49 — = ^ (cos $ — c sin 0).........................(C) a la /3a = T/?Ta (— cos 0 + c sin 0) ) „ = T/?Ta {cos (7r — 0) + € sin (7r — 0)} \ * * * ' ' The first equations of (B) and (D) express the versors of a6 and /3a in terms of 0, the angle of the quotients ^ and ^ ; P a the second equations express them in terms of (7r — 0), the angle of the products themselves, 1° (e). 19°. We have now reached the remarkable result that a versor may be expressed as the sum of a Number and a Vector. It may be objected that a number and a line are hetero- geneous quantities, and can no more be added together than a pint and a mile. A difficulty exactly analogous is presented in the Calculus of Finite Differences by the symbol 1 + A; ‘ where the number 1 appears to be added to the characteristic A, which is not a number at all, but the sign of the operation of taking a finite difference.’ 1 + A is, in fact,1 the symbol of an operator which changes any given function of x to the same function of x + 1 ’; and we learn, in that Calculus, what the proposed sum 1 + A is by learning what it does (Hamilton’s ‘ Lectures, &c.,’ p. 388). In a similar way (cos 0 4- c sin 6) is the symbol of an operator, a Versor, which has the power of turning any line upon which it operates in a plane perpen- dicular to itself, through an angle 6, positively. This symbol represents a unit-vector, as may be proved by taking its tensor, _______ ___ T (cos 0 + c sin 0) = T OA' + OA" OB TTJ« TU/3 Its square is not negative unity, as are the squares of i,j, k, because it is not in general a right versor. In the special case when 0 = \ 7r, how- ever, its square is negative unity. 20°. It has been shown in 17° that a quaternion is the sum of a scalar and a vector. It remains to prove the converse : the sum of a scalar and a vector T* is a quaternion. FlG·16· Let w and p be any scalar and any vector which50 A QUATERNION AS SUM OF A SCALAR AND A VECTOR it is proposed to add together. Let OA (a), fig. 16, be any assumed vector in the plane of the paper, to which plane p is supposed to be perpendicular at O. Operat- ing upon a with w, we obtain a vector, OB^ equal to wa. Operating upon a with p, we obtain a new vector, OC, equal to pa at right angles to a. Completing the rectangle OBDC, and drawing the diagonal OD, we obtain OD = wa + pa ; hence, w ·+ p = wa pa Wa -j- pa a a a OD , . - = a quaternion. OA 21°. It is evident from equations (10), (18), and (19) of 17° that the commutative law of multiplication does not hold good for vectors : fia ^ aft. 22°. From equations B and D, 18°, we obtain, by adding and subtracting, Sa/3 = S/Ja = I (a/? + fia)..............(1) Ya/3 = - Y/3a = A (a/3 - /3a) . . . . (2) 23°. If 0 = o or 7r, the vector parts of (A), (B), (C), (D) vanish, and the quaternions degrade to scalars. If q be a product, and 0 = o, we have /3a = — TfiTa — — ba — sl negative scalar; and for 0 — 7r, f3a = T(3Ta = 5a = a positive scalar. Hence the product of two parallel vectors is a scalar which is negative when the vectors have similar, and positive when they have contrary, directions. If q be a quotient, and 0 = o, we have - = - ; and for a a 9 = 7r, ^ ; results which confirm those of 19°, Part I. a a Conversely, if = o, the constituent vectors, a and /?, are parallel, or a =03/3. For if Ta and T/3 have actual and real values, the vector of a quaternion can only vanish when 0 = o or 7r. It follows that if a quaternion degrade into a scalar, or if q = + 03, then T<7 = T ( ± x) = 03; Yq = U ( + as) = + 1. In words, the versor of a scalar, regarded as the limit ofA QUATERNION AS SUM OF A SCALAR AND A VECTOR 51 a quaternion, is equal to positive or negative unity, according as the scalar itself is positive or negative (Hamilton). 24°. For the future (Ta)2 will generally be written T2a ; T(a2) will be written Ta2 ; ('Vq)2 will be written V2q; V(q2) will be written Vq2, &c., &c. If 6 = o and Ta = Tfi, then ¡3 = a and /3a = /32 = (T/5TJ/3)2 = T/3U/3T/3U/3 = T2/3U2/? „ = T2/3 (- 1) = - T2/3 = - b2 ; or, the square of a vector is a negative scalar, being the square of its tensor combined with the minus sign. It follows that the square of a vector must be considered as having no direction in space. Since ft2 is a scalar, V/32 = o ; S/32 = /32= - b2. 25°· If 0 = the scalar parts of (A), (B), (C), (D) vanish, and the quaternions degrade to vectors. For example, /3a becomes T/3T« . e, a vector which Sir W. R. Hamilton called the Index of the Right Quaternion fia, or IV/3a. Con- versely, if q be any quaternion, and if S# = o, then the constituent vectors are at right angles to each other. For, provided that Ta and T/3 have real and actual values, the scalar of a quaternion can only vanish when (9 = 7T 2’ 26°· S/3a, or — T/3Ta cos 6, is (neglecting signs) the area of a parallelogram whose sides are equal to OB and OA, and whose angle is the complement of BOA. TV/3a, or Tf3Ta sin 0, is the area of the parallelogram BOA. V/3a, or TjoTa sin 0 . c, is the Vector-Area of the same parallelogram, Tfia sin 0 representing its numerical value, and c indicating its sign, which is positive or negative according as the rotation of a particle round the periphery is positive or negative as seen from the term of e. Thus, Yfia = ha sin 0 . e = OABC ; Va/3 = db sin 0 (— c) = — ab sin 0 . e == OCBA.52 A QUATERNION AS THE POWER OF A VECTOR Section 3 A Quaternion as the Power of a Vector 27°. Since i, or il, is a versor which turns any vector on which it operates, say j, through one right angle positively, it is natural to define i2 to be a versor which turns the same vector through two right angles positively; P through three right angles positively . . . im through m right angles positively, m being a positive integer. In the same way, since i*1 is a versor which turns the vector upon which it operates through one right angle negatively, we define i~2 to be a versor which turns the same vector through two right angles negatively, through three right angles negatively . . . i~ through m right angles negatively, m being still a positive integer. In perfect consistency with the foregoing, we define P to be a versor which turns the vector upon which it operates through one-half a right angle positively, P through one-third of a right angle positively . . . im through one mth of a right angle positively. Similarly, is a versor which turns the vector upon which it operates through one mth of a right angle negatively. Definition.—If rj is any unit-vector and t any scalar, whole or fractional, 'positive or negative, rf is a versor which twists any vector at right angles to rj through an angle t x —, the direction of rotation depending upon the sign of t. Hence, every such power of a unit-vector is a versor ; and, conversely, every versor may be represented by such a power (Professor Hardy). Since the angle of a quaternion has been defined to lie between o and it, the value of t must lie between o and 2, If the angle of the versor in degrees be 0, then 0 = ^ 2 and t = 0, or (as it will for the future be written) cO. 7T Hence i;*' = rj±r6. 28°. It was shown in 17° (a) that the value of the versor of which turns ¡3 positively through the angle 6, is cos 0 + c sin 0.A QUATERNION AS THE POWER OP A VECTOR 53 But the versor would turn ¡3 positively through the same angle. Therefore, cos 0 + c sin 6 = ecd..............(1) Similarly, the value of the versor of which turns a a positively through the angle 0, is cos $ — c sin 6. But the versor (— e)c9 would turn a positively through the same angle. Therefore, cos 0 — € sin 0 = (— €)c9.............(2) If it be desired to give the value of this versor in terms of + e, we have only to bear in mind that the effect of turning any vector through any angle negatively by means of a versor OX, is equivalent to the effect of turning the same vector through the same angle positively by means of the versor — OX, and vice versd. Hence, (— e)cd = e~cd, and consequently, cos 0 — € sin 0 = (— c)c0= €~cd. . . . (3) But (— c)c0, not €~c9, is the versor of because it operates a through 0 positively, 4° (c). In precisely the same way, we have for the versor of /3a, cos (tt — 0) + c sin (?r — 6) = €« («■ ~ *>; . . (4) and for the versor of a/?, cos (w — 0) — c sin (ir — 6) — (— c)c (*-*). . (5) ___We maynow discard the notion of two coincident versors, OXg and OX*—#, in the direction of OX, 4° (c), which was introduced merely for the purpose of explanation. The base OX may be considered as one and the same in both cases, being expressed by tcd when it operates through the angle 0, and by (ir~6) when it operates through the angle (tt — 0), &c., &c. 29°. As in the case of quadrantal versors, i~l = so in the case of non-quadrantal versors, e~c9= For, TjP = = Ua U£ Ua . Ua = , . Ua a Ua Ua ’ Ua Ua # U/3 ' Up' But U^ = cos 0 — c sin 0 = (— e)c9 = c-c0; and U^ = cos 0 + c sin 0 == cc9.54 A QUATERNION AS THE POWER OF A VECTOR Therefore, «-<*== _L. e* But although i~l = — i, yet €~ce 7= — cc0, and — cc0 ^ (— For, €~c9 = (— c)c0 = cos 0 — c sin 0, and — €cd = — (cc0) = — cos 0 — c sin 0; and the right-hand members of these two equations can only be equal in the particular case when 0 = that is, when A t == 1. Therefore, in general, €-c0 =y£ — €cd, and — €c0 7^ (— c)c0. 30°· Let a, /3, y, fig. 17 (a), be any three coinitial, coplanar vec- tors, the angle between a and /3 being , that between ¡3 and y, x- Then, = cos 0 + c sin = ec*, 17- = cos x + c sin x = cCx, = t^l· =(C0S ^ + csin (cos X + esinx)=£<* . £«. Now, U/3' Vy = 'uy’ and (cos ^ + £ sm ) (cos X + £ sm x) = cos ( + x) + e sin (<£ + x) = €cW+*). Therefore, €<* . i« = TJ^ = cos (<£ + *) + t sin (<£ + x) = cc'*+°. . (1) Suppose that y lies between a and /3, fig. 17 (6), and con- sequently that the angle between a and y is (<£ — x). Then,A QUATERNION AS THE POWER OF A VECTOR 55 TJV^ = — UV70 since the rotations from 3 to a and from 7 P y to are contrary in direction. Hence, = cos + c sin • 9 § y = cos (<£ — x) + c sin (<£ — x) = ccC**"x). Therefore, ............................................ 1 ,cx (2) The meaning of (cc*)m may be investigated in the same way as that of rf, €c% or (ecfl)1, is a versor which turns a vector at right angles to itself once through the angle 6, positively ; (ec3)2 turns the vector twice through the angle 0 ; . . . (c^)m turns it m times through the angle 0. And since the opera- tion of turning a vector m times through the angle 0 is equiva- lent to the operation of turning it once through the angle m0, we have, 31°. Suppose that the plane of a quaternion is indeter- minate, and consequently that the versor of its vector is >/"— 1· Then the equation, = €m', becomes {(s/^I)ci)m=.(N/'-i)imi; or, (cos 0 -f sin 6 . V —l)”1 = cos mO -f sin mO , s/ — 1, which is Moivre’s formula. This formula, then, adprits of a real geometric interpretation when the symbol s/ — 1 receives the interpretation assigned to it in this Calculus. According to that interpretation, Moivre’s theorem asserts that the operation of turning a line m times successively through any (3) Hence, (1), (2), (3), the Algebraic Law of Indices holds good for versors :56 A QUATERNION AS THE POWER OF A VECTOR angle 6, is equivalent to the operation of turning it once through an angle mO. Those who wish to go further into this matter are referred to Professor A. S. Hardy’s ‘Elements of Quaternions/ pp. 50-55. 32°· If we assign any integer values to t, positive, nega- tive, or null, it will be found that if y= i, t is an even multiple of 2 ; if „ = -1, ,, odd ,, ,, \ if »> = ~ Vi „ odd number. In symbols, rj* = ± 1 ; V*t+ l = ±rj; the upper or lower signs being taken according as the number t (assumed to be a positive, negative, or null integer) is even or odd. 33°. Prom the preceding considerations we are justified in defining, that if p be any vector and t any scalar, p' = Tp.TPp;.............................(1) „ = product of a tensor and a versor ; ,, = a quaternion, 15°. From this equation we have at once Tp‘ = T‘P; Up' = U'p; x Up' = Upc* = cos 0 + sin 0Up ; Sp* = Spc9 = Tpc* cos 0 ; Yp‘ = Yp“> = Tp°° sin 6JJp; ' UYjo' = XJYpcS = ± Up; p* = /_pce = imr ± 6 ; ■ ■ (2) the upper and lower signs accompanying each other, and n being an integer, positive, negative, or null. With regard to the expression for ¿_pc\ it must be borne in mind that the amount of rotation from JJ/3 to Ua, fig. 10, admits of being increased or diminished by any whole number of circumferences, or of entire revolutions, without altering the final direction of U/3. In symbols, vce _ (inn + «>.(_ vye = (inn -6) . . . (3) 2ri7T ± 6 is the Amplitude, 0 the Angle, of pc9A QUATERNION AS THE POWER OE A VECTOR 57 In the particular case when t = 1, p is the representative of a right quaternion ; and since its angle is the general expression, p‘ = Tp‘(cos| + Up.sin|) becomes p = TpUp. 34°. It is now clear that any quaternion may be reduced to the form p* by a suitable choice of the base, p, and of the scalar index, t. The conditions are, i = Tp = T?«; TJp = UV?; 7T t > o ; t < 2. Section 4 A Quaternion in the form of a Quadrinomial 35°. Let XY, YZ, ZX, fig. 18, be three rectangular co- ordinate planes, and let i, j, k be unit-vectors along the three axes OX, OY, OZ respectively. Let OP = jo be any vector; from its term let fall perpen- diculars PL', PM', PN', on the three planes; and com- plete the rectangular parallelo- piped LL'.___ ____ _______ Then T . OL, T . OM, T . ON Fig. 18. are the Cartesian coordinates, x, y, z, of P, the term of p. Consequently, OL = xi, OM == yj, ON = zk ; and the equa- tion of OP is, Part I., 37°, p = xi + yj + zk...................(1) If the coordinate planes are not rectangular, and if a, ¡3, y are unit-vectors along OX, OY, OZ, the equation of OP becomes, p = xa + y¡3 + zy, (2)58 A QUATERNION IN THE FORM OF A QUADRINOMIAL where x, y, z are the Cartesian coordinates of P, referred to oblique axes. Since a quaternion is the sum of a scalar and a vector; if w be a scalar, any quaternion may be represented by an equation of the form, q = w + xi + yj + zTc;...............(3) which, depending as it does upon the values of the four scalars, w, x, y, z, furnishes a new reason for calling the complex quantity, q, a quaternion. From the last equation, tf = (w2 — x2 — y2 — z2) + 2w (xi + yj + zk) . (4) If / q = w = o in equation (3), and q = xi + yj + zk,..............................(5) q* = — (æ2 + ƒ + z2)...........................(6) 36°· (<*)· Suppose we have any two vectors, a = mxi + + mzk, P = nxi + ntf 4- njc. Multiplying, first a into /?, and then ft into a, we have a/3 = — (mlnl + m2n2 + mznz) -f /3a = — (mlnl + m2n2 -f- mznz) — * » j . k \ mx, m2, m3 n2, nz V. i , j , k «1, m2, m3 nun2, nz i • · (1) Hence, we see again that the commutative law of multipli- cation does not hold good for vectors.1 (b). The multiplication of vectors, however, obeys the distributive law. Let, a —· /ji 4“ ^2»1 4 P = wiji + m2j + ra3&, y = nxi -f n2j + nzk, be any three vectors. 1 This section was written a considerable time before I saw Herr Dillner’s article on quaternions in the ‘Mathematische Annalen,’ vol. xi., for 1877.A QUATERNION IN THE FORM OF A QUADRINOMIAL 59 Then, a + /? = (¿i + ^i)i + (l2 + m2)j 4* (lz + m3)k ; and (a), (a + P)y = — {ft + + (^2 + w2)n2 + (¿3 + ^3)^3} 4- y j» Q\n\ 4* hn2 + ^3^3) 4- * > j > & hi hi h nli n2i n3 — (m^! + m2n2 + m3n3) + i i j i k mlt m2, ra3 ; U\i H2j w3 „ = ay + ¡3y. It can be proved in the same way that y(« 4- fi) = ya 4- y/3. Therefore the multiplication of vectors is a doubly distribu- tive operation in the case of three vectors. (c). The associative law of multiplication also holds good for vectors. Taking the vectors a, /3, y of (6), we have, d/?.y = i i j i k 1’ — Sm 4- ^1 , ¿2 y h mx, m2, m3 («1» + n2j + njc); = — — rioj'Zlm — nzk%lm i i j i & hi hi h mj, m2, m3 (nxi 4- n2j 4- njc) ; = — n^lrn — n2j'Um — njc%lm 4-^1 — 1» —h j hi hi h 4-^-2 k, —1, —i hi hi h + n3 —jy ii— hi hi h mlt m2, m3 m„ m2, m3 ml9 m2, m3 Expanding the three determinants and rearranging the terms of the whole, we get a/3 . y = — — ljc%mn °1 i —i 4“ h —k, —1, i i»—*> —1 m2, to3 TO, , to2)to3 4- ¿3 OT„ TOj, TO3 n2, tl3 w,, n2, n3 Tlj, 12-2> W360 A QUATERNION IN THE FORM OF A QUADRINOMIAL a8 . y = — — l2j%mn — l3k%mn i , j > k i , j 9 k i > j > k + hi mu ra2, m3 + hj mu m2, m3 mu m2, m3 nu n2i n3 wl> n2i W1 > n2i n3 = (hi + hj + W \ “ ^mn + i , j ? k m j, m3 ™2> nz , „ = a . fiy. The multiplication of vectors, therefore, is an associative operation in the case of three vectors. The method by which it has been proved that the distribu- tive and associative laws apply to three vectors can obviously be extended to the multiplication of any number of vectors. 37°· By the aid of the distributive principle we can now find the values of (a + /3)2, (a — py, (a + 0)(a — 0), and a/3 . /3a. (a+/3)2=(a+/3)(a + /3)=a2 + a/3+/3a+/32=a2 + 2Sa/3+/32 . (2) (a - py = a2 - 2Sa/3 + P2...........................(3) (a + P)(a—P) = a2—a/3 + /3a—/32 = a2—(a/?—pa)—/32 = a2 - 2Va/3 - p2 . ..... (4) a/3 . /3a = (Sap + Va/3)(Sa/3 - Va/3) ; „ = S2a/3 - SapVap + Sa/3Va/3 - V2a/3 ; „ = S2a/3-V2a/3..................................(5) „ = (— TaT/3 cos (9)2 — ( — TaT/3 sin 0 . «)2 ; » =T2«j3....................................... (6) By the associative law, a/3 . /3a = a . /32a = a . a/32 (since /32 is a scalar, 24°) ; „ = a2P2.........................................(7) but this last expression is riot to be confounded with (a/3)2. 38°. If a, ft, y be any three vectors, in general, For and a f' P “ y = ap-'y, p-'y^yp-1; ButA QUATERNION IN THE FORM OF A QUADRINOMIAL 61 therefore, aP V ^ ayP 1; or, If, however, P7^ P y = P, then, a f> — aP. / ¡3 ’ for “/3 = aP~'P = (n°) aPP~l = =£· Again, for a y’P yp’ IP II 1 I =,, 36°, a . 7->y . > = a/3-1 = a = ay. p P py But Py=£yP‘, therefore, Py ^ y P and yP^yP’ In such an expression as the denominator must be yP treated as one quaternion ; so that, if we equate the fraction to g, we have ay = qy/3, and q = ay/J^y"1 (Prof. Tait). 39°. Prom equations (1), 36° (a), SajS = S/3a = — + ^2n2 + ^3^3) ; Ya/3 = - V/3a = ¿ , Í , h m1} m2, m3 J ^2 J „ = (m2n3 — m3n2)i + (m3nl + (m^ — m2nj)k. Returning to the simpler form, q 3= w + ad -f yj -f zk, m,n3)j we have at once S q = w .... Yq = xi + yj -f zk. (1) (2)62 A QUATERNION IN THE EORM OF A QUADRINOMIAL Also, since (TV#)2 = — (Yq)2 = x2 4 y2 4 z2, TYq = s/x2 4 y2 4 z2............... UYo = ^ + yj + TYg iy/a;2 H- 2/2 + is2 Further, since T2g = S2# — Y2q, 37° (5) and (6), Tg = s/w2 4 x2 4 y2 + z2 . . . jj jl — u> + xi + yj + zk T q V w2 + x2 + y2 + z2 SU? TVU? Sg_______________w_________ Tg if T?' = T?> and Uq' = U? ;) „ =„ „ Sq> =Sq, „ Yq' = Yq f ‘ * W Tg' = Tq, and JJq' = JJq, if q' = q, ;) sq' = s?, „ Yq' = v?, „ „ = „ I · ' (6). On the equality of the tensors of two such quaternions as fia = 8y, ^ = -, there is nothing to be said ; the tensors are P ^ positive numbers obedient to the rules of Arithmetic. (c). Definition. Two versors are equal, anc? only equal, ?ƒ the rotations they communicate to the vectors on which they operate, m planes perpendicular to themselves, are similar in direction and equal in amount. And the converse. As in the case of vectors, Part I., 7°, the phrase ‘ similar directions,* was defined to mean ‘ parallel directions with the same sense * ; so in the case of versors the phrase is defined to mean ‘ in coincident, or parallel, planes towards the same hand, as seen from the same side of the two planes.* The phrase £ equal in amount * means 4 through angles equal in magnitude.* This definition may be expressed in symbols as follows : U q = Tty, l /n if z?=z^and,UV? = TJV, ÜY?' = V ; 0) becomes, if €c* = rf*, 6 = (j>, and c = y ; (2)64 EQUALITY AND INEQUALITY OF QUATERNIONS 41°. If tfl and vfi are coplanar quaternions, and if £»7 = vp, then, obviously, = pv, for their tensors and angles are equal, and the direction of rotation is the same for both quater- nions. For the same reasons, if ^ then ^ ay p 0 g o Further, - = A For, let ~ be made to slide and revolve pa a in the common plane until its origin coincides with that of S -, and a and y are collinear. Then, /3 and 8 will be collinear, 7 and since - =-, \ or, Part I., 20°, ^ = X c a b a pa 42°. Any two quaternions, considered as geometric fractions, may be reduced to a common denominator. Let the two quaternions be in the plane LMQP, and — in the plane PQSR, fig. 19. Whatever be the planes of the quaternions, or however they may be posited in these planes, by causing the quaternions to slide about and rotate in their own planes (without turning them over), they can be made to assume the positions shown in the figure, where their origins are coincident, and the divisor vectors are collinear, in the line of intersection of the two planes, PQ. Let E be any point in that line. Produce OB, and draw EF and EG parallel respectively to AB and CD. Then, OF OB OG ODEQUALITY AND INEQUALITY OF QUATERNIONS 65 Let another vector, OH, be drawn in the plane LMQP, making the angle HOE = EOF, and of such a length that OF : OE:: OE : OH. Then we have, OE = OF OB . OH — OE OA9 and, therefore, OD OB OG OF _ OG ± OF ( OC ± OA OE± OE OE ; OB = OG OE OG . OF OF 3 OF OG OE ___ ^ ^ OD OC OD OC OA OB OA OE OG ' OE (1) (2) OE OE OH OG OH Any two quaternions, then, such as and ^5 may be reduced to the form and or to the form and OE OE OH OE, without undergoing any change in value. 43°. It follows that no two diplanar quaternions can be , tj, OB OD equal. For suppose —■= mi OF OG 1 Then -- = and conse- OE OE quently OF = OG, which is contrary to definition, since the two vectors have not similar directions. Conversely, if two quaternions are equal, they are coplanar. 44°. If q and qf are equal quaternions, so that Sq + Yq = $q' + Yqf ; then, by definition, Sq = Sq\ and Yq = Yq'. More generally, if an equation involves any number of scalar and vector quantities, the sums of the scalars and of the vectors on either side are respectively equal. For example, let x + wa + nfi = y + z + ty. Then, ma + nft = (say) 18, and x -f IS = (y + z) + ty. But (x + ZS) and {(y + z) + ty) are quaternions, 20°. Therefore, 40°, S (x + ZS), or S (x + ma + nfi) = S \(y + z) 4* ty), Y (x -f ma + nfi) = Y {(y + «) + ty); therefore, · x = y + z ; ma + n/3 = ty. ¥66 THE VARIOUS KINDS OF QUATERNIONS CHAPTER Y THE VARIOUS KINDS OF QUATERNIONS 45°. Collinear Quaternions. Quaternions whose planes intersect in, or are parallel to, a common line are said to he Collinear. For example, the quaternions OB . OA and OD.OC, fig. 19, 4?°, are collinear ; and OL . ON, OM . ON, fig. 18, 35°, are also collinear, whatever be the angles YZ, ZX, ZY. Since the versors of collinears are each perpendicular to the common vector, it follows that if q, qf, q", &c., be collinear, Uq, Vq', Uq", =£ = T£(cos0-£sin0) = T£.(-t)‘'=T£.t-‘* . . (3) cl cl a a The versors of reciprocals are reciprocal, ccd and c~c* being reciprocal, = IT?"1 = and U?-'U? = UtfU?-* = 1; . . (4) or, the versor of the reciprocal is equal to the reciprocal of the versor. 47°· Opposite Quaternions. If any two opposite vectors, ¡3 and — ¡3, be divided by any one common vector, a, the two unequal quotients thus formed, P ancl ^-P} are called opposite quaternions. Accordingly, — q a a is the opposite of q. Since, 1° (i), -e+p=-p+i=°=o, a a cl a and a =zl®._ _i· a * a a ¡3 ft the sum of any two opposite quaternions is zero, and their quotient is negative unity, g -i + i ——1· (1) ,, Opposite quaternions, fig. 20, have a com- mon plane, equal tensors, supplementary r angles, and opposite axes, "v0, T(-?) = Ti; B'i UV (- q) = - VVq = ~ f--() FIG· 30· Hence, if q = ~ (cos 0 + c sin 0) — . then, -q = f=T?{cos(*~0)-‘sin (tt-0)|=T“ . £-« > .. (3) F 268 THE VARIOUS KINDS OF QUATERNIONS 48°. Let OA, OB, fig. 21, be any two vectors. From 0 draw OB' = OB in the plane AOB, making L AOB' == AOB ; and draw BB', cutting OA produced in A'. Let OB' = ft'. (a). The unequal quotients, ^ and are called Conjugate a a Quaternions, and if ^ = g, = Kq, read ‘ conjugate of qj a a Conjugate quaternions have a common plane, equal angles and tensors, and opposite axes : ¿Kq= ¿q; TKq = T.q\ UVK^-UV,.^-} · · · - W Hence, if q == ^ (cos 6 + € sin 0) = · *c9, then, Kq = K~ = T~ (cos 6 — € sin 6) = . c-ci. . (2) The versors of conjugates are reciprocal, since ecd and c ~cd are reciprocal, and the product of the versors is positive unity : UKq = A = lA \Jq q (3) UKq .TJq = TJq. UK? = lj From the foregoing it is evident that a/3 and ¡3a are con- jugate quaternions. (b) . Since afi = Ta/? (— cos 0 — c sin 0), and Ka¡3 = pa = T/3a ( — cos 0 4* « sin 0), we evidently have SK? = Sq ) . . . (4) VKq=-Yq] · V ' Hence, we have as general expressions for a quaternion and its conjugate, q = Sq + Vq,..................(5) Kq = Sq-Yq;....................(6) whence, q + Kq = 2Sq..................(7) q - Kq = 2Yq..................(8) (c) . If ¿_q = o, Yq vanishes in (5), q degrades to a positive scalar, say x, and (6) becomes K* = .....................(9)THE VARIOUS KINDS OF QUATERNIONS 69 Similarly, if q — ir, q in (5) degrades to a negative scalar, say — x, and (6) becomes K (-*) = —*.....................(10) If ¿q = 7T 2’ S vanishes in (5), q degrades to a positive vector, say y, and (6) becomes Ky=-y...............(11) Since, 47° (3), (- q) = - Sq - Yq,.....(12) K(-q)=-Sq + Yq..........(13) If, therefore, ¿_q = ^, — vanishes in (12), — q degrades to a negative vector, say — y, and (13) becomes K(-y) = y......................(14) From (9), (10), (11), and (14), it is clear that, (1) The conjugate of a scalar is the scalar itself; (2) „ „ „ vector is its opposite : K(±a) = ±x; K(±S)= +8 . . .(15) (d). By adding and subtracting equations (5) and (6), it is seen that while the sum of a quaternion and its conjugate is a scalar, their difference is a vector. (e). The most important formulae of the last three sections are collected here for facility of reference : Quaternion = q = ** = (cos $ + c sin 6) . = T^ . cc*. . (j) fj ±p p Reciprocal = q~1 = (cos 0 — e sin0) = T- . e~c*. . (l) a Ta a Opposite = ~ |cos(»-e)-€sin(»-e)J . (m) Conjugate = ^ (cos 0 — e sin 6) = T^. c~c9.. (n) 49°. Miscellaneous Theorems. (a). The reciprocal of the reciprocal, the opposite of the70 THE VARIOUS KINDS OF QUATERNIONS opposite, or the conjugate of the conjugate, of a quaternion is the quaternion itself : 1: = ?; - (- q) = q; kk? = q· (b). Let y] be any versor. Then, since K(,) = - V, K(v«) = (- vr= rc>; or, the conjugate of the versor of any quaternion is equal to versor of the conjugate (n). Hence, KTJ? = UK? = 17^ = A, 46° (4) . . (1) Since TKq = T#, and UK# =1 : Ug, we have q = 1q . Uq ; K? = Tq : IT? ; . . . . (2) whence, by multiplication and division, qKq = T2q ; q : Kq = U2g.............(3) / \ Ha____— IJa T/? Ua . —H^_a”1 ... a Ta JJ/3 Ta '-U£ Ta ’ T/J ~/3~l * ‘ W Hence, Ka/J = /?« = £ = K^, .... (5) (d) . The conjugates of opposite quaternions are themselves opposite ; or, K(-q)=-Kq; an equation which is a particular case of a more general formula, K xq = xKq...................(6) where x is any scalar. This may be proved' by supposing that the vectors OB, OB', fig. 21, are multiplied by any common scalar; or, that both are cut by any parallel to the line BB'. (e) . The conjugates of reciprocals are reciprocal; or, For, suppose the two triangles AOB, AOB', fig. 21, to revolve inwards round O in the plane B'OB until the points B, B', coincide in D, a point in the line OA produced. Then FOD and EOD represent respectively the two triangles after the revolution. From B and B' draw lines parallel to EDTHE VARIOUS KINDS OF QUATERNIONS 71 and FD, cutting OD produced in C ; circumscribe a circle to the triangle ABC ; and with O as centre and OB as radius describe the circle BDB (/). If we are given such an equation as ^ = K ^; or, - = K we can infer—first, that a, /?, y are coinitial and coplanar ; secondly, that Ty = Tf3 ; and thirdly, that a bisects the angle between j8 and y, or, that a (produced either way if necessary) bisects the join of the terms of J3 and y at right angles ; or, again, since the angles of incidence and reflexion of a ray of light are equal, that the ray y is the reflexion of the ray — ¡3 (O being supposed to be a point on a plane mirror whose sur- face is perpendicular to a). (g). Since L OBA = ¿ OB'A = L ODE = z. OCB, fig. 21, it follows—first, that AB and BC are antiparallel, or that the triangles AOB and COB are inversely similar (the triangles DOE and COB are directly similar); and, secondly, that OB is a tangent to the circle ABC at the point B. Hence, the circles BDB' and ABC are orthogonal, because a tangent72 THE VARIOUS KINDS OF QUATERNIONS to BDB' at B would be perpendicular to OB, which is a tangent to ABC. (h). Again, OA : OB :: OB : OC. Therefore, OB or OD is a mean proportional between OC and OA, and C and A are inverse points with respect to the circle BDB'. If, therefore, OD = vOA. = va, where v is a scalar > o ; then, oc = ^2 = ^ = ^oa. OA OA Consequently, the equation, OC OB' may be written, v2a T K§|,(7) K- a ft. (8) an equation which expresses that AB and BC are antiparallel, or that the triangles AOB and COB are inversely similar, but expresses nothing more. Now, in order that this relation should hold good, it is only necessary either that (1) T . OB should be a geometric mean between T . OA and T . OC; or, that (2) T . OB = T . OD. If, then, O, A, JD, C be fixed points, while B is a variable point and OB = OP = p, it is evident that the locus of P is the surface of a sphere with centre O and radius T . OD = vTa. Equation (8) then becomes — = Kp;or,?K£ = v2..................(9) p a a a (i) . If, then, we meet with an equation of this form, we can infer that the locus of P is the surface of a sphere with centre O and radius vTa. Further, if we take a point C such that OC = v2a, the sphere will be a common orthogonal to all the circles APC that pass through the fixed points A and C ; because every radius of the sphere is a tangent, at the variable point P, to the circle APC, AP and PC being antiparallel. (j) . Since qKq = T2q,V2 = T2£; aTHE VARIOUS KINDS OF QUATERNIONS 73 and the first equation of (9) becomes rjr2P __ j^P a p a T2P. T- U- = TK £ UK P = T £ KU^, (6) (1); or, 17? = KU^. p a (k), Since, fig. 21, BC _ BC __ OB _ OD __ v AB ED OA OA 1’ we have OB = OP = vOA ; BC = vAB. . . . (10) From the first equation we have at once, Tp = vTa......................(11) From the second, T (p — v*a) = vT (p - a).............(12) Since, ____ AB = p - a, and T(BC) = T(- BC) = T(CB) = T(OB - OC). (l). Article (h) contains the solution of the problem of Apollonius of Perga : given any two points, C and A, in a plane, and a ratio of inequality, 1 : v ; to construct a circle BDB' in the plane such that the lengths of the two straight lines AB and CB, or AP and CP, which are inflected from the two given points to any common point, B or P, of its circum- ference, shall be to each other in the given ratio. Cut AC externally at O in the duplicate of the given ratio of sides, so as to have OC = v2OA. Take OD, a geometric mean, to OA, OC ; and, with O as centre and OD as radius, describe a circle. This is the locus of all points for which CP = vAP. Paragraphs (e) to (I) are chiefly from Sir W. R. Hamil- ton's * Elements.'74 THE POWERS OF QUATERNIONS CHAPTER VI THE POWERS OF QUATERNIONS 50°· Let <7 = p\ L q = and UV# = c. Then, by 33° (2), we have TJ'p = €*, and q* = (p‘)n = (T*p . IPp)n = (Tq. €*)" = Tty . €wi = Tty . c"*, q11 = Tty (cos nO + c sin w0)...............................(1) From this equation we have at once, Tqn = Tty ; ttyn = e™6 == (ec*)n = Uty; ^ S^n = Tnq . cos n$ ; Yqn = Tty . sin nO . e. } (i) (2) 51°. If rc = 2, g2 = T2q (cos 2(9 + c sin 2(9) ... . (3) As this is the only power of a quaternion with which we will have to do in the following pages, it is desirable to inquire particularly into its nature. The first question that arises is, has q* two square roots like an ordinary algebraic quantity ?' Let OA, OB, fig. 22 (1) and (2), be the unit-vectors of any two vectors a and /?, inclined to each other at an acute angle in (1), and at an obtuse angle \ in (2). Draw OC making /_ CO A = <£ in (1), and = ^ in (2) ; and produce AO to meet the circle in A'. Then, if TT OA TT/ , - OA OA' r* = OB’U(-''; OB OB Since q2 = T2q . Uty, if q2 has two square roots, either Tty or Uty must have two square roots. But as Tq is alwaysTHE POWERS OF QUATERNIONS 75 positive, T2q can have only one square root, namely + Tq. If, therefore, q2 has two square roots, U3^ must have two square roots. What is U2g ] Since T-r OC OA u*“5a_0B· U = OC OA __ OC OA OB OB' OC Now, has two geometric square roots. 013 For, first, OC _ OC OA _ OA OA _/0A\a OB OA ' OB OB ' OB VObJ ' Secondly, OC 00 OA' OA' OA' _ /OAV /_ 0A\* OB OA' 'OB OB ' OB \0B / ~ \ OBj ' Therefore, Therefore, q2 has two square roots ; or, s/q2 = ± q = . Ug or Tq ( — U#), whatever be the form of q. Were the angle of a quaternion unlimited in magnitude, either of these real and unequal square roots might be used at will in calculation. But as the angle is defined to lie between the limits o and ir, we must discriminate between them, and select as the square root (or the principal square root) of the quaternion, q2, that one of the two which enables us to con- fine the angle of q2 within the prescribed limits. If ¿_q is acute, fig. 22 (1), Z. ( — nr. In this case, consequently, s/q2 = + q. For a similar reason, if Z q is obtuse, fig. 22 (2), s/q2 = — q. That one, therefore, of the two opposite quaternions, q and — q, whose angle is acute is the square root of q2.76 THE POWEBS OF QUATERNIONS In symbols, Lq-> UVg2=UVg;g2=T2g . <·**; if ¿g= J z = i /. „ „ =S Iq + VSq.78 ADDITION AND SUBTRACTION OF QUATERNIONS Therefore, 44°, (Sg-, + Sg2 + . . . Sgn) = H (ql + q2 + . . . qn); (V?1 +Vg2 + . . . Yqn) =V (qi + q2 + . . . qn) ; or, SSg = S2g ; SVg = Y2g......................(1) Similarly, ASg = SAg ; AYq = VAq......................(2) In words, S and V are distributive symbols. Taking the conjugates of the constituent quaternions above, Kg, = Sg, - Vg„ Kg2 = Sg2 - Vg2, Kgn = Sg,, - Vgn, we have 2Kg=2S?-2Y?=SS?-Y2g=SQ-YQ==KQ=K2?. . (3) Similarly, AK# = KA q..................(4) Therefore, K is a distributive symbol. Also, since KSg = S? = SK?,................(5) and K Vq = — Y q — YK q, .... (6) it follows that K is commutative with S and Y. 54°. Let any two qua- ternions, ql = q2 = be fX v reduced to a common de- nominator, OA, fig. 23, OA not lying in the plane OBC, but being drawn towards us from O. Let the reduced quaternions be OB and OC nr P A y OA OA Complete the ___ gram OCDB ; diagonals CB and OD = 8 ; and draw, OF = CB = CO + OB = /? — y. Then, Fig. 23. or — and a a parallelo- draw the — t£_±_£. ,8 _T. OD a T. OA’ T(g,+i2)=T(f + *)=T m , r„ -W? I T/3+Ty_T.0B+T.0C_T.0B+T. T71 + A?2_ a+ a“ Ta T^A TjOAT BDADDITION AND SUBTRACTION OF QUATERNIONS 79 Consequently, Tg, + Tq2 = T (gr, + y2), if, T . OB + T . BD = T . OD. But T . OB + T . BD > T . OD, Euclid I., 20. Therefore, Tqx + Tq2 > T (qt + q2). Let qx + c/2 = w, and let q3 be any third quaternion. Then, similarly, Tw + Tg-3 > T (w + q3) = T (qx + q2 + q3)· But Tq{ + Tg2 > Tw. Therefore, a fortiori, T T (q{ + g2 + g3). As this process may be carried on to any extent with similar results, we may infer that, generally, ..................(1) It may be similarly proved that TAg ^ ATg......................(2) If L BOC=o, that is, if Uy = Uy3; then t(T+£)=TZ+ t£ \a a/ a a If BOC=7r „ „ Uy=-U/3; „ TA+^)=T^~Tf. \a aj a a In general, if the quaternions, qu q2 . . . grt, bear scalar and positive ratios to each other, i.e., if they are coplanar, with versors similarly directed ; then, TSg = STg. 55°. In iig. 23, let the angles AOB=4>, AOC=x, AOD=0, BOD^, DOC = 0, X + o-2 > or, (<£ + x) + a- > 20 ; and 4* x > or. Therefore, 2 ( + %) + cr > 20 + cr, and + X > 0 ; ¿ L (z(?i+ q2); therefore, a fortiori, Z <7i + L· $2 + / qz > Z. (qi + #2 + £3)· As this process may be carried on to any extent with similar results, we may infer that SZtf^ZS?....................(1) It may be similarly proved that A Z. q ^ L Ay.................(2) 56°. Let / BOC = 0, fig. 23. Then, OD OA °— + OA 00 OA = q + q', and T (q + q') U (q + q') = T?Ug + Tq'Vq'. If, therefore, U (q + q') = JJq + U^', we must have T(q + q’) = Tq = Tq'; or, OD = OB = OC. Let OB = OC, and we have OD2 = 20B2 (1 + cos 0), and OD = 20B . cos In order, therefore, that OD = OB, we must have cos ~ \ ; or, 0 = 120°. Evidently, then, U (q + qf) = JJq + JJq', O _ only when Tq = Tq', and 0 = BOC = —, i.e., o in a special case. In general, therefore, U (q H- q') ^ Uq + Uq'. More generally still, U ^qjL%Vq......................(1) Similarly, AUq ^ TJAq.....................(2) The result of 53° to 56° is, that the symbols S, Y, and K are, while T, , and U are not, distributive in the addition and subtraction of quaternions.81 CHAPTER VIII MULTIPLICATION AND DIVISION OF TWO QUATERNIONS Section 1 Diplanar Quaternions 57°· Before proceeding further it is necessary to explain the meaning of certain forms of expression that will be met with in the present and succeeding chapters. Sqxq2 means the scalar of the product qiq2· Similarly, K.qxq2 means the conju- gate of the product qxq2. It does not mean the conjugate of qx multiplied into q2> which will be written Kqx . q2i or, ~Kqx(q2), or (Kqx)q2. And so on for the other symbols. Points and brackets should never be omitted if their omission is likely to lead to any misapprehension. The product of any two quaternions is a quaternion. For, let the quaternions be thrown into the form, ) Yvxv2= I (vxv2 - v2vx).)..............w Again, 62° (a), "Kvxv2 = Kv2Kvx = ( — v2) (— vx) — v2vx . . (3) 67°· Suppose the plane of vx to be at right angles to the plane of v2l and the direction of rotation to be such that vx = Tvx. i; v2 = Tv2 . j. Then, vxv2 = Tt^Ti^ . V = T^jTv2 . Jc,. . . , (1) v2vx = Tv2Tvx .ji = — TvATi;2 . A; . . . (2) Therefore, v2vx .= — vxv2;.................(3) and Lv2vx=: /_vxv2 = 25°, and (1) and (2).86 MULTIPLICATION AND DIVISION OF QUATERNIONS Further, the versors of vxv2 and v2vx (k and — k) are perpendicular to the versors of both vx and v2 (i andj) ; or, the plane of vxv2 (and consequently the plane of its opposite, v2vu or — is perpendicular to the plane of vx and to the plane of v2. Hence, the product of any two right quaternions in rectangular planes is a third right quaternion (v^) a plane rectangular to both, which is changed to its own opposite ( — vxv2) by reversing the order of the factors (Hamilton). In symbols, Yv]y2 = YvlVv2. Section 4 On Circular Vector-Arcs 68°. Let 0 be the centre of a sphere of unit-radius. Then any arc AB of any great circle of the sphere may be OB regarded as the representative of the versor —For the UA plane of the versor is the plane of the arc; the angle of the versor is measured by the length of the arc ; and the direction of rotation is indicated by the direction in which the arc AB is drawn—from A to B, fig. 24. Definition.—Two vector-arcs are equal, and only equal, when the origin and term of the first can be brought to coincide simidtaneously with the origin and term of the second, by sliding the first backwards or forwards on its own great circle. Thus, if on sliding (either way) the arc AB round the great circle of a unit-sphere, shown in fig. 24, the point B coincides with D when A coincides with C ; then, AB = CD. Two consequences follow from this definition. First, no two vector-arcs of the same great circle are equal, unless the direction of both, as seen from either pole of the common great circle, is towards the same hand. Secondly, whatever their length, no two vector-arcs of different great circles can be equal, except in one particular case. This case occurs when both the arcs are great serni- Fig. 24.ON CIRCULAR VECTOR-ARCS 87 circles. All great semicircles are equal vector-arcs, since they OA all represent versors of the form ——— = — 1, and the plane of — 1 is indeterminate. - OA 69°· Let any two arcs, C'C and AA', of different great circles bisect each other in B, fig. 25. Join A and O, C' and A', by arcs of great circles, and let the versors of any two quaternions, reduced to the form ? =->?' = £> a p Tt OB Q tt , OC ^ l? = oa = AB;U9=ob=bc· Then, U^ = 5i-§i = S = AC=BC+AB where BC is said to be added to AB. Similarly, (1) TT TT , OB 00 OA' LsU5=oaob“ob 5§“o^"c'a-:ba'+c':b"(2) The multiplication of versors is thus reduced to the addition of circular vector arcs. 70°. Unlike the addition of rectilinear vectors, and of quaternions, the addition of diplanar vector-arcs is not commutative; BC + AB ^ AB + BO. For BC+AB = AC;and AB + BC = BA' + UB^CA.'. But AC 7^ C'A', although the two arcs are of equal length. For, if C'C and AA' are both less than great semicircles (as shown in fig. 25), or if one of the two is a semicircle and the other less than a semicircle, in both cases AC and C'A' belong to two distinct great circles, and are therefore unequal by definition. Were C'C and A A' both semicircles, AC and88 ON CIECULAE VECTOE-AECS C'A' would both belong to the same great circle, of which B would be a pole; but they would have contrary directions, and wouid therefore be unequal by definition. In every case, therefore, AC ^ C'A', and, consequently, AB + BC^BC + AB; or, the addition of diplanar vector-arcs is not commutative. 71°. We now see why (in general) q'q qq', 60° (6). For CA'^AC; therefore, JJqJJq' 96 JJq'JJq ; therefore, qq’ ^ q'q. We also see why qq' = /.q'q, 61° (1). For Z TStf = L (UqOq1) = z ~ = ¿COA'; Z W? = Z (UjTJi) = Z gj = Z AOC. But Z C'OA' = Z AOC, because C'A' and AC are equally long. Therefore, Z ^'I'l — L ; or, L fM' = ¿.q'q. 72°. The addition of coplanar vector-arcs, however, is commutative ; for, evidently, fig. 25, BC-f-C^ = C7C = C7B4-Ba . . . . (1) These equations show that the multiplication of coplanar quaternions is commutative, since they are equivalent to OC OB_OB OC. OB ‘ OC' OC' * OB ’..............W or, TJqf . Vq = Uq . TJqf. Hence, qq' = q'q, a confirmation of 64°.ON CIRCULAR VECTOR-ARCS 89 73°. For the same reason that AB represents the versor BA represents the versor But, if UA (JJd 5z-u*6£--i-rrUK»-4iI(S>· Hence, if a vector-arc represents the versor of any quater- nion, the revector-arc (or the arc reversed) represents the versor of the reciprocal, or of the conjugate, of the quater- nion. Consequently, fig. 25, CA represents 17-^ = UKq'q : qq BA „ ui = UKq ; „ ul= UK?'. CB But CA = BA + CB; therefore, i_—1 1 q'q q * g'’ Kg'g = KgKg', a confirmation of 62°. 74°. If C'C and AA', fig. 25, are both great semicircles, AB (= Ug) and BC (= Tig') will be quadrants, i.e. q and q' will be right quaternions; and C'A' and AC will belong to the same great circle, but will have contrary directions. Therefore, since C'A' and AC are equally long, C'A' is the revector of AC ; and since AC = TJg'g, TW = C'A' = XjA = UKq'q ; q’q or, qq’ = ;...................(1) ” ~ q’q Equation (1) is simply equation (3) of 66°, in different symbols, v2vl =90 ON CIRCULAR VECTOR-ARCS If the semicircles C C and AA' cut each other at right angles, C'A' and AC will be quadrants of the same great circle with contrary directions, i.e. qq' and q'q will be right quaternions, and, consequently, U^' = UK q'q = KVq'q = - VU q'q, 66° ; or, qq' = —q q, which is a confirmation of (3) of 67°, v2vx = — v,v2. 75°. It remains to show how either of the two unequal quaternions, q'q and qq , may be geometrically transformed into the other. Let ABC be any spherical triangle, O being the centre of the unit-sphere; and let the versors of any two quaternions, reduced to the form ^ and X be XJq = U^' = a p OA OB Let P be the Positive Pole of AB, i.e. that one of its two poles round which rotation from A to B, or OA to OB, would appear to be right-handed to an observer standing upon the surface of the sphere at P. Let Q be the positive pole of BC, and S the positive pole of CA; S being, consequently, the negative pole of AC. Then, joining the points P, Q, and S by arcs of great circles, we have OP=§|=JJq; OQ=^|=W; OS=H=W?.73°;.. (1) and, from the known properties of the polar triangle, Lq = AOB = APB = 7r — QPS ; or, L QPS = 1r- L q,, . (2) Lq'= BOC = BQC =ir — SQP; or, SQP = - L q'· ■ ■ (3) Further, since the angle of a quaternion is equal to the angle of its conjugate, or, ^ ^ q'q, q'q = COA= CSA = 7r — PSQ; or, ¿PSQ=tt-/_q'q . . (4) Let us now pass from the triangle PQS to a third triangle, PQR, where R is the point upon the sphere diametrically opposite to S, and is consequently the positive pole of AC. Then, since the versors of conjugate quaternions are opposite unit-vectors, and since OS = UK^'g, (1), OR = TJq'q (5)ON CIECULAE YECTOE-AECS 91 Calling the angles of the triangle PQR, P, Q, R, we have ZP = tt-QPS =_Z<7, (2); / Q = 7r-SQP=Zg', (3); ) (6) Z R = PSQ = 7T — z q'q, (4) In fig. 26 the arcs C'C and AA' bisect each other in B, as in fig. 25, and the triangle PQR is derived from the original tri angle ABC in the manner just described. Let R' be the posi- tive pole of C'A'; join R' with P and Q by arcs of great circles ; and draw OR'. Then, OR'=^=U??',69“(2)...(/) Since the angle between the fig. 26. perpendiculars to two planes is the supplement of the angle between the planes, ZQOR = 7T - zC = tt - zC' = ZQOR', Z POR = 7T - zA = tt - ZA'== Z POR'. Therefore, QR = QR', and PR = PR' (in length) ; and from the equality of the triangles PQR, PQR', it follows that Z QPR = Z QPR'. But, (6), „ =zP =Z?; therefore, Z RPR' = 2 Lq........................(8) Since PR = PR', a small circle, described with P as its positive pole, which passes through R, will also pass through R'. Therefore, OR (= Uq'q) may be transformed into OR' (=LW) by the Conical Rotation of OR round OP through the angle RPR' = 2 /_ q. In symbols, TT / OB OC OB OC OA qq OA OB ÔA OÂOB = u q (W q) u?-1 ; or, q(q'q) q~l = qq'................(9) It is evident that the symbol q ( ) q “1 is an operator which produces a positive conical rotation of the versor, or axis, of the operand quaternion (which is written within the parentheses) round the versor, or axis, of the operating quaternion, q, through an angle = 2 /_q, without altering the angle or tensor of the operand q'q (since Tqq' — Tq'q, and92 ON CIRCULAR VECTOR-ARCS L qq' = L q'q)- This rotation is positive, because it is right- handed as seen from P, the term of JJq. Were it negative, the operator would be written q “1 ( ) q. Thus, u , = OCOB^OAOB OCOB q q OB OA OB OA “OB OA ? = U q-'(Uqq')Vq; or, q~l (qq')q = qfq..................(10) Regarding vectors as right quaternions, it follows from the preceding argument that, if qPq~x=P'·,.................(11) then p is the vector generated by the positive conical rota* tion of ¡3 round the axis of q through 2 ¿_q. Evidently Tp = T p Finally, if a¡3a~l = P ]...............(12) then p or OB', fig. 21, is the vector generated by the posi- tive conical rotation of ¡3 round Ua through twice the angle of a considered as a right quaternion, that is, through 7r.93 CHAPTER IX FORMULAE 76°. Let a{, a2 · · · anbeany coinitial vectors. VyjSa = V (Sy£ + Yy$) a - V . aS/8y — V . aVyp = V . aS/87 + V . aV£y „ = V. a(Sj8y + Vj8y) =Ya&7. By extending this process we obtain Y («1^2 · · · a«) ^ -f- ^ (ana7t — l · · · al)> · · · (1) according as n is even or odd. For example, Vaj3= -Y/3a. Sa/?y = S . a (S/?y + Y/3y) = S . aY/3y = S . Y(3y (a) = S (S/?y + Y/3y) a = S/?ya· By extending this process we obtain S(a1a2 . . . an)=S(a2a3 . . . ana1)=S(a3a4 . . . ana1a2)=&C. . . (2) 8a/3y = Sya/3 = Sy (Saj8 + Ya/3) = S . ySayft -f SyVa/2 ; therefore, Sa/?y = SyVa/J. And Sy/?a = SyV0a = - SyVa£ ; therefore, Sy/?a = — Sa/3y. By extending this process we obtain S (04012 . . . an) = ± S (anan_1 . . . 04), . . (3) according as n is even or odd. For example, S0a = Sa/?. It is unnecessary to write S . yVa/? above, instead of SyVa/3; for, if the expression has a value different from zero, it cannot mean Sy . Ya/3, because Sy = o. The product of any number of vectors in space is generally a quaternion ; for a/2y8 = a/3 . yS = qxq2 = qs. If the num- ber of vectors be odd, one of them may be treated as the representative of a right quaternion.94 FORMULAE The product of any even number of coplanar vectors is generally a quaternion whose axis is perpendicular to their plane. The product of any odd number of coplanar vectors is always a vector in the same plane. Since cyclical permutation is permitted under the signs S and T, it is obviously permitted under the signs SU and /. j or, SU (aja2 . . . an) = SU (a2a3 . . . a^a,]) = &C. . . (4) L (cqa2 . . . an) = L (a2a3 . . . c^a,) = &C. . . (5) 77°. By 22°, 2Y/?y = £y-y/3. Multiplying both sides by a and taking the vectors, 2Y . aV/3y = V . a (py — yfi) = V (apy - ay/?), „ = Y (a/?y — ay¡3 + /?ay — /?ay), „ = Y. (a/3 + Pa) y — Y . (ay + ya)/?, ,, = 2ySa/? — 2/JSya Or, Y . aY Py = ySa/? — /?Sya...........................(1) From a mere inspection of (1) it is evident that Y.aYfiy is perpendicular to a and coplanar with ¡3 and y. If equa- tion (1) be given in the form 8 = ySa/? — /?Sya, it is easy to show that a is at right angles to 8. For, multi- plying both sides of the equation by a, and taking scalars, we get, Sa3 = SaySa/3 — Sa/3Say = o. Therefore, a and 8 are at right angles. 78°. Py = S0y + Y/?y. Multiplying a into this equation, and taking the vectors, * Y apy = aS/3y -f Y . aY py „ =: aS/?y — /?Sya + ySa/? . . . (1) This equation is of the form, 8 == la + m/3 + ny. Yafiy, therefore, is the intermediate diagonal of the parallel- opiped of which the three coinitial edges are aS/?y, — /2Sya ; and ySa/?, Pt. I., 37°. 79°· Y · 8S . yYa/2 „ = yS/?Yay3-/?Syay3. But S/?Ya/2 = o, because ft and Ya/? are at right angles. Therefore, Y . Ya/3Y/?y = — /3Sya/2 . . . . (1) We have, therefore, aSapy = Y . YySaYay ; /?Sa£y = Y .YyySYySa ; ySaySy = V · VayVy/3...............(2) 81°. Y . VaftVyp = pSaygy - ySpa/3, - Y. Yay8Vyp = V .VypVaft = ySSypa - aSyGyp. Adding pSayfty = aS/Jyp + /3Syap + ySaySp, . . . (1) a formula expressing a vector, p, in terms of three given diplanar vectors, a, ft, y. pSayfty is the intermediate diagonal of the parallelopiped of which the three coinitial edges are aSftyp, /2Syap, and ySa/?p. 82°. Assume, pSaySy = xV fty + 2/Yya + *Va/3; it is required to determine the values of x, y, z. Multiplying a into the equation, and taking scalars, SapSay&y = icSaYfty + 2/SaYya + z&aV aft „ = xSafty + 2/Sya2 + zSa2/? „ = xSafty. Therefore, x = Spa. Similarly, y = Sp/?; z = Spy. Therefore, pSaySy = SpaYySy + Sp/3Yya + BpyVaft. . (1)96 FORMULAE 83°· Sa/?y8 = S . (Sa/?y + Va/3y) 8 „ = S . (Va/Jy) 8 „ = S . (aSySy — ySSya + ySa/3) 8 = SaSSy9y - SayS/38 + Sa/JSyS 84°. s . Va/3VyS = S . (a/3 - Sa/3) (y8 - Sy8) „ = Sa/5y8 — Sa/iSyS = SaSSjSy - SayS/38 .... (1) (1) =s 85°. S (VapVpyYya) = S [Ya£ . Y (Y^yYya)} „ = - S (Ya^ .ySa/3y), „ = - S (yYa0 . Sa/?y) „ = — Sya/3 . Sa/?y = - (Sa/?y)2. 86°. By 54°, Ya/?y = aS/?y — /JSya + ySa/J = (say) a' — /?' + y'. Y/3ya = /3Sya — ySa/J + aS/?y = a' + /?' — y. Yya/3 = ySa/2 — aS/?y -f /JSya = — a' + /?' -f /. Therefore, S .YafiyVI3yaYyaP=&(a'-pf+ y') (a'+ 0'-yf) (-a'+ 0'+y>) - /?', y' a', /?', - y' - a', /?', y „ = 4Sa/y8/y/= 4S . (aS£y . /3Sya . ySa/3) „ = 4Sa/SS/3ySyaSa/2y. In expanding the determinant, the cyclical order, a', /?', y', must be preserved. 87°· Bet a beany vector, q (=/?y) any quaternion, and let Yq = & Then, aq -f qa = a (Sg + Y#) -f (Sg + Yq) a = 2aS# + aYq -f Yq. a ,, = 2aS^ + «8 + 8a = 2aSg + 2Sa8 „ =2 (aSq + S . aY/3y) = 2 (a$q + 8af3y) „ =2 (aSg S/?ya) „ =2 (aS? + S . qa). 88°· ag'a :=: a + Y = - sin = i 9 i sin y = %Yil = jVj* = kYk1. H98 FOKMUUE Since S . jlk = o, j'k = - K . fk = - Kk . Kf = kj~\ and k zzzfkj1. 91°. If q and r be any two quaternions, (qrq~iy = grty”1. For, (qrq~1)2— qrq~x , qrq~l= qr2q~l; (qrq~l)*= qr2q~* . qrq~x— qr%q~x (qrq x)l=qrlq l. Similarly, (jkj~vf =j&j~l, and (ktj8kj~*k~t)v= Wj'tej^kr*, 92°. The proofs of the following formulae are left to the reader * (1) .' S . Va/3V/?y = S . a/3V/?y. (2) . S (a+ /?)(/? + ?)(? + a) = 2Sa/?y. (3) . Y (aVjiy + fiVya + yVa/3) = O. (4) . Ya/Iy + Yya/2 = 2ySa/?. (5) . a V(3y + /3Vya + yVa/3 = 3S a/3y. (6) . Y2a/? = S2a/? - a2£2. (7) . S . YydyVya = y2Sa£ - S/?ySya. (8) . S (Va/3Vy8 + YayVS/? + YaSY/?y) = o. If a be any unit-vector and £ any scalar, f9). cr* = Sa* - aSa*- 1. (10), aYa* = Sai+1. (11 \ iVf = Y£f; jVtt = Yt ; kYi* = Yf. (12). /VV = - Vkl; kVf = - Vi*; iVU = - Y/.99 CHAPTER X INTERPRETATION OF QUATERNION EXPRESSIONS 93°· The symbol S . a x means that a is to be multiplied into some given expression, and the scalar then taken. S . x a means that some given expression is to be multiplied into a, and the scalar then taken. And similarly for Y . a x, ............................(1) But TaTft sin 0 is the area of the parallelogram AOB, and Ty sin = OP is the altitude of the parallelopiped 01). Therefore, Bafty = volume of the parallelopiped OD . . (2) whose three coinitial edges are a, ft, y. 95°. As a confirmation of equation (2) of 94°, we may deduce the value of Safty in the form of a determinant by making use of the trinomial form of the three vectors. Let a = x,i + y j + *i£, /? = *a» + yd + »2h y=x3i + yd + Then, Sa/3y = S . (as,i + 2/, j + 2,h) (xsi + yj + zjc) (x3i + yj + z.Jt) „ ——Xl (y2z3 - y3z2) - yx (z2x3 — z3x2) — z, (x2y3 — x3y2) V Vu *i X2> Vi) z2 ^3» 2/3, «3 It will be observed that the sign of this determinant is negative, while the signs of (1) and (2), 94°, are positive. To explain this difference of signs, let OB = a, OA -- ft (fig. 27). Then the axis of aft will be a unit-vector, y = — c, and Bafty = ByVaft = TaT ft sin O&yy = TaT ft sin 0 . Ty (— sin ) „ = - TaT/3Ty sin 0 sin . = — volume of the parallelopiped OD. It is clear that the change of sign is due to the change in direction of the rotation from ft to y round a. In 94" this rotation was negative and the volume positive. When we change the names of a and ft, this rotation becomes positive and the volume negative. In other words, the sign of the volume is positive or negative according as the pyramid OABC is positive or negative, Part I., 30° (c). We can now see why the sign of the above determinant is negative. The result was obtained by resolving each of the vectors, a, ft, y, into three other vectors in the directions of i, j, k. The sign of the volume, therefore, depended upon the sign of the pyramid OIJK (fig. 11), and the sign of this pyramid is, and must always be, negative by definition.INTERPRETATION OF QUATERNION EXPRESSIONS 101 (2) We may, therefore, write generally, Safiy = it volume of parallelopiped OD „ = ±; 6 x volume of pyramid OABC In the case of Sy/Sa everything is, of course, reversed, because Sy/?a = — Sa/?y. But the rule of the pyramid holds good. 8y/3a is positive or negative according as the pyramid OCBA is positive or negative. 96°. What is the geometric meaning of the symbol Yafiy, when a, /?, y are the successive sides of a triangle ? Let a, /3, y repre- sent the sides of a vec tor-triangle ABC, % 28. Circumscribe a circle to the tri- angle, and let the tan- gents at the points A, B, C meet in T,, T2, T3. Let the angles of the triangle be A, B, C, and let ’U’y'J-2 = V· m Then, since L T3BA = L C, U/Ja = — TJyr* ; TJyTJjSa = Vypa = - UyUyr2 = - Uy2Ur2 = Ur2 ; y/?a = c&aUr2. But Sy/3a = o, 93°, therefore, y/3a = Yy/3a = Yaf3y, Therefore, Yafiy -■= abcJjT2................(1) The product of the three coplanar vectors, a, /?, y, then, represents a vector along the tangent to the circumcircle at B, the origin of a, whose tensor is the product of the sides of the triangle, and whose direction represents the initial direction of motion along the circumference from B through C towards A [the point Tj is not the term of r2]. Were the direction of rotation round the triangle negative, we should have (since yfta = Yyfta = Va/?y = a/?y), (- y) (- ft) (—<*)= — yPa = — aPy = a6c (— Vts), a vector equal in length to a/?y, but drawn from B in the contrary direction.102 INTERPRETATION OF QUATERNION EXPRESSIONS Similarly, V(3ya = bca . Ut3 Yya(3 = cab . Urj. } * 97°. What is the geometric meaning of S = a/?a-M Multiplying a “1 into the equation, a" !8 = (3a “1 ’ therefore, @ = K -. a a (2) Therefore, 49° (/), a, (3, 8 are coinitial, coplanar vectors ; a bisects the angle between (3 and 8 ; and T8 = T(3. See 75° (12). 98°. ^ j8 = ap...................(1) is the equation of a Point. For, since (3 = Sap + Vap, if we equate the scalar and vector parts, we have, o = Sap ; (3 = Yap. From the first equation, p ± a; from the second, (3 X a, ¡3 J_ p· Multiplying a-1 into the given equation, P = a-!/3 ; or, since ¡3 1 a and a-1/? is a right quaternion, p = Va-'yS = - VKa-'/S = - Y/Sa"1 = - v£ a Therefore, p is a constant vector, and the locus of P is a point, the term of — Y a The system V^ = y£;S£=S^,....................(2) a a a a also expresses that the locus of P is a point. For V^ = o; s£^£ = 0; a a therefore, S + Y = o ; a a a » P = AINTERPRETATION OE QUATERNION EXPRESSIONS 10B 99°· Y p= o, or Yap = o,................(1) a is the equation of the indefinite straight line OA. = £......................(2) a a is another form of the equation of OA. For, if q = A, a o=zq -Kq=z 2Vq, 48° (8). Therefore, ¿_q = z. ^ = o onr. a Tj£ = ..................(3) a a is the equation of the indefinite straight line OB in the case of the positive, or OB' in the case of the negative sign (OB' = - OB). If y = OC be a third vector such that L BOC = Z. AOB, U2 £ = Ul......................(4) a a is the equation of the indefinite straight line BB. For, since the angle AOC is bisected by p or — p, 51° (9). Vi = V^orV^ = o, . ... (5) a a a is the equation of an indefinite straight line drawn through B parallel to a. What locus is represented by Yap = /? ?.................(6) From inspection, /3 _L a, ƒ? 1 p· Y . a”1 X. Ya"1/? = Y . a”1 Yap = pSaa“l — aSpa-1 = p — £Ca, where # is an indeterminate scalar. But since /3 _L a, Ya-1/? = a-1/? = y = OC, some vector perpendicular to a and /3. Therefore, p = y -f xa, and the locus of P is the indefinite straight line through C, parallel to a.104 INTERPRETATION OF QUATERNION EXPRESSIONS The system of equations, Scp = o ; Sc/? = o ; S/?p = — c (constant) . . (7) where c is a given unit-vector, expresses that the locus of P is a straight line. The tirst equation restricts p to a fixed plane through the origin J_e, which, by the second equation, contains /?. From the third equation, if Tp = sc, and T/? == 6, 6 being the variable angle between the two vectors, we have — xb cos 6 = — c, /i c x cos 6 = _. o If, then, we take on /? (or /? produced) a point C such that 0G=i> . the sought locus will be the straight line through C J_ ¡3. If c = o, S/?c = o ; Sftp = o ; Sep = o, .... (8) expresses that the locus of P is the indefinite straight line through O, J_ /?. 100°. uv£ = uv£ .......................(1) a a is the equation of that part of the plane AOB which lies on the same side of the indefinite straight line OA as the point B. If OA' == - ÒA, T (p + a) = T (p a),...........(2) being equivalent to AP = A'P, is the equation of a plane through O perpendicular to a. S p = o, or Sap = o,..........(3) a expresses the same locus ; as also does K?=-p......................(4) a a To deduce (3) from (2), T (p + a) = T (p — a), rp p -j- a_ijt p —■ a T(?+ai) = T(?-i), 1 + 2S? + T*q = 1 - 2S? + T% 89°, S? = Sp = o. aINTERPRETATION OF QUATERNION EXPRESSIONS 105 To deduce (4) from (3), o = 2S£ = S£ + V£_V'> + S? = ?+ K£. a a a a a a a s£ = 1..........................(5) a is the equation of a plane through A perpendicular to a. For it is equivalent to S ------ = o, which shows that a (p — a) la. Sap = - a2,...................(6) which is equivalent to Sa (p — a) = o, gives the same locus. Sap = 1.....................(7) is the equation of a plane through A', the term of a"1, per- pendicular to a-1 or a. For it is equivalent to . = Sop - 1 = s 8a- sp- O................(2) a give a similar locus, the radius of the circle being in this case vTa. If c be any given unit-vector, Tp = Τα ; Sep = o...............(3) is the equation of a circle with O for centre and Τα for radius. S/o (p — 2α) = O ; UVap = c .... (4) is the equation of a circle passing through O, with A for centre. The system of equations, p = αίβ ; Τα = 1 ; Sa^ = o . . . . (5) represents a circle with O for centre and OB for radius, in a plane perpendicular to OA, t being a variable scalar. 102°. What locus is represented by ΥαρΎρβ = Ύ2αβ ? Since Υ2αβ is a scalar, Yap and Υρβ are parallel; and consequently p, a and β are coplanar. Therefore, p = xa + yfi, where x and y are scalars. V.aX, Yap = yYaβ. V. xft Υρβ = χΥαβ.INTERPRETATION OF QUATERNION EXPRESSIONS 107 Therefore, Y2a/3 = VapVp/3 = xyY2aj3 ; » xy = i ; j, p = ira 4“ — æ the equation of a Hyperbola, a and ft being unit-vectors along the asymptotes taken as the axes, and x and y the Cartesian co-ordinates. 103°. P = a*P; Ta = 1 ; Sa/3 < o ; represents a plane ellipse. For, f tir . · tlT \ r\ _ r\ t/TT n · tlT p = (cos — + asm - J y3 = [icos ^ + aySsm Let ^ = <9, and Va/? = ÔC = y. Then, taking vectors, p = p cos 0 + 7 sin 0, the equation of a plane Ellipse of which O is the centre, ¡3 the major and y the minor semi-axis, and 0 the eccentric angle. lor. P2 = — 1, or, Tp = 1...............(1) is the equation of the unit Sphere. Tp = Ta.......................(2) is the equation of a sphere with 0 for centre, which passes through A. (s ey - (v ?y=i....................(3) gives the same locus ; for, 1 = S2? - V2? = (Sq + Yq) (Sq - Yq) = qKq = T\ = (V)2; whence, Tp == Ta. T (p — a) = Ta..................(4) is the equation of a sphere passing through 0, with A for centre. (p - a)2 = (/? - ay...............(5) or, T(p-a) = T(yS-a)...................(6) being equivalent to AP = AB, is the equation of a sphere108 INTERPRETATION OF QUATERNION EXPRESSIONS passing through B, with A for centre. O may be any point in space. S - = 1, or, S = o, or, Sftp = p2 . . . (7) P P is the equation of a sphere with OB for a diameter ; for since (/? — p) is at right angles to p, the angle OPB is always a right angle. -^ = Ka. . ..................(8) a p gives the same locus as (2). For, multiplying by -, 1 = - £ = - K- = T2- ; therefore, Tp = Ta. P a p p p 105°. The system of equations, Sfiasl ; S^=l 0) expresses that the locus of P is a circle, namely, the Circle in which the plane through A, perpendicular to a £ = 1^ intersects the sphere with OB for a diameter == 1^. For since P must lie at the same time upon the sphere and in the plane, its locus is necessarily the only line which the sphere and the plane have in common—the circle of inter- section. 106°· p = a7?a-‘, or, p = ac6fia~eS9 where t and 0 are variable, is also the equation of a Circle. What circle ? Comparing these equations with (9) of 75°, it is clear that p is the result of the positive conical rotation of j3 round a through an angle = 20 = tir = 2t i.e. twice t times the angle of a considered as a right quaternion. The locus of P, therefore, is a circle upon a sphere with T/3 for radius. 107°. su £ = su £ a a is the equation of one sheet of a Cone of revolution passing through B, with O for vertex and a for axis.INTERPRETATION OF QUATERNION EXPRESSIONS 109 The other sheet is represented by SU P = — SU @; a a both sheets by (su y- = (su py. 108°· If we multiply together the two equations of 105° we get s£s£=1,.................(1) p a u the equation of the Cyclic Cone, discovered by Appolonius of Perga. It is an oblique cone, and has a circular base. Equation (1) is evidently satisfied when the two equations of 105° are satisfied. Therefore every point of the circle re- presented by the equations of 105° must lie upon the locus represented by equation (1). But this equation remains essentially unchanged when p becomes xp} x being any scalar, positive, negative, or null. For, s£s?P=1s£.a;SP = s£sP = 1. xp a x p a pa Equation (1) therefore represents the cone (prolonged both ways) with O for vertex which has the circle of 105° as its base. It becomes a cone of revolution when A (the term of a) coincides with the point in which ft cuts the plane of the circle. Any plane parallel to the plane S - = 1 will obviously cut either sheet of the cone cyclically. But there is another and a distinct series of planes which also cut the cone cyclically. For therefore, And P*s£ = S . = S/3p = Sp/2 = S ■-£- ; P P P s£ = is-£-,. p p S P = Spa'1 = Sa-V = S = p2S 1 = S^SP = S£ . p2Sa~- = S^-S-P-,, t»·*1 p2 p p -1 r-l P P Therefore110 INTERPRETATION OF QUATERNION EXPRESSIONS an equation which expresses that the locus of P is a cyclic cone with O as vertex, whose base is the circle in which a sphere with a-1 for a diameter ^ S — = 1 ^ is cut by a plane through B' (the term of /3_1), perpendicular to = 1^. 109°. TV P = TV £, or, TVap = TVa/3, or, V* P = V*£ . . (1) a a a a is the equation of a Cylinder of revolution passing through B, with a along its axis. For, if / £ = <£, and/. @ = 0,the given a a equations are equivalent to Tp sin = T/3 sin 0 ; or, if D and Q be the feet of the perpendiculars let fall from B and P respectively upon a, PQ = BIXIll CHAPTER XI ON THE DIFFERENTIATION OF QUATERNIONS Section 1 General Principles 110°. When we speak of a variable quaternion, we mean, in general, a quaternion whose tensor, angle and plane are variable. Hence, if q be any variable quaternion, then and Uq, Sq and Yq, Lq and UV# are, in general, variable quantities. When the plane is variable the quaternion and its differen- tial, which is obviously a quaternion also, are diplanar. If the plane happen to be constant, while the tensor and angle vary, the quaternion and its differential are coplanar. The method of the ordinary Differential Calculus, in so far as it involves the commutative law of multiplication, is inapplicable to vectors and quaternions. To show this, let us examine some simple case of ordinary differentiation, say, f(x) = X2. Then, f(x + Air) = (as + Ax)2, f(x + Ax)—f(x)_____(x + Ax)2 — x2 __ 2xAx + (Ax)2 , v Ax Ax Ax „ = 2x + Ax ; and at the limit, when Ax = o, diMss.2x. ax This result, it will be observed, depends upon our having, as in (a), (as + Ax)2 = x2 + 2xAx -f (Ax)2.112 ON THE DIFFERENTIATION OF QUATERNIONS But in the case of the equation ƒ (?) = where q is any quaternion, in general, (q + Aq)2 y^q2 + 2q\q + (Aqf ; because, in general, q and dq are diplanar quaternions, and the commutative law of multiplication is applicable to coplanar quaternions only. In fact, (q + Aq)2 = (q + Aq)(q + Ag) == q2 + Ag . q + qkq + (Aq)2. In consequence of this peculiarity of quaternions, it becomes necessary to frame a definition of differentials which shall not involve the commutative law, and which shall also remain true for quaternions which degrade to vectors or to scalars. Definition.—Simultaneous Differentials are the Limits of equimultiples of simultaneous and decreasing Differences. Conversely, if any simultaneous Differences, of any system of Variables, tend to vanish together (according to any law), and if any equimultiples of these decreasing Differences tend all together to any system of finite limits, then these Limits are Simultaneous Differentials of the related Variables of the System (Hamilton). In symbols, let q, r> *........................(i) denote any system of connected variables ; let A?, Ap, As.......................(2) denote a system of their connected (or simultaneous) differ- ences, so that q + Aq, r + Ar, s + As..................(3) shall be a new system of variables satisfying the same law as the old system (1). Then, in returning gradually from the new system to the old one, the simultaneous differences (2) can generally be made to approach together to zero, since evidently they may all vanish together. But if, while the differences themselves decrease indefinitely together, we mul- tiply them all by some common but increasing number, n, the system of their equi multiples, nAq, nAr, nAs......................(4) may tend to become equal to some determined system of finite limits, a, b, c (5)ON THE DIFFERENTIATION OF QUATERNIONS 113 When this happens (as it can be generally made to do by a suitable adjustment of the increase of n to the decrease of Aq, &c.)y the limits thus obtained are the Simultaneous Differentials of the related variables, q, r, s . * and we have a = dq, b — dr, c = ds.....................(6) A quaternion decreases as its tensor decreases, and it decreases indefinitely when its tensor tends to zero (Hamilton). 111°. As an algebraic illustration of the foregoing prin- ciples, let us investigate the differential of x2, where a? is any scalar. Let y =. x*....................... (1) Then y + Ay = (x -f Ax)-, and Ay = ‘lxAx -f Ax2; . . . . (2) where Ax2 represents (Ax)2. A (x2) will be represented by A . x2 ; (dx)2 by dx2 ; d (x2) by d . x2. If n be an arbitrary multiplier, say a positive whole number, nAy = 2xnAx -f n~l (nAx)2...............(3) Conceive, now, that while the simultaneous differences Ax and Ay tend together to zero (always, however, remaining connected with each other and with x by equation (2)), the number n tends to infinity, in such a manner that nAx tends to some finite limit, a. This will be the case if we oblige Ax to satisfy always the condition, Ax = n~la, or nAx = a................(4) W^e then have nAy = 2xa -f n~la2 = b + n~la2 ; if b = 2xa ; where b is finite, because x is supposed to be finite. But as n increases indefinitely, n~la2 decreases indefinitely, a being given and finite ; and the limit of n~xa2 is rigorously zero. We therefore have at the limit, nAy n= b....................(o) Since, then, a and b are the limits of equimultiples of simultaneous and decreasing differences, we have dx = a ; dy =.b = 2xa ;.............(6) or, dy = d . x2 = 2xdx..................(7) It will be observed that the use of the word ‘ limit ’ has been extended so as to include the case of constants. A constant is here regarded as its own limit.114 ON THE DIFFERENTIATION OF QUATERNIONS 112°. As a geometric illustration of the foregoing principles, let us investigate the differential of an area, say a rectangle. Let ABCD, fig. 29, be any given rectangle, and let BE q M P and DG be any arbitrary, but given and finite, increments of its sides AB and AD. Complete the rectangle AF, which will thus exceed the given rect- L angle by the gnomon CBEFGDC. R Let I be a point upon the diagonal of CF such that the line Cl is an arbi- i trary but given submultiple of the line CF ; and through I draw HM and KL parallel to AD and AB. These parallels intercept, on A D and AB produced, equisubmultiples DK and BH of the two given increments (DG and BE) of those given sides; for, obviously, DK Cl BH DGr CF BE* Let the point I gradually approach C. Then the lines BH and DK, and the gnomon CBHIKDC, or the sum of the rectangles CH, Cl, CK, will all indefinitely decrease, and will tend to vanish together ; remaining, however, always a system of three simultaneous differences (or increments) of the two given sides, AB, AD, and of the given rectangle, AC. The increments of the sides, being constant, are their own limits ; and since (by construction) they are always equi- multiples of the simultaneous and decreasing differences, BH and DK, we are justified by the definition, 110°, in taking BE and DG as the simultaneous differentials of the sides AB and AD ; provided that we take the limit of the equimultiple of the gnomon CBHIKDC as the simultaneous differential of the rectangle AC. What is the equimultiple of this gnomon, and what is the limit of the equimultiple ? The equimultiple of the gnomon is evidently — . CBHIKDC = BE (CH + CI + CK) = BE (BC . BH + CP . CQ + DK . DC) BH BH BH =|| (BC . BH) + ^ (CP . CQ) + (DK . DC) „ = BC . BE + DO . DC + CP . CR M = CE + CG + CL. Now, the limits of CE and CG are these rectangles them- selves, since they are constant; while the limit of the rect-ON THE DIFFERENTIATION OF QUATERNIONS 115 angle CL (i.e. its value at the instant when the point I becomes coincident with C) is exactly zero. Therefore the differential of the area, or rectangle, ABCD is the sum of the rectangles CE and CGr. These two examples show, first, that differences and differentials must not in general be confounded together; secondly, that differentials are not necessarily small: thirdly, that the differentials of quantities which vary together according to a law need not be homogeneous, it being sufficient that each separately should be homogeneous with the variable to which it corresponds, and of which it is the differential, as line of line and area of area. Section 2 Differential of a Vector 113°· Let us apply the foregoing principles to the differen- tiation of a vector. The equation P =ƒ(*),.....................(1) where t is an independent and variable scalar, generally repre- sents the vector of a point, P, of a curve in space (fig. 30). If Q be another point of the same curve, we have OP + PQ = OQ, or, P P + Ap = f (t + A<), · . (2) where Ap and At are the simul- taneous differences of p and t. Subtracting (1) from (2), Ap = ƒ (* + A£) — ƒ (t). . (3) Let Ap, or PQ, be the nth part of the vector PR, = (t), depends upon the single scalar, t (Hamilton). 114°. As an illustration of these principles, let us differentiate P = where a is a given and constant vector. P =/(<) = “<*. p + Ap=|(t +Ai)2, Ap = |{(f+ Ai)2 - i2} ·ON THE DIFFERENTIATION OF QUATERNIONS 117 Let nAt — u. a given and constant scalar. Then, Ai = n and Hence, (4), and at the limit But therefore, 4'> = ?K‘+;),-,'! = crn = nkp = ua^t + ua n r = uta. r = dp \ u = dt i dp = df (¿) = d . ^¿2a = atfcfa. Section 3 The Differential of a Quaternion 115°· Let Q = F(?,r . . . ), and let dq, * . . be any assumed (but generally finite) and simultaneous differentials of the variables q, r . . . . , whether scalars, vectors, or quaternions. Then, 110°, the simultaneous differential of their function, Q, is equal to the following limit, dQ = lim . n {F(g + n~'dq, r+n^dr,... ) — F(q, r, ...)} *. (1) n = oo where n is any whole number, or other positive scalar, which increases indefinitely (Hamilton). If the function Q involves only one variable q, or Q =ƒ(?)=ƒ?; then, dQ = dfq= lim . n {f(q + n~ldq) —fq} . . (2) n-cc 116°· As an illustration, let Q = q2> Then, dQ, = lim . n {{q + n~ldq)2 — q2}, «=00 ), = .. · »{(? + n-'dq) (q + n~ldq) — q*}, „ = „ . n (n~lqdq + n~ldq . q -f n~2dql), „ = „ . (qdq + dq .q + n~ldq2). Therefore, d . q2 — qdq + dq . q 0)118 ON THE DIFFERENTIATION OF QUATERNIONS This expression cannot be further reduced, the quaternions q and dq being, in general, diplanar. In the special case when they happen to be coplanar, we have, 64°, dq , q = qdq, and d , q2 = 2qdq. It will be observed that n~ldq2 vanishes at the limit, because n~lTdq2, or n~lT2dq, vanishes. If q degenerate to a vector, d , p2 = pdp + dp . p = 2Spdp (22°) . . . (2) 117° Section 4 Miscellaneous Examples y = sin a?. dy = lim . wfsin (x + —^ — sin aA w = oo \ \ nj ) (. dx t . dx . \ sin a?cos--l·- cos x sin----sm x). n n ) But lim . n cos — = n, n = qo n . dx 7 „ n sin - = dx. n and Therefore, dy = cos xdx. 118°. y = cos x. = lim . n ( cos f x -f —— cos x^ %=oo V V ™ / y (€¿07 . . dx \ cos x cos---sin x sin--cos x 1 n n ƒ „ = — sin xdx. 119°. y = rar, where m is a constant scalar.ON THE DIFFERENTIATION OF QUATERNIONS 119 - 1 1 Let mn= 1 -f , so that z =------. Z mn — 1 Then, since l0g(1 + ^ = 2log(1 + ^)’ we have iogm + ;) dx n dx dx mtl — 1 n Therefore, treating dx as a constant, dx lim . logM (l + 1Y = log,,, Z — 00 v ~ / lim . n~ qo i(mn- l) Consequently, lim n = oo Therefore, i(mn _ l) ! (¿a; l0gm « f * > lim . n \mn — 1J = logemdx ; log* m. and dy = mx loge racfc. 120°. ? = rf, w here rj is a constant unit-vector. = lim . n(rf+ n — 77*), )> — » . nrf / da; Un 5 J = )J ^COS Now, lim . cos - irdoc____, 2™ ~ 1 . irdx , x Tri/.x ' Sm 2w = (arc) ^)n‘120 ON THE DIFFERENTIATION OF QUATERNIONS Therefore, dp = ^ rf+l dx. Had we taken x = c6, the result would have been some- what simpler. Then, dp = lim . n ji7cOO —77^ ' n = go v » = „ .rn\vn -lj „ = „ . r)c6n(cos + 77 sin — — l'j \ n J ,, = rjcd7]d0 = 77 ^ + This result is identical with that above ; because, since sc = cO, ¿9 = I & = ? c 2 ’ and c(i+i) „*+1 77v 2rf rj = rf . We have here a case in which the plane and tensor are constant, and the angle alone varies. It will be observed that a unit-vector in the first power can only vary in direction, its tensor and its angle (as a versor) being constant by definition. 121°. Q = qr. dQ = lim . n{(q + n~ldq) (r -f n~ldr) — qr) n — 00 „ . n {n~ldq . r + n~lqdr + n~2dqdr) „ = „ (dq . r -f qdr + n~ldqdr). Therefore, c?Q = d . qr = dq . r + qdr. If c be any constant quantity, evidently d . cq = cdq ; d . qc = dq . c. 122°. ? = j8* where /3 is a constant vector. cfy = dp = d (TfteTJpt) = d (6*U)9*) „ = dbx . TJ0* + bx . dUjS*, 121°, „ = (119° and 120°) bx log, bdx . U/3* + 6* . - U/3X + I.£c .4 .. = (loge6 + |U^^.ON THE DIFFERENTIATION OF QUATERNIONS 121 123°. Q = g"1. By 46°, gg_1 = i; therefore, 121°, qdq~l + dq . g-1 = o. q ~ 1 X, q~ lqdq ~1 + q ~ xdq . q ~1 = o, and d . q~l = — q ~Yd,q . q~\ 124°. Q = g*. Squaring, Q2 = g, dq = QcfQ -f dQ . Q = q'd . q* + (d . q') qK q~- X and X Kg*, q~h (dq) Kqi = (d . gl) Kg* -f g~* (d . g*) g*Kg* „ = „ + Tg . g~* (d. g!). But, 49° (6), (1), Tg . g“* = Tg-'TJg-* = Tg'UKg* = Kg*. Therefore, g~* (dg) Kg·1 = (d . g*) Kg* + Kg* (d . g!) „ = 2Sg* (d . g*) - {(d . g*) Vg* + Vg' (d . g*)|. . (1) Again, dq = d (q'q'!) = (d . g*) g* + g* (d . g*) „ = 2Sg* (d . g*) + {(d . g*) Vg* + Vg* (d . g*)} ... (2) Adding (1) and (2), dq + q ~ * (dq) Kg* = 4Sg* (d . g'), and d.g* = ^±lAi$l^!. 4Sg* 125°· The symbols S, Y, and K are commutative with d. For q = Sg + Y g. dq = dSg + dVq. But, since dq is a quaternion, dq = S dq + Ydq. Therefore, dSg = Sdg ; dVq = Ycfg, 44°, Again, Kg = Sg — Yg ; therefore, cfKg = c?Sg — dYq = S dq — Ydq = Kt/g.122 ON THE DIFFERENTIATION OF QUATERNIONS 126°· To find dTq and dTp. T2q = qKq ; 2Tg TV? . ¿UV? + ¿TV? . UV? =zdVq = Vdq, 125°, dWa-Jdl-^J-i.VVq dV V? ” TV? TV? _MuV?-S^.UV?,126<>, ” V? ÿ V? ¿UV? = V^ . uv? ; V o (1) or, Since dVYq TJVq ¿UV? _ ■yVdq UV? v? ■ a unit vector, it follows that g dUV? UV? • (2) and, consequently, that dUVq±.TJYq. Therefore the diffe- rential of the axis lies in the plane of the quaternion. If the plane (and consequently the axis) of the quaternion be constant, the quaternion and its differential are coplanar, and dJJVq vanishes. Conversely, if dUY? = o....................(3) the quaternion and its differential are coplanar. Equation (3) is the condition of coplanarity of a quaternion and its differential.m SCALAR AND VECTOR EQUATIONS CHAPTER XII SCALAR AND VECTOR EQUATIONS! Section 1 Scalar Equations 130°. If we have two given equations, Spa = o ; Sp/3 = o; . . . , , (1) it is evident that p, being _L to both a and /?, is parallel to Va/3; or, p = ccY ap...................(2) It is equally evident that x, being an indeterminate scalar, the vector p is indeterminate. p would still be indeterminate were the given equations Sap = m ; S ftp = n. For, in this case, S (na — m/3) p = o, which shows that p is perpendicular to (na — ra/2), but shows nothing more. The conclusion is, that a vector cannot be determined from two scalar equations ; a conclusion that might have been arrived at from the consideration that a vector depends upon three scalars. 131°. A vector can always be determined from three scalar equations. For, let the three given equations be Sap = l; Sj3p = m ; Syp = n...........(1) Then my — n/3 — ySftp — /?Syp = Y . pVfiy ; na — ly = Y . pYya ; 1/3 — ma = Y . pY a/3. Therefore, Y . pTVfiy = Imy — nl/3y Y . pmVya = mna — Zmy, Y . pnVa/3 = nip — mna ;SCALAR AND VECTOR EQUATIONS 125 and V . p (IV/3y + rnVya + nVaft) = o. Therefore, p || (IVfiy + mVya + nVaft), or, S . aX. p = x (IVj3y *f mVya -f nVa/3). Sap = scSa (IVfty 4~ mVya + 7lVa¡3) „ = icZS . aVpy + a?mS . aVya + „ = xlSafiy. But Sap = l therefore, X = S a£y ; therefore, IV /3y + mV ya + nVa/3 This value of p satisfies the three given equations, but no other value of p will satisfy them. For, suppose the three equations to be satisfied by p\ and p2. Then Sag] = O j Sap2 = O , S . a (pt — p2) = o. S · /5 (pi — P2) = °» s · y (pi — P2) = °- Therefore the vector (px — p2) is at once perpendicular to a, to ft, and to y. But no real and actual vector can be per- pendicular at the same time to three diplanar vectors, which a, /?, y are supposed to be. Therefore Qq — p2) vanishes ; therefore pj = p2. Therefore, the three given equations can be satisfied by cne, and only one, value of p. The principle that no real and actual vector can be at once perpendicular to three diplanar vectors may be put in symbols as follows : If Sacr = o ; S/?cr = o ; Syor = o ; then 132°. A vector, p, cannot be eliminated by fewer than four equations. If we are given only three equations, Sap = h ; 8/3p = l; Syp = m ; we have, 131° (2), SapVfiy + S/?pVya + SypYa/3 — pSa/3y = o . . (1) an equation into which the vector p enters once. Now, suppose we are given a fourth equation, SSp = n. Then, if we multiply (1) by 8 and take scalars, we get SSYySySap + S8VyaSft> + SSYaySSyp - SSpSa/Sy = o, SapSfiyh — S/3pSy8a + SypSSayft — SSpSa/?y = o, AS/?yS — lSy8a + mSSafi — nSajSy = o . . . (2) an equation into which the vector p does not enter. Section 2 Linear Vector Equations 133°. The general form of a linear vector equation is defined to be p = S/3Sap + Y . qp,* ..... (1) where p is an unknown vector, q a known quaternion, and a = (a, + a2 + . . an), fi = (ft + ft + . . ft), known vectors. The symbol <£ stands for ‘ function/ and p is some vector coinitial with p. Similarly, if or be any other vector, <£o- = 2/3Sacr + V . qa.............(2) * For proof, see Molenbroek, pp. 188-191.SCALAR AND VECTOR EQUATIONS 127 If we interchange a and /?, and introduce K# instead of q, we have 4>'p = 2aSftp + Y . (Kq) p .... (3) <^/cr = 2aS/3f(r = Scr<£p. Spficr = 2SpaS/?o* + S . pV . (K^) p = 2SpaS/?o- + Sp(K^)or = Sp(f>'cr . . . (1) Functions which, like and <£', possess this property are called Conjugate Functions. The function is said to be Self-conjugate when, for any two vectors, p and cr, Sp = Sp<£cr.................(2) 135°· Since /JSa (p + + or + . . ) = <£p + xp = xp....................(2) d(j>p — lim . n {<£(p -f w -1 dp) — <£p} = lim . ^ n - oo n=x>n = lim . cf>n ~~ = $dp.............., (3) n - oo n Since <£p is a vector, d . p*f>p = pdp + dp . p.........(4)128 SCALAR AND VECTOR EQUATIONS 136°. Let 4>p = 8...................(1) where and 8 are given. Then it is defined that ...................(2) ~x is a function which possesses properties corresponding to those of . As a matter of convenience we write : M) = \ Ac. ; \ Ac. «/>(/)“18 = <£p ~ 8 · <£_1<£p ~ c/)_18 = p . . . (3) According as m J w, , <£2, &c., are operators which may alter both the length and direction of any vector upon which they operate. «¡¡>2 is not to be confounded with the square of ,—(<£)2. 137°. We cannot enter here into the general theory of vector equations ; suffice it to mention a simple method for their solution suggested by Dr. Molenbroek (“ Theorie, =s........................(!) Then we have at once S . Ap = SA8 \ S . p,p = Sp.8 ; S. vcf>p = SvS . . (2) where A, /x, v are any three noncoplanar vectors. But, by 134°, we also have S. pfiA = SAS ; S . p'/x = S/x8 ; S. pcj>'v = SvS . . (3) Therefore, 131° (2), ___SASV . ^V + S/xSV . <£>V<£'A + Sv8V . <£'A v As an example, let Vap/3 = y· Then, 78°, p = y = Yap¡3 = aS/3p — pS/?a -f* fzSap ; and, 133° (3), <£'p = fiSap — pSKfia + aS /3p „ = jSSap — pS/?a + aS/?p = cj>p.SCALAR AND VECTOR EQUATIONS 129 Since a, y8, y are any three vectors whatever, we may select them to represent respectively the A, /*, v of equation (2). Hence, SA8 = Say ; S/tS = S/?y ; SvS = y2. 'A = 'a — a = Va2;8 = a2^ 'ix = <£'/? = <¿>0 = Va/J2 = /32a ; 'v — <£'y = <£y = V ay/8. V . = /32V . aYay/3=/32Va (ay/J - Say/J) = a2/32Vy/3 - /32Sayy3 . a. „ = a/32Sa/3y - a2/32Vy3y ; V . = a2V(Vayy3. /3)=a2V{(ay/3 - SayyS)/?} „ = a2/32Vay - /?a2Say/? = y3a2Saj8y - a2/32Vya ; V . = a2/82 Vy8a = - a2/82Va/3. S . ¡xfiv = S (a2/3 . /32a .Vay/3) = a2j32S . /3aVay/3 „ = a2y32S/?a (ay/J — Say/3) = a2/32Sa/5Sa/3y. Therefore, (4), = Sya («A»Saj3y - a'P^Py) + Sfry (ffa2Saj3y - a^Wya) - 7* . W ^ a2$2Saj8Saj8y (a$2Sya + $a*S0y) 8a0y - (SayVfly + S&yVya + 8yy\a0) " ~ ai32Sa/3Sa/3y a20i(a-lSya + fi-'SPy) Sa/ty—a2/32 , ySajSy otlo ” = arfi-SafiSafly ’8 (1)’ a-,Sya+j8_1SSy—y, " “ Sa3 K180 PLANE TRIGONOMETRY CHAPTER XIII ILLUSTRATIONS IN QUATERNIONS Section 1 Plane Trigonometry 138°. The simple relation between the three vector- sides of a triangle, fig. 31, a + P 4- y = o, conducts at once to the three funda- mental equations of plane trigono- metry. For, since — y = a 4· /?, y2 = a2 + /32 + 2Sa/?, or, c2 = a2 + b2 — 2db cos C . . (1) It will be observed that ACF is the angle between a and /?, and that (tt- ACF) = Z.C. Operating upon — y = a + ¡3, with S. y x, — y2 = Sya + Sy/3, Hence, c2 = ca cos B + be cos A ; or, c = a cos B + b cos A. . . . (2) Operating with V . ¡3 x, - Y/3y = V/3a + Y/32 ; or, Yy/3 = Y/3a ; or, TYy/2 . U Yy/2 = TY/?a . T7V0a = TV/3a . UYy/? ; Therefore, TYy/3 = TVfta; or, c sin A = a sin C..............(3) It will be observed that UYy/2 = TJV /3a.PLANE TRIGONOMETRY 131 139°. Let L c = 90°. Then, — a = ¡3 + y -1 = P + y a a Taking the scalars, - 1 = si? + Sr_ = - cos CBE = - * cos B a a a a or, cos B = — ...............(1) € Taking the vectors, o = v/3 + V 7 = TV^ . uv£ + TV t-. TJV? a a cl a a a = tv£- - UV^ - TV — . UY £ a a a cl „ = ^- — C sin CBE = b — c sin B ; a a or, sin B = ................... (2) c 140°. To find the sine and cosine of the sum of two angles. Let a, /3, y be three coinitial, coplanar unit-vectors, ft lying between y and a; and let L AOB = 0, L BOC = <£. Then, Z = cos (d + ) + c sin (6 + <£). But therefore, Z. P P = cos -f c sin cj>. = cos 0 + e sin 6. 1L — 7.P· a f3 a ’ cos (9 + ) + e sin (0 + ) = (cos ^ + e sin ) (cos 0 + e sin 0) „ = cos cos 9 + e (sin 0 cos 9 + cos sin 9) — sin sin 9. Equating successively the vector and scalar parts, sin (0 + <£) = sin 6 cos cj> + cos 6 sin . . (1) cos (0 + <£) = cos 6 cos — sin 6 sin c/> . . (2) To find the sine and cosine of the difference of two angles, k '2,132 SPHERICAL TRIGONOMETRY Let the angle between y and a be if/. Then, 2 = cos (\j/ — 0) — c sin (if/—0), y = cos 0 + c sin 0, a — = cos if/ — € sin if/. But 7 ay Therefore, as above, sin (yfr — 0) = sin if/ cos 0 — cos if/ sin 0 . . . (3) cos (if; — 0) = cos if/ cos 0 ■+■ sin if/ sin 0 . . . (4) Section 2 Spherical Trigonometry 141°· Let ABC, fig. 32, be any spherical triangle, its angles and sides being, as in the case of a plane triangle, A, B, C, a, b, c. Let the^sphere be a unit-sphere, with O as centre, and let OA = a, OB = ¡3, OC = y. Let L, M, N be respectively the positive poles of AB, BC, AC ; L, M, and N corresponding to the points P, Q, R of 75°. Let OL = A, OM« y, ON = v. We evidently have, = ft sin c sin a sin B. Since .....................(1) a p aSPHERICAL TRIGONOMETRY 188 we have, by 60° (1), s2 = s£ s £ + s. V? v£; . . . (2) a p a pa or, substituting the above values of these symbols, cos b = cos c cos a + sin c sin a cos B ; . . . (3) one form of the ordinary fundamental equation of spherical trigonometry. 142°. Taking the vectors of equation (1) of 141°, V*.= s£ v£ + s? + v. v^ Y-, 60° (1), . (1) a a /3 f3 d Pa W>W and substituting the values of these symbols given in 141°, v sin b = p cos c sin a + A sin c cos a + ft sin c sin a sin B . (2) Let p = OP be any unit-vector. Dividing each term of this equation by p, taking scalars, and rearranging the terms, sin c sin a sin B cos PB = sin b cos PN — sin c cos a cos PL — cos c sin a cos PM (a) Let P coincide with M. Then, cos MB = o ; cos MN = cos C ; cos ML = — cos B ; cos MM = 1 ; and sin b cos C = cos c sin a — sin c cos a cos B. . . (3) 143°. It is evident that if A and M be joined by the arc of a great circle, this arc will cut B C at right angles in a point P, ; or AP, is the arcual perpendicular let fall from A upon BC. Let BP2 and CP3 be respectively the arcual perpendiculars let fall from B on CA and from C on AB. Let the point P, equation (a), coincide with A, and we have cos AB = cos c ; cos AN = o ; cos AL = o ; cos AM = — sin AP! ; and sin APj = sin c sin B. 'j C3· -i i . tv tv sin c sin a . t, I Similarly, sin BP2 ==----—sinB L . . (1) sin CP3 = sin a sin B J Since VayS == A sin c and Vy&y = p sin a, Y . Ya/3Vfly = sin c sin a YXp = — j3 sin c sin a sin B.134 SPHERICAL TRIGONOMETRY But, 80° (1), V . YafiVpy = - £Sa/?y. Therefore, Sa/?y = sin c sin a sin B . r . (2) It may be similarly shown, by using Y. YfiyYya and Y. YyaYa/?, that sin a sin 6 sin C = Sa/?y = sin b sin c sin A. . (3) Therefore, sin a : sin b : sin c = sin A : sin B : sin C . . (4) The equations of (1), therefore, become sin APj == sin b sin C 1 sin BP2 = sin c sin A > . . . . . (5) sin CP3 = sin a sin B J 144°. Let the point P, equation (a), coincide with IT the centre of the small circle inscribed in the triangle ABC ; and let the arcual perpendiculars let fall from I upon the sides cut the sides, BC in Pl? CA in P2, AB in P3. Let r = IP! = IP2 = IP3. Then, bearing in mind that the arcs (of great circles) IP3, IPj, and PaI, produced, pass through L, M, N respectively, we have, from equation (a), sin c sin a sin B cos IB = sin b cos IN' — sin c cos a cos IL — cos a sin c cos IM. Since IBP3 is a right-angled triangle, we have, by equation (3) of 141°, sin c sin a sin B = Sa/?y, (2) of 143° ; cos IB = cos r cos P3B = cos r cos (s — 5), where s == \ [a + b + c) ; cos IN = cos (" — r) = sin r ; cos TL = cos (| -f r) = — sin r ; cos IM = — sin r. Therefore, cos r cos (s — b) Safiy = sin r (sin b -f sin (c + a)} „ =2 sin r sin s cos (s — b) ; ^__ Sa/?y____sin a sin b sin C 2 sin s 2 sin i ■ (1) 145°. If the points L, M, N, fig. 32, be connected by aresSPHERICAL TRIGONOMETRY 135 of great circles, we obtain a triangle exactly corresponding to PQR of 75°. Consequently, zL = zP = Lq = ¿M = zQ= z?'= Z 00- OB’ and ZN = ZB = (» — Lq'q), Z?'? = Z§^ = (-- ZN). Now, Therefore, OC OB =OC . OBOA OA ’ _ VC(W_N) = v2 . v-c rr= v2 . V — c^:= i „ vcN jjbcMXcL = - 1................. As a verification, multiply (2) into OA. Then - OA = vcVMAcL. OA = vcVM · OB = i/N. OC; „ = V^-COA) . OC = — OA. Equation (2) may evidently be written, — OA OC OB (1) (2) OC OB OA 146°· Since equations (1) and (2) of 145° are perfectly general, we may write for any triangle, ABC, /?CVA = yc^~C)....................(1) ycCpcBacA = _ 1.....................(2) the fundamental quaternion equation of Spherical Trigonometry. As the left member of (2) reverses the direction of any vector it operates upon, by causing it to revolve successively through the three angles of the triangle in a certain order, it is evident that the sum of the vector-angles of a spherical triangle, taken in a certain order, is equivalent to 7r. If the radius of the sphere becomes infinitely long, the triangle ABC becomes plane, and y = ft = a. Consequently (2) becomes f3cBacA = yc(A + B + C>== — 1 = yC7r · Therefore, A -f B -f C = 7r (Dr. Odstrcil).136 SPHERICAL TRIGONOMETRY 147°· Reverting to (1) of 146°, (cos B -f ft sin B) (cos A -f a sin A) = — cos C + y sin C, or, cos A cos B -f a sin A cos B + ¡3 cos A sin B -f /3a sin A sin B = — cos C + y sin C................(1) Taking the scalars, cos C = — cos A cos B — sin A sin B . S/?a „ = — cos A cos B 4- sin A sin B cos c . . (2) another form of the fundamental equation of Spherical Trigono- metry. ' 148°· Taking the vectors of equation (1) of the last article, y sin C = a sin A cos B + ¡3 cos A sin B + sin A sin B. V/?a (1) Let P be any point upon the sphere, and let T be the foot of the arcual perpendicular let fall from P upon AB; PT being considered as positive when P and C lie upon the same side of AB, but negative otherwise. Then, dividing each term of (1) by p = OP, and taking scalars, sin A sin B sin c sin PT = sin C cos PC —sin A cos B cos PA—cos A sin B cos PB (b) If P coincide with B, sin C cos a = cos A sin B + sin A cos B cos c . . (2) 149°. Let Q be the centre of the small circle circumscribing the triangle ABC ; let QA = QB = QC = R ; and let QT be the arcual perpendicular from Q upon AB. Let the spherical excess be E, and A -f B -f C = 2S ; so that 2S = A + B + C = tt + E. Then, if the point P, equation (b), coincide with Q, that equation becomes sin A sin B sin c sin QT = sin C sin QC—sin A cos B cos QA—cos A sin B cos QB (1) Since QA = QB = QC, it is easy to see that 2 / QBT = A + B - C = 2 (S - C), and Z QBT = S - C ; and since BTQ is a right-angled triangle, we have, by (4) of 143°, sin QT = sin R sin QBT = sin R sin (S - C).SPHERICAL TRIGONOMETRY 137 Also the right member of equation (1) is cos R [sin C — sin (A + B)} =cos R {—2 cos v sin (S — 0)} „ =2 cos R sin sin (S — C). Therefore equation (1) becomes sin A sin B sin c sin R sin (S — C)=2 cos R sin \ E sin (S — C) sin A sin B sin c and cot R = ■ 2 sin \ E (2) 150°. Let CC' be a perpendicular from C upon the plane AOB ; to find y' = 00', the orthogonal projection of y upon that plane. __ y = y — C'C = y + A. sin COB. But sin COB = sin CP3 = sin a sin B, (5) of 143°. Therefore, substituting for y its value in equation (1) of 148°, we have , a sin A cos B + 8 cos A sin B — A sin A sin B sin c , A sin c sin a sin B y — —----------------------------------H---------------- sin O sin c _ a sin A cos B + 0 cos A sin B ” “ sin 0 .............................K } 151°. The three arcual perpendiculars, AP^BPs, CP3, let fall from the corners of a spherical triangle upon the opposite sides are concurrent. CP3 and APj must intersect in some point P, OP being consequently thejine of intersection of the planes C0P3 and A0Pt. Draw OP3 = p3 and OPj = p1# Then, y', the ortho- gonal projection of y on the plane AOB, and a', the orthogonal projection of a on the plane BOC, lie respectively along p3 and pj. Let p3 = yy, p} = za ; and let L, M, N be respec- tively the positive poles of AB, BC, AC. Then, by 79°, x OP = Y . VypaVp!*!. Vy/»3=yVn-((^^ 150°)yVya-in AcosB+cos Asin B sin C v cos A cos B / , t, , , AX „ = — ^—-----(a tan B -f v tan A) . . . . (1) sin C 7 y a __ zVa'a___ z Yy0a sin B cos C + Vya cos B sin C 1 sin A z cos B cos C sin A (v tan C + A. tan B) . . . (2)138 SPHEKICAL TEIGONOMETKY Therefore, (1) and (2), Y . VyPsVPla = Acos2Bc°s 0 ( tan B tan 0 + ft A tan2 B + v\ tan A tan B) ' sin 0 sin A „ = \yz cot A sin 2B cot 0 (a tan A + /3 tan B + y tan C) = x OP ; ... - yz cot A sin 2B cot C or, putting p for -------------------, OP = p (a tan A + tan B + y tan C). If the expressions for the vectors along the lines of inter- section of the planes AOP! and BOP.2, BOP2 and COP3, be worked out, they will be found to be of the form : qq = q (a tan A + ft tan B + y tan C), OR = r (a tan A + tan B + y tan C). Therefore,^, q, r being scalars, OP, OQ, OR are parallel; and, consequently, the terms of their unit-vectors coincide. There- fore the three altitudes, which pass through the terms of these unit-vectors, are concurrent. 152°. What is the geometric signification of the symbol (3a~ly, when a, /3, y are vectors drawn from the centre of a unit-sphere to the corners of a spherical triangle, ABC 1 Let the sides, as usual, be a, b, c, and let cos a = Sy/3~l = — Sfiy = l; I COS b — Say”1 == — Sya = m ; > . . . . (1) cos C = S/?a~1 = — Sa¡3 = n. J Let it be supposed that l, m, n are each greater than zero, or that each side of the triangle is less than a quadrant. Let 8, e, £ be three vectors, such that 8 = Y/fo^y = VycT1/? 1 € = Yy/?-la = Ya/^y S .... (2) £ = Vay-‘£ = Y/3y~'a J Then, 78°, Yj3a~ly = /?Sa_1y — a-1S/?y + yS/?a-1 „ = aS/?y — y8Sya — ySa/?, since a is unit-vector, ,, = m/3 + ny — la ; and similarly for Yyfi~la and Yay-1/?. Hence, 8 = m/3 + ny — la, 1 € = ny + la — m/3, >...... (3) £ = la -f- m/3 — ny. JSPHERICAL TRIGONOMETRY 139 To find the lengths of these vectors, o2 = — T2c) = (m/3 + ny - la)2 = - (l2 + m2 + n2 - 2lmn). It will be found that e2 and £2 have each the same value. Therefore, T8 = Te = T£ = s/{l2 + m2 + r*2 - 2Imn) = (say) t . . (4) This common length, t, is less than unity. For, let V — Sct/3y = — Sa_1/?y = S/3a ly. Then, since t = TS = TV/?a-1y, we have t2 + v2 = (TY/3a~ly)2 + (S/3a-1y)2 = (S/Sa^y)2 - (V/fo^y)2 = T2(3a~ly = 1. Now, v is different from zero, because the three vectors are diplanar. Therefore t < 1. Dividing the three vectors by their tensor, t> we obtain three unit-vectors : OD = r1» = US; OE = rxc = Ue ; OF = = U£ whose terms are the corners of a new triangle, DEF, upon the sphere. We have now to inquire what relation this new triangle bears to the original triangle, ABC. By (3), c + £ = 2la ; Therefore, Z-1€ + l~= 2a ; „ OA bisects the angle between the unit-vectors OE and OF. Consequently, the point A lies upon and bisects EF. Similarly, B „ „ „ „ FD, C „ „ „ „ DE. Fig. 33 shows the two triangles. To establish a relation between the sides of the two tri- angles, let EF = 2a! ; FD = 2b' ; DE = 2c'. Then ¿Uc + =e-{-£= 2la = 2a cos a, (1). Dividing across by a and taking scalars, ,/cUc , ,UA_0________________140 SPHERICAL TRIGONOMETRY But therefore, S — = S —, since EF is bisected in A ; 21S = 2 cos a ; a or, cos a = t cos a! 1 Similarly, cos b = t cos V >................(5) cos c = t cos cf J To establish a relation between a, j3 . . e, C From the equality of the angles EOA, AOF, &c., we have at once ac = £a ; ££ = 8/3 ; yS = cy . . . . (6) If we write the first equation as ca = a£, 41°, and multiply both sides into a-1, we get e = a£a-1 ; a confirmation of 97°. Had we multiplied by a-1, instead of into a”1, £ = a-1ca. From the second equation of (6), multiplied into ft \ Substituting the above value of £, 8=/3a-'(a/3-' = P<% = q Ypa~ ly = TV/?a" ly . UVjSa" !y = tU8, 152° (4) and (2). Therefore, fia~ly = v + ¿US..................(2) Equating the scalar and vector parts of (1) and (2), •cosAE = tf...................(3) sin |E = v. But v, or Sa/?y, is positive or negative according as rotation from a to /? round y is negative or positive, 95°. In general, therefore, sin t>E = + ?;= + &a/3y .... (4) This equation is the quaternion expression for Keogh’s theorem : The sine of half the spherical excess is the volume of the parallelopiped, the three edges of which are the radii drawn from the centre of the sphere to the middle points of the sides of the triangle (DEF). Bearing in mind that by 152° (3), *(« + {). ) /?= hm-ur 4- xv l 0=£m-'(£ + S), [ . y = + «);) '■ (5) we have, (4), sin JjE = Sa/?y _ S(e + *)(£ + 8)1S + e) _ S8eC 8 Imn ±lmn • (6) and, (3), i-c 4 lmn± • ("> therefore, to JE - **4 4 Imnt . (8)SPHERICAL TRIGONOMETRY 145 But, since m = — Sya, (1) of 152°, 4Imn = — 4?i£Sya = — S (2ny. 2Za) = — S (S -f c) (c + £), by (5), „ =~S(€2 + cC-f^ + Sc)=^-S(^+^ + Hby(4)ofl52°. Therefore, if _ i!M_ _ ^ 2 ~ klmnt t3 - iS (e£ + £S + 8c) _ SUScf » 1 _ SUC£ - STJ£8 - S USV ’ . . (9) . . (10) a general expression for the tangent of half the spherical opening at O of any triangular pyramid, ODEF, whatever be the lengths of its edges, T8, Te, Tf. Applying equation (9) to the triangle ABC, we have at once, tan JE = _______Sapy_____= _.sin a sin h sin 0--_ by (4) of u3o . . . (11) 1 _ S/3y — Sya — Sa/9 1 + cos a + cos b + cos c 154°. In (20) of 152° we obtained a versor whose angle was half the area of the triangle DEF, (1) of 153°. To obtain a versor whose angle is the area of this triangle, we have only to take the negative square of equation (1) of 153°. Then - (£a- V)2 = - (sin JE + US cos IE)2, or, ~‘i~ = cos E - US sin E ... . (1) afiy Articles 152°, 153°, 154° are almost entirely from Hamilton. 155°. The Chordal Triangle of a spherical triangle is the plane triangle formed by joining its comers by cords of the sphere. Let ABC be a spherical triangle such that Z C= A + Z B; then the chordal triangle is right-angled at C. For, draw an arc of a great circle from C, cutting AB in D, so that Z DAC = Z DCA. Then, with the usual notation for a spherical triangle, but calling the three equal arcs, DA, DB, DC, t, we have 1 + cos 2t = 2 cos2t ; and o = sin2£ (cos CDB + cos CDA) ; L146 SPHERICAL TRIGONOMETRY adding, 1 -f cos 21 = (cos2£ + sin2£ cos CDB) + (cos2tf + sin2£ cos CDA) „ =cos a + cos 6, 141°, (3) ; — cos 21 H- cos a +■ cos 6 = 1= OC2 = — y2 ; Sa/3 — S/ly — Sya + y2 = O ; s (a — y) (P - y) = 0 > therefore, (chord) CA J_ (chord) CB, and the chordal triangle is right-angled at C. 156°. To find the angle at which two opposite sides of a spherical quadrilateral meet when produced, in terms of the sides and diagonals (Gauss). Let ABDE, fig. 33, be the quadrilateral, and let BA and DE produced meet in L. Let 0 be the angle between o- and r, the axes respectively of fia and eS ; the angle between the planes OAB and ODE, that is, the angle at L, being consequently (tt — 0). Then, 84°, S . V/?aVeS = S£8Sac - S/3cSaS = cos BD cos EA — cos BE cos DA. Now, V/3a = 7 a-‘ 08-y)· <* _ P—y ____________________ /3 — a y ¡3—a —y AB * CO whatever the vectors a, ¡3, y may be. Now, in Pt. I., 29°, it was defined that (OABC) = OABC AB CO’ where O, A, B, C are any four collinear points. Let O, A, B, C now be any four points, and?^ being, consequently, qua- ternions. Definition. <0ABC>=!i.....................(4) is the Anharmonic Quaternion Function of the group of four points, O, A, B, C, or of the (plane or gauche) quadrilateral OABC (Hamilton). We may therefore write equation (3) as, = (OABC)..........(5) If OA', OB', OC' be the reciprocals of OA, OB, OC, fig. 35, (OABC)=Kg=|J=KgJ.....................W In the particular case when A', B', C' are collinear, (OABC) is a negative scalar.150 THE CIRCLE ____160?. If OA', OB' be the reciprocals of any two vectors, OA, OB, fig. 35, then A'B' || OT, the tangent at O to the circle passing through 0, A and B. For, by 96°, OT = /? (a — /?) (— a) = a?p - Wa. B'A'—a-1 — P Ta +~T/3~a?b* Therefore, B'A' || OT. If a circle be circumscribed to OA'B', the tangent to this new circle at 0 will be parallel to BA. 161°. If any three coinitial vectors, OA, OB, OC, be chords of one common circle, the terms of their coinitial reciprocals, OA', OB', OC' are collinear, fig. 35. For it has been shown, 160°, that B'A' || OT, the tangent at 0. And it may similarly be shown that C'B' || OT. Therefore A', B', C' are collinear. The indefinite straight line A'B' is evidently the locus of the terms of the reciprocals of all the vector-chords of the circle which have O for origin. Conversely, if the terms of three vectors, OA', OB', OC', are collinear, their coinitial reciprocals, OA, OB, OC (if not parallel), are chords of one common circle, or the points O, A, B, C are coneyclic. Let /_ OAB = 6 ; BCO = . Then, since A', B', C' B'C' are collinear, is a negative scalar, — t, and (OABC) = B'A' OA AB AO AB Therefore, U BC CO CO CB = K = K (-*) = -*; B'C' B'A' co> CB; =<(-§§> £§=D(-i)=* OC CB > €C6 = 0 -f- cj> = 7T. andTHE CIRCLE 151 Therefore, the quadrilateral OABC is inscribed in a circle, or the points O, A, B, C are concylic. It is clear that for any circular group, O . . . C, we have UA-° = AB U oc CB (1) the upper or lower sign being taken according as the quadri- lateral OABC is uncrossed or crossed. And conversely, if we are given such an equation as (1), connecting a group of points that are not collinear, we know that the group is circular. 162°. Let (OABC) = k|^=_1. From this equation it follows that A', B', C7 are collinear ; that O, A, B, C are concyclic ; and that A'B7 = B'C'. Con- sequently, /3" = i(v~l + «-')................0) The vector ¡3 is defined to be the Harmonic Mean between the two vectors y and a. Multiplying 2ft into (1), 2 = P(y~' + a~‘); and» P = _i l _i- y 1 + a 1 Therefore, ¡3__ 2 _ 2 _ 2a # /.)\ y (a-1 + y~l)y oTly + 1 y + a 9 P= 2 = 2 - 2r . . (3) a (a 1 + y *)a 1 + y Ja y-f a From (2) and (3), 2a 0 2y ■------y = p =------------a y + a y + a (4) If E be the middle point of the chord AC, fig. 35, y -j- a — 20E = 2c,. . . . . . (5) and -y = ^==^ a....................(G) € € Therefore, as in algebra, the harmonic mean between any two vectors is the fourth proportional to their semisum and themselves.152 THE CIRCLE By (5) and (6), ft - « = \y-h(y + «)= (7)5(2«-a) - ; ft — £__j _ 2a a2_____A — a\ 2 — €— € £2 _ V £ / Therefore, §2 = 12! and EC2 = EO . EB, ... (7) EO EO2 or, EC is the mean proportional between EO and EB. Conversely, if any three vectors, EO, EB, EC, be in continued proportion, and we draw EA = CE ; the points O, A, B, C will form a Circular Harmonic Group. The points A, P, B, P', fig. 34, are an example of such a group. If (OABC) = -1, (OBAC) = = 2K — = 2 ; v ' A'B' A'B' ' and (OBAC) = 22 ^ ' BA CO Therefore, taking tensors, OC . BA = \ (OB . CA). Similarly, CB . AO = „ „ Therefore the rectangles under the opposite sides of an in- scribed quadrilateral are each equal to one half the rectangle under its diagonals, if (OABC) = — 1. Let F be the cross of the tangents at O and B, fig. 35. Then it is easy to prove that F lies upon AC produced. Simi- larly, the cross of the tangents at A and C lies upon OB produced. Therefore the diagonals, OB and AC, are Conjugate Chords,—each passes through the pole of the other. 163°. If ABCD be a quadrilateral, plane or gauche, / vb^P)-AB cd-ABBC CDDA_AB.BC.CD. da ( ' BC D A BC BC DA DA BC2 . DA2 ; or, v2 (ABCD) = AB . BC . CD . DA = continued product of the sides,.....................(1) where v2 = BC2 . DA2 is a positive scalar.THE CIRCLE 158 If the quadrilateral be plane and inscribed in a circle, the anharmonic function, (ABCD), 159° (4), is a scalar, m, which is positive or negative according as the quadrilateral is crossed or not. Hence, in this case, (1) becomes AB . BC . CD . DA = ± mv2 = ± t . . (2) In general, the product of the successive sides of an w-gon inscribed in a circle is a scalar if n be even, and a vector- tangent to the circle at the initial point of the n-gon if n be odd. 164°. Someof the equations of the circle have been given in 101°. If OE = c be any given unit-vector ; OK = k, the vector of any given point in the plane through E 1 t; and KA = a, a constant vector in that plane ; then (p-«)» = «»; s e = l.................(1) is the equation of a circle passing through A, with K for centre. If T/c = c, these equations become t>* - 2Sκρ = c2 - a2; S 2= 1 . . . . (2) € If O be on the circle, p2 — 2Sκρ = o ; UVkP = η..............(3) where η is some other given unit-vector. The first equation of (3) may be written : s p (p — 2k) = o; therefore the angle of a semicircle is i π. 165°. Let OD = δ be a diameter of the circle, fig. 35 ; and let DO produced cut C'A' in D'. Then, 161°, OD' = δ-1 ; and since A'D' || OT, if τ be the vector along OT, γτ(δ-ι~α~Ι) = ο. Therefore, Υτδ ”1 = Υτα ~1, or, since τδ-1 is a right quaternion, τδ"1 = Υτα"1, Κτδ"1 = δ-V = ΚΥτα-1 = - Υτα"1, and δ = Υτα”1.......................... the expression for the diameter passing through O.154 THE CIRCLE 166°· It has been assumed that the tangent is perpendicular to the radius drawn to the point of contact. This may be shown by differentiating the equation, p2 - 2SkP = o, (3) of 164°. Then 2S pdp — 2Skg?p = o, S (p — i<)dp = o. Therefore the radius drawn to the point of contact, p — k, is perpendicular to dp, the vector-tangent. 167°. Let O A = a be the vector of any point upon a circle whose centre is O, and let p be the vector of a variable point upon the tangent at A. Then, vj being a given unit-vector perpendicular to the plane of the circle, Sa (p — a) = o, UVap = rj,\ /■« v or, Sap = — a2, UVap = rj, ƒ ’ ’ ’ W are the equations of the tangent at A. If the circle be a great circle of a sphere, equation (1) is the equation of a tangent to the sphere at A. Sap = — a2......................(2) is the equation of the tangent plane to the sphere at A, since it represents all the tangents that can be drawn at A. 168°. From O, the centre of a given circle, draw a straight line, cutting the circle in A, to a given external point, E ; let T, T' be the points in which the tangents from E touch the circle ; and let TT' cut OA in D. Let OA = a, OD = 8, OE = e, and let p be the vector of a variable point in TT'. Then D and E are inverse points with respect to the circle, and OD . OE = OA2 ; OD = e Therefore, S = T8TJ8 = -Uc = a%................(1) e e2 But o = SS (p — 8), or, S8p = — d2. Substituting in this last equation the value of 8, (1), °L e2’ Sep = — a2. andTHE CIRCLE 155 Hence, if rj be a given unit-vector perpendicular to the plane of the circle, Sep = - a2 ; XJYcp = V............(2) are the equations of the chord of contact of the two tangents from E. If the circle be a great circle of a sphere, Sep = - a2...................(3) is the equation of the polar plane of E. 169°. The join of the points of intersection of two circles is perpendicular to the join of their centres. Let the circles intersect in A and A' ; let their centres be K and K; let KK' = y, AA' = 8. Then, (KA - y)2 == KA2 = KA/2 = (KA' - y)2, and S (KA . y) = S (KA' . y) ; S (KA' - KA) y == o, Sy8 = o. Therefore, y _L 8. Section 5 Conic Sections 170°. To find the locus of a point such that the ratio of its distances from a given point and a given straight line is constant. Let F be the given point and QR the given straight line, fig. 36. Let P be any point, and let PQ and FR be perpen- dicular to QR. Let FP = p ; FR = p, RQ = yv ; PQ = xfi. This last assumption limits the locus of P to the plane con- taining QR which passes through F. Then, if e be a scalar representing the constant ratio, e=_Tp_ ^==_e! T . PQ ' T2. PQ zV p2 —. FP + PQ = FQ = FR + RQ, p + xp = p + yv ; Spp + wp2 = p»2 + 2/Spv „ = p2, since Spv = o. S . p x.156 CONIC SECTIONS Therefore, But *V == (z^2 — ®w)2> xi _ _el. 35 “«V’ therefore, /¿2/o2 = e2 O'*2 ^W)2................(1) the general focal equation of a Conic Section, which is an ellipse, <■ parabola, or hyperbola according as e = 1. The point F is a focus, and QR is the directrix. 171°. To find the points in which the conic cuts p, or FR, fig. 36. M Let XfJL be substituted for p in the general equa- tion, and we get a= ,or· 1 d- 6 Therefore, 1 - e FM = ep ; fm' = ep 1—6 M'M = M'F + FM = (r*— + T-T—) /* = \1 — 6 1 + ej I — el hence, //,= 1 - 62: 26 ■ M'M = 1 -62 26 -T . M'M . L>. Let T . M'M = 2m, and 1 — P* T/j. = FR = 1------------m. T . FM = T . FM' = 1 + e e Tp = (1 — e) m. 1 _ gT/x = (1 + e) m. Let C be the middle point of MM'. Then T . CF = m — (1 — e)m — era. T . CR = era + 1 - 62 -m = —. e T. MR = — — m = ---------~m·, e eCONIC SECTIONS 157 Collecting these results, we have, for the ellipse and hyper- bola, OF = em = CF' ; CR m FM' = (1 + e)m = F'M ; FM = ±(1 -e)m = FM'; MR = ± 1 —- m ; the positive signs being taken for the ellipse, the negative for the hyperbola. For the parabola, 6 = 1, and x = i or go. Therefore the point M', and consequently C, is at infinity ; and FM = = MR. 172°. To transform the focal equation of the ellipse and hyperbola into their central equation. Let the focal equation be written, H-iW = eiW- sMiPi)2> as we are about to change the origin, and thereby change the meaning, of //, and p. Let CM = n ; CP = p, fig. 36. Then, - 1 — e2 Mi = fr =—— m; Pl = FP = FC + CP — p — ep.. Substituting these values of ^ and p{ in the general focal equation, we get - *2);............(1) the general equation of a central conic. This equation may be written, 1 _ mV — e2SVp _ mV — «2Sp/*S/xp _ - eVs«A x- m4(i-«2) p\ m4(1-«2)7'158 CONIC SECTIONS But the quantity in brackets is of the same form as the right-hand member of equation (1), 133°. Therefore, if we put uL2a — e2pS pp} we get, as the equation of a central conic, Sp<£p = 1. . Let p be parallel to /a, or xp = /a. Then If p Jl /a, or S/a/3 = o, cf>p = 9P p _ p V(1 -e2)' (2) (3) Hence the coinitial vector p is parallel to /> when p is parallel to either the major or minor axis of the ellipse, or the transverse or conjugate axis of the hyperbola. 173°. To find the points in which the conic cuts the line NN' drawn through C perpendicular to MM', fig 36. ppp2 — e2S2/Ajo = p.4( 1 — e2). But in this particular case S/Ap = o. Therefore, P2 = (! — e*)...................(1) Let n be the tensor of p, and we have n2 = m2 (1 — e2) and 2 m2 — n2 e = ----2— m£ (2) (3) Obviously, n2 in (2) is positive for the ellipse and negative for the hyperbola. Further, in this latter case, since e> 1, n = 4; m \/(l — e2) gives an imaginary value for n; or, the conjugate axis does not meet the hyperbola in real points. CN = CN' = n, the square root of n2, without regard to sign ; and NN' is called an axis of the curve. The vector CN will be designated by v, for both the ellipse and the hyperbola. 174°. It has been shown that, if p be the vector of a planeCONIC SECTIONS 159 curve, dp is a vector successive to p, along the tangent at P, 113°. Differentiating the equation Spp = 1 we get G?Sp<£p = Sc? (pp) = S (dp . (ftp) + S (p . dp) = o. But S (dp . p) = S (p . 4>dp) ; and S (p . dcj>p) = S (p . <£c?p) ; therefore, S pdp = o. Now, since o?p is parallel to the tangent at P, if CD (= 7r), tig. 36, be the vector of any point upon the tangent, 7r = p + xdp and c?p = - (7T — p). x Therefore, Sp<£ (ît —■ p) = o = S (7r — p) <£p, . . . (1) Or, S pTT = S7Tp = S pcf>p =1. . . . (2) Equation (2) is the general equation of the Tangent of a central conic. Since 7r — p is parallel to the tangent, equation (1) shows that <£p = CY is perpendicular to the tangent, or parallel to the normal at the point of contact, fig. 36. 175°. The locus of the middle points of parallel chords of a central conic is a straight line. Let any number of chords be drawn parallel to any given diameter, 2y, and let cr be the vector of the middle point of any one of them, 2xy. Then the vectors of the extremities of this chqrd, a + xy ; ( (a- + xy) = + xy, 135° (1), (2), So-y + Syy. Therefore the locus of the middle points of all chords of a160 CONIC SECTIONS central conic parallel to any diameter, 2y, is another diameter, say 2?, which is at right angles to <£y, or parallel to the tangent through either extremity of 2y. 176°· By the last article, if vx be a diameter which bisects all chords parallel to /ij, S = o. But = S/x^vi = o ;.................(1) therefore bisects all chords parallel to vx ; and as v, is parallel to the tangent at the term of /xj, so is parallel to the tangent at the term of vx. Diameters which possess this property are called Conjugate Diameters. 177°. The system p = VacSfji * Ta = 1 ) Sa/x O, . . . . (1) represents a plane ellipse, 0 being the eccentric angle and fjL and Ya/x (= v) the major and minor axes, 103°. p = Yac*/x = Y (cos 0 -f a sin 0) /x = Y (/x cos 0 + a/x sin 0) „ = /x cos 0 + v sin 0.....................................(2) Differentiating (1), = ........................................(3) du = ValCwa€0fi = Ya (cos 0 + a sin 0) yn = Y (a2/x sin 0 + a/x cos 0) = — /x sin 0 + v cos 0 . k (4) Since the value of ^ in (3) (which is parallel to the conjugate d0 of p, 176°) is the value of p in (1), when 0 becomes (0 + \ ir) ; it follows that any two expressions for p in which 0 differs by r, represent conjugate diameters. For example, if we sub- stitute (0 -f \ir) for 0 in (2), we obtain (4). 178°. From (2) and (4) of 177°, 4TYp = 4TY (/x cos 0 + v sin 0) (— /x sin 0 + v cos 0). d0 „ = 4TY/IV. In words, the area of the parallelogram circumscribing an ellipse and touching it at the extremities of conjugate diameters is constant.CONIC SECTIONS 161 179°· Let DD' be any diameter of a central conic. Then, if E be any point upon the conic, DE and D'E are Supplemental Chords. ' ____ Let C be the centre of the conic ; CE = p ; DD' = 28. Then, ___ _______________ DE = 8 + p . ED' = 8 — p, and S . (8 + p) <£ (8 — p) = S . (8 + p) (<£S — <£p) = SS<£8 — S 8<^>p + Sp<£8 — Sp^>p. But — S8<£p + Sp<^>8 = — Sp<£8 + Sp<£8 = o ; and S88 — Sp<£p = 1 — 1 = o. Therefore, S . (8 + p) <£ (8 — p) == o. Therefore, if € and £ be two diameters parallel respectively to the supplemental chords (8 + p) and (8 — p), Se£ = O. Therefore diameters parallel to supplementary chords are conjugate. 180°. From T, any point exterior to a central conic, draw tangents touching the conic in P and B. Let C (as usual) be the centre of the conic ; draw PB, cutting CT in Q, and let CT = 7r. Then the equation of the tangent, 174°, Sp(f)7r = 1, is satisfied by the values of p for the points P and B. And this equation is also the equation of a straight line, 99°. It must, therefore, be the equation of the straight line passing through the points P and B—i.e., the equation of the chord of contact. Writing o- for p (to avoid confusion with the p of the conic), we have So-7T =1.....................(1) as the equation of the chord of contact, or of the Polar of the point T. 181°. To find the locus of T, the cross of the tangents, if the chord of contact always passes through a fixed point, A. Let o- be the vector of the point A. Then, since A lies on the chord of contact, S (T7T = 1, and Sttc/mt = 1.162 CONIC SECTIONS Therefore, 99°, the locus of T is a straight line perpendicular to <£o·—i.e., parallel to the tangent at the point where CA produced meets the conic. 182°. The equation, p = ta -f t~l/3 . . (1) represents a hyperbola, 102° ; a and ft being unit-vectors along the asymptotes CY and CZ respectively, and t and t ~1 the Cartesian co-ordinates, fig. 37. For the tangent at G, r = dp = (a — t ” 2/J) dt, or, %=*-·-'»■ · <2> If it be the vector of a variable point upon r, 7r = p-= ta +1~ lj3 + Xa — xt 2¡3 - (t + x) a + t~2 (t—x) y3. ctt If 7T = CY, t~2 (t — x) = o ; or, x = t, and CY = 2ta. If 7T = CZ, t + a? = o ; or, x = — t, and CZ = 21~ l/3. Therefore, GY = CY - p = ¿a - t~ >y8 = p - CZ = ZG . . (3) Therefore the intercept of the tangent between the asymp- totes is bisected in the point of contact. Obviously, any diameter, CG produced, bisects the inter- cept between the asymptotes of all lines drawn || the tangent at its vertex ; or, UQ = QW. 183°. From (2) of 182° it follows that, if any diameter, CG, be the intermediate diagonal of a parallelogram whose coinitialCONIC SECTIONS 16S sides, CA, CB, lie along the asymptotes, the other diagonal, AB, is parallel to the tangent at G. For, since CA + CB = CG = ta + t ~1 /?, we have CA = ta ; CB = t~lfi, and BA — ta — t~l/3. But, 182° (2), = a - r2 P = r1 (ta - r'/3) = r'BA. Therefore, BA || dp. 184°. It is evident that if we complete the parallelograms CGYD and CGZD', DD' and CG will be conjugate diameters ; or, the asymptotes have the same direction as the diagonals of parallelograms whose adjacent sides are any pair of conjugate diameters. 185°· Any diameter of a hyperbola, CG, bisects all chords parallel to the tangent at its vertex (G) ; for example, RP which is cut in Q by CG. For, let CP = va + v~]/3. Then, va + v-'/3 = CQ + QP=æ (ta +1~1 P) + y (ta -1~'/3) 182° (1), (3). Therefore, t (x + y) = v = .. , x - y and x2 — y2 = 1. Therefore, for every point, as Q, determined by x, there are two points, It and P, determined by the two corresponding values of y, which values are equal with opposite signs (Hardy). Since UQ = QW, 182°, and PQ = QR, it follows that UP = RW, or, the intercepts of the secant between the hyperbola and its asymptotes are equal. 186°. The area of the triangle formed by the asymp- totes and any tangent is constant. For, since CY = 2ia, CZ = 2t~l/3, we have at once, Y (2ta . 21 ~ lfü) = 4Va/? = constant vector-area. 187°. The general focal equation of the parabola is, (1) of 170°, p.y = o** - sw)’,......................(i) M 2164 CONIC SECTIONS which may also be written, 4 v'} =1· Let p = ...............(2) /* and the equation of the parabola becomes SP (4>P + 2/t-1) = 1...............(3) Operating on equation (2) with 8 . fix, S fuj>p = S fip — S fjp = o (4) Therefore <£p, which is coinitial with p, is _L the axis of the parabola. S.pX, Spp = ^ = 1 _ 2SM->p. . (5) /* 188°. Differentiating equation (3) of 187°, o = S (dp . p) + S (p . dp) + 2Sprldp „ = S (p . dp) + S (p . <£dp) + 2&fi~1dp „ = 2S pdp + 2Sp“1e?p, Or, SpC?p + Sp-1d!p = o. If 7r be the vector of any point upon the tangent, 7r = p + xdp, and dp = —P. X Hence, - {Sp<£ (tt — p) + Sp-1 (7r — p)} = o, x and Sp7r — S pp + Sp_l7r — Sp-1 p = o, . . (1) the focal equation of the tangent. Since, equation (5) of 187°, Spp = 1 — 2Sp-1p, and Sp<^>7r = S7r<£p, S7r(<^p + p-1) + Sp_1p = 1, .... (2) another form of the equation of the tangent. Equation (1) is evidently equivalent to S (ti- — p) (0p + p"1) = o. But (7r — p) is a vector along the tangent. Therefore, <£p + p (3)CONIC SECTIONS 165 is a vector along the normal ; and if p + p~l).................(4) 189°. Let PA be _L /x (= FR), fig. 38, and let equation (2) of 187° be written, p = FA + AP = pr !S/xp + p.2<£p. Since FP = p, and p _L p, (4) of 187° ; this equation gives us FA= /X_1S pp y AP = MP · · (1) Since tt in equation (2) of 188° is any vector drawn from the focus to the tangent, it may repre- sent FT = xp. Substitut- ing this value of tt in that equation, we have x =1 — S /x_1p. Hence, 7T = FT = Xp = p — pS/X ]p. MT = FT — FM = \p — /xS/x xp — Jr/x p 1S/.xp „ = (by (1)) FM - FA = AF + FM; therefore, MT = AM........................(2) Since FT = p—/x-1S/xp, we have (FT)2 = (p. - /x'1 Sw)2 = + SV A1 „ _ W = ^ (!) of 187o Therefore, FT = FP........................(3) Consequently, if PD be a line parallel to the axis, meeting the directrix in D, the tangent PT bisects the angle FPD ; and FD is perpendicular to, and is bisected by, the tangent.166 CONIC SECTIONS By (3) of 188° (p + /a-1) is a vector along the normal. Therefore, _ ___ ______ + /*-') = PNT = PA + AN „ = — mVp + (!)· Therefore, (y + yP·) p = (— y + zy2)fi-1; » 2/ = -p2; y = «p·2; ,, « = - 1 ; „ _ NAj=/^=FR...........................(4) Further, since NA + AP = NP, NT == fx + p?p = FR + AP == FR + RD = FD, . . (5) and ___ ____ \ (p + />i2^) = |FD = FB = FM + MB. Therefore, MB = \ /¿2<£p,..................(6) and MB is parallel to and equal to half AP. It follows from (4) that the subnormal of a parabola is constant. The figure FDPN is evidently a rhombus. 190°. To transform the focal equation of the parabola to the equation with the vertex for origin ; let the focal equation be written, Pi2Pi2 = (Pi2 — Sp.Pi)2, as we are about to change the meaning of both p and p. Then, ___ px = FR = - 2MF = - 2p, Pl = FP = FM + MP = p — p. Substituting these values of px and px in equation (1) of 187°, p2 — 4S/xp — p~2S2pp = o, . . . . (1) the equation of the parabola with the vertex for origin. This equation may be written, Sp (p — p~l&pp — 4p) = O. Let (j>p = p — p~l$pp,...............(2) and we have, for the equation of the parabola with the vertex for origin, SP (4>P — 4p) = o...................(3) As before, Spp -= o ;...................(4) but, in this case, Spcf>p = (<£p)2...................(5)CONIC SECTIONS 167 The equation of the tangent will be found to be S7rp — 2/x) — 2Spp = 0, | · · · v / and the vector along the normal p — 2p.....................(7) As a verification of equation (1) ; let i and j be unit- vectors along the axis and tangent at the vertex, and let p = mi. Then p = xi + yj ; Spp = — mx ; and, y2 = 4ra#,.......................(8) the Cartesian equation. 191°. To find the locus of the intersection of the normal and the perpendicular upon it from the focus, FS, fig. 38. Since FS_LPNT, if FS = 0-, Spo-=P + /t"1), by (5) of 190°. . (2) S . p X. Sp,o- = Spp — \p2 (Spu/>p + 1). But Spp = o ; therefore, Spp = Spa -h \p2.......................(3) S . p X . Sper = p2 — (Spp + S/x-1p). Subtracting (1) from this equation, p2 - o·2 = |p2 (Spp + Sp-'p). Now, (5) of 187°, Sp+P = ft=^!S. p Therefore, p2 - cr2 = ¿P2 P? 2S/x (t 1 /q\ » = -—1—> by (3)· Multiplying this equation by p2, and transposing, p,V2 -f- ^----2/xS/xcr = iX2p2=.^A_2p2Spp + S2/xp, (1) of 187°, „ = p*-2p\Spp. S7r = <«’. THE CISSOID 169 Substituting these values of S/jwt, S7rp and &/ip in (6) of 190°, \2 __ 1 - 2x ('*P)2 Squaring (1), (^)2=S-v· Therefore, 7r2 = 2fl?7r2 + 4x2/x2. Substituting the value of # from (2), /X27T2 a (S/A7T — 7T2) S/X7T j the equation of a cissoid, as will be shown in the following article. Section 6 Various Curves 193°. The Cissoid of Diodes is the locus of a point, P, fig. 39, where the radius-vector of a circle, OP, is cut by an ordinate so that OQ = SB. ___ ___Let C be the centre of the circle ; OC = a ; OQ = ya ; OP = p ; OR = xp. Then, since OR — OC =~CR, (xp — a)2 = CR2 = — a2, and the equation of the circle is xp2 — 2Sap = o . By similar triangles, OQ _ OP OS OR’ 2/“ _ P . 2a — ya Xp ’ hence, (1 + x) y = 2 . From (1), 2S ap * P2 ' Again, p = OQ + S . ax. Sap = ya2 ; and y = Sj?· U a2 Fig. 39.170 THE CYCLOID Substituting these values of x and y in (2), we get 2a2/o2 = (/>2 + 2Sa/o) Sap ; (3) the equation of the cissoid. The cusp of the cissoid, O, coincides with the vertex M ; the point B coincides with R ; and BT, the tangent to the circle ORB, coincides with the directrix of the parabola, 192°, Consequently, the vertex being the origin,__ OC = a(in the foregoing calculation) = ^MR = — i/x, in 192°. Substituting this value of a in (3), we get /x2p2 = (sm> - p2) spp ; the equation there deduced. 194°. The Cycloid is the path described by a point on the circumference of a circle which rolls in a fixed plane upon a fixed straight line, called the base. Let APR be any position of the generating circle, fig. 40, A being the point of contact of the circle with the base, AB. On the base take the point O, such that AO = arc AP. Then, obviously, O is the position of the tracing point, P, when in contact with the base. Draw OD and PQ perpendicular to the base ; draw the diameter of the circle, AC, and CP, AP ; and suppose OP to be drawn. Let CA = c, /. PCA = 6 ; and let a and ¡3 be vectors in the directions OB, OD, such that Ta = Tj3 = c. Then, _ OP = OQ + QP. OQ = OA — QA „ = (arc AP — c sin 0) Ua „ = (cO — c sin 0) Ua „ — (0 — sin 0) a ; QP = (c — c cos 6) U/3 „ = (1 — cos 0) ¡3. p = (0 — sin 0) a + (1 — cos 0) /?, . . . (1) Therefore, the equation of the cycloid. For the tangent at P we have dP = n dO K cos 6) a + p sin 0. (2)THE LOGARITHMIC SPIRAL 171 Let PA = v. Then v = OA — p — Qa — (0 — sin 0) a — (1 — cos 0) ¡3 „ = a sin 6 — (1 — cos 0) ¡3. ^v<40 ~ s{a sin0 — (1 — cos 0)£} |(1 — cos 0) a + jBsin 0} » = s { — c2 sin 0 (1—cos 0) + aj3 sin3 0—/3a (1—cos 0)2 + c2 sin 0 (1 — cos 0)} „ = o. Therefore, vJL or AP is normal to the cycloid at P . . (3) du Let us assume Hamilton’s formula for the vector of curvature, where p and p" are the first and second derivatives of p, and the origin of 7r is the centre of curvature. Then P'3 = (¿j) = {(1 — cos 0) a + 0 sin 0}3= — 2c2 (1 — cos 0) J(l-cos 0) a+/3 sin 0} ; p" = a sin 0 + 0 cos 0; "Vp,fpf = c2 (1 — cos 6) c, where c is a unit-vector coinitial with and _L a and (3, such that rotation round it from a to /3 is positive ; ir = — 2 {a sin 0 — (1 — cos @) fi} = — 2v. . . (4) Therefore the length of the radius of curvature is twice the length of the normal, PA. 195°. If a vector, p, revolve round its origin in a plane, and if equal angular motions of p correspond to equal multiplica- tions of Tp ; then the locus of P is a Logarithmic Spiral, and its equation evidently is /> = a<0;....................i1) Ta 5 1 ; Sa/J = o. For, let OB, OC, &c., fig. 41, be a number of equiangular rays, BOC = COD = DOE = &c. = 0 ; let OB = /3 ; and let a be a vector perpendicular to the plane BOE, drawn- from O towards us. Then OB = p0 = a°/3 = /3, OC =Pl= ac0/3 = ac6T/3 . Uac*Uft OD = p2 = o,™(3= a**Tp . Uac(*+*}U/3, OE == p3 = a86''/? = aSc6T/3 . JJac{9+m V¡3; and ultimately, op = p = dr*p = a%172 THE LOGARITHMIC SPIRAL The scalar exponent may, of course, be either positive or negative. Suppose, for instance, that, as in fig. 41, t = J ; OA = 21 ; OB = 10 mm. Then <*>= L QOR = L POQ = &c. . . = FOG = ¿_ GOII = 30° ; and TPl = OC = TaiTyS = 10(21)* = 10 y/2 ; Tp2 = OD = TaiTp, = 10y/2y/2 ; Tp3 = OE = TalTp2 = 10 v/2\/2\/2 ; and so on. For the points L, M, &c., we have p = or‘/3 ; Tp\ =OL = = i0 . and so on dp = (atfUa* + da1 . Ua*) /?, „ = (121°) (log a + |Ua) a?pdt = (log a + |Ua) pdt, . . (2) the vector-tangent of the spiral. Hence, pdp = p2 log a. dt -f papdt; Spdp == p2 log a. dt (since Spap = o) ; Vpdp = pap . dt = — ~p2adt ; TVprf/1 = 7TP2 dt 2 'THE HELIX 173 Therefore, if /_ pdp = 0, tan 6 = TY(,d<- = --3**__________= -1-; Spö?/j 2p2 log a. dt 2 log a • · (3) or, the angle between the vector of any point upon the logarithmic spiral and the tangent at that point, is constant. It will be observed that in this expression for tan 0, a is the tensor of a versor perpendicular to the plane of the spiral. In the ordinary polar equation, tan 0 = log a a is a constant line in the plane of the spiral. 196°. A Helix is the path generated by a point, P, upon the rim of a very thin circular disc, fig. 42 (a), which revolves (like a wheel) at a constant velocity, and at the same time moves at a constant velocity along a very thin, straight, and fixed wire, OA4, which passes through its centre, the plane of the disc being always perpendicular to the wire. The point P is supposed to take the same time to rotate through one quadrant of the circle that the centre of the disc takes to move in a straight line from O to A„ or A, to A2, &c. Fig. 42 (a) shows the position of B, the term of a given vector ft, before motion has taken place. Let a be a unit-vector in the direction OA,, and let T . OA, = T . A^2 = &c. = c. Then OP, = OA, + A7P„ or, pi = ca + a/3 ; or, OP2 — OA2 + A2P2 ) p2 = 2ca -f a2/3 ;174 and, ultimately, THE HELIX • · (1) p = eta + «*/?>............. Ta = 1 ; Sa¡3 — o, where t is a variable scalar. It is evident that this is a helix upon the cylinder, TYap = TVa/?,.......................(2) of 109°. The mathematical pitch of the screw is the ratio of the rectilinear velocity of the centre of the circle to the angular velocity of P. The mechanical pitch is the rectilinear distance passed through by the centre of the circle in the time that P takes to make one complete revolution of a circle, that is, OA.,|· dp = cadt + t+xfidt ; .... (3) Sa-'dp = cdt ; Ya-'rfp = T^a‘(} ; . TVa-'dp = T/3 = ~ b, since Ta = 1 ; 2 e (4) , /i , , -\j TVa }dp 7rbdt itb /KX tan 6 = tan /a ldp = 01— , , r- = ·— , = —. (5) ^ H Sa-'dp 2cdt 2c v ' The inclination of the tangent, therefore, to the axis of the cylinder on which the helix is traced, is constant. If ÿ - 0, cot = irb 2c (6) is the expression for the constant angle at which the helix cuts any circle traced upon the cylinder. 7rb is the semi- circumference of the cylinder ; 2c = OA2, that is, half the interval between the two spires, B and P4. If a series of vectors be drawn from O, in both directions, parallel to a series of tangents, the locus of the vectors will obviously be a cone of revolution whose equation is, SUa-1(/p = it cos 0...............(7) The normal plane is the plane perpendicular to the tangent at the point of contact. Let BPP', fig. 43, be a helix, and let CD be the normalTHE HELIX 175 plane at P to the tangent PT. Let the plane cut the axis of the cylinder in Q ; let OA = a, OQ = x, a = y, PT = r. Since r is J_ the plane CD, s (y — p) dp = o . , (8) is the equation of the normal plane. Since the equation of the helix is p = eta + a*/?, with the conditions Ta = 1 ; Safi = o ; p2 = (eta + a*/3) (eta + afi) = — cH2 4- eta} + 1 fi + ctatfia + a^afi, „ = — C2t2 eta11 (afi + fio) + VVfi, „ = cH'2- T*a% „=fi*-cH*...............................................(9) Therefore, Spr^p = — cHdt. But, (8), Spdp = Sydp = a;Sac?p = — xedt, (4). Therefore, x — ct; y = eta ; Say = eta2. Also, (1), Sap = Sa (eta + a*/3) ,, = eta2 + Sa< + lfi = eta2. . . . (10) Therefore, Say — Sap = Sa (y — p) = o \ )> y-p = PQla.......................(II) In words, PQ is perpendicular to the axis of the cylinder, or is a normal to the cylinder. It follows that the locus of all the perpendiculars let fall from the helix upon the axis of the cylinder is a screw-surface, or helicoid, bounded by the surface of the cylinder, and con- taining the helix itself ; its equation being of the form, p = eta 4- uatfi,.................(12) where t and u are independent variables. Or, reverting to the conception of a moving circle, we may say that u is a function of the velocity of rotation of the circumference of the circle, and t is a function of the velocity of translation of its centre, the two velocities being absolutely independent. To gain a definite idea of the shape of this surface, we have only to imagine that the cylinder is upright, and that a corkscrew staircase is constructed round the axis. The (smooth) bottom surface of the staircase is a helicoid. Fig.43.176 THE PLANE Section 7 The Plane 197°. Three given coinitial vectors, a, /3, y, terminate in a plane; jfco find the perpendicular from the origin upon the plane, OD = 8. We have at once Sa8 “1 = 1 ; SjffS-1 = 1 ; SyS-1 = 1 ; therefore, 131° (2), £___ ______Sa/3y_____ V (Py + ya -f aft) Since SY (Py + ya + aft) = a scalar, Y (/3y + ya + a/?) |j 8. 198°. To find the condition that four points shall be co- planar. Let the vectors of the four points be a, /?, y, p. If the point P lie in the plane passing through the points A, B, C, which are supposed to be non-collinear, then S · (a p) (P p) (y p) = °, or, S/?yp + Syap -f- Sapp — SaPy = o. The geometric meaning of this equation is obvious. As a verification, let p = xi +yj 4- zk; a — x{i + yj + %xk ; tkc , &c. Then o = x(y1z2 - 2/2-1 + 2/2^3 - 2/3*2 + 2/a*i ~ 2/1^3) + &C.; or, X , . y = 1 * s , 1 » V\ > -1 : , 1 x2 , 2/2 , *2 ! . 1 xs > 2/3 > ^3 , 1 For another solution, see Part I., 38°. 199°. The intersection of two planes is a straight line.1 For let OA = a, OB = p lie in one plane, and OC = y, OD = 8 lie in the other. Then, 79°, the intersection of the two planes is Y . Ya/3Yy8, that is, a straight line. 1 As the point of intersection of two lines is called their cross, I venture to suggest that the line of intersection of two planes might be called their cut.THE PLANE 177 200°. The condition that three planes shall intersect in one common straight line. Let the equations of the three planes, P1? P2, P3, be respectively, 8ap=p; S/3p = q; Byp = r ; and let cr be a vector along their common line of intersection. Then, since a is perpendicular to Pt, it is perpendicular to cr. Similarly, ¡3 and y are perpendicular to cr. Therefore, S aj3y = o, and pV f3y + qVya + rV a/3 = o. This is the condition that the three planes should be parallel to the same straight line. 201°. To find the condition that four planes shall meet in a point. Let the equations of the given planes be Sap = h ; Sj3p = l; Syp = m ; SSp = n. For the point of intersection, the variables of the first three equations must have a common value, __ hY /3y + lYya + mV a/3 ^ Saf3y But since the fourth plane passes through the same point, this value of p must be also a value of its variable. Therefore, substituting this value of p in the fourth equation, we get n = _ ASSV/fy + mVya + mSSVajS Saf3y ’ or, hSj3y$ — ¿SySa + mSSa/3 — nSaf3y = o ; the sought condition. Section 8 The Tetrahedron 202°. To find the diameter of the sphere circumscribing a given tetrahedron. This will be given in the Section on the sphere, 211°. N178 THE TETRAHEDRON 203°. If two pairs of opposite edges of a tetrahedron, OABC, be at right angles, the third pair will be also at right angles. Let OAj_BC, OBJ.CA. Then Sa(y-/3) = o, S/3 (y — a) = o. Subtracting, Sy (a — /?) = o, that is, OCJ.AB. 204°. To find the condition that the perpendiculars from the corners of a tetrahedron, OABC, upon the opposite faces shall be concurrent. Let xY f3y and yYya, the respective perpendiculars from A and B upon the opposite faces, intersect in P. Then, since (ft — a), Yfiy and Yya are coplanar, Y . YfiyYya is per- pendicular to ft — a. Therefore, S(0-a)V . Y£yVya = o, S (/3 - a) ( - ySa/3y)= o, (Sya — Sfiy) Sa/?y = o. Therefore, o = Sya — S/?y = Sy(a — /?), . . . . (1) that is, OCj_AB. We get corresponding results in the other cases; therefore the sought condition is that the opposite edges shall be at right angles. 205°. If the opposite edges of a tetrahedron be at right angles, then the sums of the squares of the opposite edges are equal. For, by (1) of 204°, we have S/3y = Sya, T¡3 cos BOC = Ta cos AOC, ^ OB2 + OC2 - BC2 _ ^ A OC2 + OA2 - CA2. 20B.0C 20C.0A consequently, OB2 + CA2 = OA2 + BC2. We get corresponding results in the other cases ; therefore, Ac., &c. This is another form of the condition of 204°. 206°. The sum of the vector-areas of the faces of a tetra- hedron, OABC, is zero. OBC = £V0y ; OCA = ¿Vya ; OAB = \Ya/3.THE TETRAHEDRON 179 As we have taken the positive areas of these three faces as seen by an observer standing upon the point O, i.e. from the outside of the tetrahedron ; we must take the positive area of the remaining face from a corresponding point of view —exterior to the solid. Accordingly, ACB = *V (y - a)03 - a) = - ABC); and the angles of these four anharmonic quaternions are equal by (2), (Hamilton). 209°. Let ABCDE be any pentagon, plane or gauche,THE SPHERE 181 inscribed in a sphere, and let the two diagonals AC, AD be drawn. We then have the three equations, AB . BC . CA = AT, AC . CD . DA == AU, AD . DE.EA== AY, where AT, AU, AY are three tangents to the sphere at A. Multiplying the three equations together, AB . BC . CD . DE . EA = ^ v A? · AY . . (i) AC2 . Al)2 Now, the product AT . AU . AY is some fourth coplanar vector AW, which, being coplanar with the three tangents, is itself a tangent at A. Therefore the product of the five suc- cessive sides of a pentagon, plane or gauche, inscribed in a sphere, is a tangential vector drawn from the point at which the pentagon begins and ends. In general, the product of the successive sides of any n-gon inscribed in a sphere is a quaternion whose axis is normal to the sphere at the initial point of the w-gon, when n is even ; and is a vector tangential to the sphere at the same point, when n is odd. 210°· The last equation, 209° (1), may be written : OA . AB . BC . CP . PO = where P is a variable point upon the sphere. Hence, taking scalars, O = Sa (P - a) (y - fi) (p - y) ( - p), or, p2Sa/3y — a2Sf3yp + /22Sya.jo + y2Sa/3p ; . . (1) the equation of a sphere, 0 being a point upon its surface. To verify the equation, we have only to observe that it is satisfied by assigning to p the successive values, o, a, /?, y. Hence, if the vector of any point in space, in terms of three given diplanar vectors, a, ¡3, y, be p xa + yj3 + zy ; the equation p2 = xa2 + yfi2 -f zy2 expresses that the vectors a, ¡3, y, p terminate in the surface of a sphere which passes through 0. ___Let OA, OB, OC, &c., be any chords of a sphere, and let OD be a diameter, fig. 35 of 161°. Then, 160°, a -1 — S”1 = t15182 THE SPHERE a tangent at O ; ¡3“1 — 8 ”1 = t2, another tangent at O; &c., &c. But these tangents are coplanar; therefore a~1 — 8~ ¡3~l — 8-1, &c., are coplanar; therefore — y~l— a-1, )(y-'-a->)(p-1-a-‘) = o. . (2) is the equation of a sphere passing through the five points O, A, B, C, P. It is not difficult to transform (2) into (1). Since S - ■~~~ -- = o, 9 SSp-1 = 1) ^ and p2 = S8/> j................w are equations of the sphere. If the origin be any point in space, k the vector of the centre, and a any given vector radius, T (p - k) = Ta (p _ k)2 = a2 = C (*) are also equations of the sphere. If Tk = c, this last equation may be written : p2 = 2S*p = c2 - a2.............(5) 211°. By 82°, 8Sa ¡3y = SSa V/3y + SS/3Vya + SSyVa^. But, by (3) of 210°, SSa = a2 ; SS/3 = £2; SSy = y2. Therefore, cs a2Vl3y + /?2Vya + y2Yaft # /1N 8="----------' ···(!) the expression for the diameter of the sphere circumscribing the tetrahedron OABO. 212°. The equations of the tangent and tangent plane have been given in 167° ; that of the polar plane in 168°. 213°. To find the equation of the curve formed by the intersection of a plane and a sphere. Prom the centre, O, of the sphere let fall a perpendicular OD — 8 upon the given plane. Let p be the variable vector of the sphere ; a a vector in the given plane drawn from D toTHE SPHERE 183 meet p in the line of intersection whose equation is required. Then, if a be the radius of the sphere, f>2 = — a2 ; but p = S + o-; therefore, (8 + o·)2 = — a2, and a2 = — (a2 — d2). The locus, therefore, is the circle, cr2 = — (a2 — d2), S8o- = o, with D for centre and s/ a2 — d2 for radius. 214°. To find the curve in which the spheres intersect. Let the equations of the two spheres be, 210° (5), p2 - 2Sk,p = Cj2 - a,2, p2 - 2SK2p = 022 - a22. For every point of the curve of intersection the variables must have one common value. We may consequently equate the two equations, thus obtaining s («, - K2)„ = i{ («i2 - V) + ((V - S(K8-ic1)p = i(Oi-C3)> „ P3; where Pl _L (k{ — /c2), P2 _L (k2 - k3), P3 J_ (k3 - k ). Now, (C2—Oj)V(k2—k3) (te3-Kl) + (C3^Ca)y(K3~Kl)(Kl-fc2) + (Cl-C3)V(xl-K2) (Ks-tc»)*=o.184 THE SPHERE Therefore P1? P2, P3 intersect in one common straight line, 200°. 216°. To find the locus of a point, the sum of the squares of whose distances from n given points is constant. Let p be the vector of the sought point; al9 a2 ... an the vectors of the given points. Then (p — a,)2 + (p ~ a2)2 + &c. = 2 (p — a)2 = — C. But (p — a!)2 = p2 — 2Sajp + a^, (p — a2)2 = p2 — 2Sa2p + a22, (p — an)2 = P2 — 2Sawp + qn2. Therefore, S (p — a)2 = wp2 — 2Sp2a + 2a2 = — C, n \n J \n) n n =(£)'-:<*··+0>-0'· Therefore, by (4) of 210°, the locus of P is a sphere the vector of whose centre is —, its centre consequently being the mean point of the n points. Section 10 The Cone 217°. To find the equation of a Cone of Revolution whose vertex, 0, is the origin. Let a be a unit-vector along the axis OA, and p the vector of any point upon the surface of the cone. Then Sap = — Tp cos 0 y 0 being the angle POA. But 0 is constant. Therefore, c2p2 = S2ap....................(1) 218°. Had we written the equation of the plane, 105°, as Sap = a2, ..................(1) and the equation of the sphere as S/?p = p2, (2)THE CONE 185 we should, by multiplying these two equations together, have obtained the equation of the Cyclic Cone in the form : «V - SapS/fy = o, .................(3) instead of S^- S @ = 1, 108°. a p This cone has for its base the circle, Sap = a2 ; SjSp = p2...............(4) Let us take a vector with the direction of ¡3 and the tensor of a,—TaTJ/3 = OB'. The equation of the plane through B' perpendicular to this vector is O = s . TaU/?(p - TaXJ/5) = |S/?(p — or, ......................(«) Again, let us take a vector with the direction of a and the tensor of /3,—T/3Ua = OA'. The equation of a sphere through A' with OA' for a diameter is O = Sp (p - T/JUa) = Sp (p - ba a), or, Sap = | p2......................(6) Multiplying (5) and (6) together we get Saps/3p = g j8V = (- J2) f-2 = “V- Therefore, (Sap = a2 ; S/?p = p2) and (S/?p= “/32; Sap=“,,2) are two different circular sections of the same cyclic cone. Every section of this cone by a plane parallel to the plane Sap = a2, is a circle. For, let c be any constant scalar. Then, 100°, (8), Sap = ca2 is a plane || to Sac = a2. Substituting this value of Sap in (3), we get p2 — cS/?p = o. We therefore have for p, Sap = C2a ; p2 ·= cS/3p.186 THE CONE The locus of P, therefore, is the intersection of the plane through the term of ca_La, and the sphere with c/3 for a diameter, i.e. a circle. Similarly, any plane parallel to the plane, Sftp = ~ /32, is a circle. Therefore the sections of a cyclic cone made by planes perpendicular to either of the Cyclic Normals, a and ¡3, are circles. Both series of planes are consequently perpendicular to the plane OAB. If, then, AD be the intersection of the plane, Sap = a2, with OAB, and AE be the intersection of OAB with a plane through A parallel to S¡3p = f ¡3; b AD and AE are evidently anti- k parallel, fig. 46. The two series of circles are consequently called the Fig 46 antiparallel (or subcontrary) sec- tions of the cone. The equation of a plane through O parallel to the plane (1)» Sap = o;.......................(7) the equation of a plane through O parallel to the plane (5) is S/?P = o...............................................(8) These are the equations of the Cyclic Planes. If the cone, in the first instance, be supposed to have for its base the circle Sap = a2 ; S/?p = p2, it is clear that the cyclic plane (8) is a tangent at the vertex to the circumscribing sphere (2). If, in the second instance, the cone be supposed to have for its base the circle = Sop = |p*. the cyclic plane (7) is a tangent at the vertex to the circum- scribing sphere (6). Since a2p2 = SapS/?p == TapT/3pSUapSU/3p, STJapSU#) = T ^ = constant ; .... (9)THE CONE 187 or, the product of the cosines of the inclinations of any variable ray, p, of an oblique cyclic cone to its two cyclic normals is constant; or, the product of the sines of the in- clinations of the same ray to the two cyclic planes is constant. If a || (3, then ¡3 = m2a, where m is a constant scalar > 1, and the equation of the cone becomes - 2 P2 = S2ap ; m2 which is the equation of a cone of revolution, 217° (1). 219°. Of a system of three coinitial and rectangular vectors two are confined to given planes ; to find the locus of the third (Professor MacCullagh). Let 7r, |o, or be the three rectangular vectors. Then, by condition, StT/O == o........................(1) Spcr = o..........................(2) Sott = o..........................(3) For the given planes we also have Sa7r = o S p9 = o. (4) (5) It would be impossible generally to eliminate two vectors with less than six given equations. But in the present case, as the tensors are not involved, five are sufficient. From (3) and (4), 7r = xVacr ; » (2) ,, (5), P = yVfto·. Substituting these values of ir and p in (1), and, 92° (7), ÍC7/S . VCLffVfia = O, (T2Sa/3 - Sa<7S£<7 = o ; the equation of a cyclic cone. Before concluding, the reader must be reminded that it is not by such geometric applications as the foregoing that the merits of the quaternion method can be adequately illustrated. Its simplicity and power can only be fully shown by physical applications, which can find no place in this, or any other, elementary book. For such applications the reader is referred to the works of Sir W. R. Hamilton and Professor Tait; to188 THE CONE the ‘ Utility of Quaternions in Physics/ by Mr. A. McAulay ; and to Dr. Molenbroek’s paper, ‘Over de Toepassing der Quaternionen op de Mechanica en de Natuurkunde' (Muller, Amsterdam). A perusal of these works will convince most readers that quaternions are ‘ the natural language of metrical geometry and of physics ’ (Clifford).INDEX The references are to the Pages. Q. stands for quaternion ; Y. for vector. Angle-Bisector, Expression for, 20 Angle of a Q., Definition of the, 31 Anharmonic Q., Function, Defini- tion of, 149 Anharmonic Ratio, Definition of, 13 Apollonius of Perga, Cyclic Cone of, 109 Apollonius of Perga, Theorem by, 73 Arc, V., Definition of, 86 Area, V., „ 51 „ of spherical triangle, 143 Axis of a Q., 46 CENTRE, Circum-, of plane tri- angle, 21 Centre, Ex-, of plane triangle, 21 „ In-, „ „ 20 „ Mid-, „ „ 21 „ Ortho-, „ „ 21 Central Conic, General Equation of, 158 Circle, Equations of, 106, 153 „ Circum-, of spherical tri- angle, 136 „ In-, of spherical triangle, 134 Circular Harmonic Groups, 152 Cissoid, The, 169 Condition of Equality of two Vs., 2 Condition of Equality of two V. Arcs, 86 Condition of Equality of two Ver- sors, 63 Condition of Equality of two Qs., 63 Condition that four planes meet in a point, 177 Condition that four points may be coplanar, 176 Condition that three vectors may be coplanar, 99 Condition that four vectors may terminate in a plane, 28, 176 Cone, Right, Equation of, 184 „ Cyclic, „ 109,185 Conic, General Equation of a, 156 Conical Rotation, 90 Conjugate Functions, Definition of, 127 Conjugate of a Q., 68 „ „ Scalar, 69 „ „ V., 69 Conjugates, Isogonal, 22 „ Isotomic, 22 Curvature, Radius of, Expression for the, 171 Cyclic normals of a Cone, 186 „ planes „ „ 186 Cycloid, The, 170 Differential, General Defini- tion of a, 112 Differential of a Q., 117 „ „ V·, 115 Euler, Theorem by, 82 Function, Anharmonic Q., 149 Functions, Conjugate, 127 „ Self-conjugate, 127190 INDEX Gauss, Theorem by, 146 Harmonic Group, Circular, 152 „ Mean, Definition of the, 151 Helicoid, The, 175 Helix, The, 173 i,j, h, Properties of, 37 Index of a right Q., 51 Isogonal conjugates, 22 Isotomic „ 22 Keogh, Theorem by, 144 Logarithmic Spiral, The, 171 MacCullagh, Theorem by, 187 Mean point of triangle, 19 „ Harmonic, Definition of, 151 Moivre’s Theorem, 55 w-gon inscribed in circle, 153 „ „ sphere, 181 Normal, Cyclic, of a cone, 186 „ of Ellipse, 159 „ „ Hyperbola, 159 „ „ Parabola, 165 „ Plane of a Curve, Defini- tion of, 174 Plane, Equations of, 104 „ Radical of two spheres, 183 Points, Inverse, with respect to circle, 72 Polyhedra, Sum of V.-areas of faces of, is zero, 179 Quadrilateral, Spherical, A property of, 146 Quaternion, Amplitude of a, 56 „ Angle „ 31 „ Definition „ 30 Quaternions, Collinear, 66 „ Conjugate, 68 „ Coplanar, 31 „ Diplanar, 31 „ Opposite, 67 „ Reciprocal, 66 „ Right, 37 ternions, Condition of equality of two, 63 Radical plane of two spheres, 183 Radius of curvature of the cycloid, 171 Radius of curvature, general ex- pression for, 171 Radius Vector, Distinction between a, and a Vector, 2 Reciprocal of a Q., 66 „ „ V., 41 Rotation, Conical, 90 „ Positive and Negative, Definition of, 34 Scalar, Definition of a, 8 Section, Anharmonic and Har- monic, of Vs., 13 Sections, Antiparallel (or Subcon- trary) of cone, 109,185 Segments, Six, of triangle, 15 Signs of geometric figures, 15 Sphere, Diameter of, circumscrib- ing a tetrahedron, 182 Sphere, Equations of, 107, 181 Spiral, Logarithmic, 171 Tensor, Definition of a, 8 Theorem by Apollonius of Perga, 73 ;, „ Euler, 82 „ „ Gauss, 146 „ „ C. Keogh, 144 „ „ MacCullagh, 187 „ „ Moivre, 55 Triangle, Chordal, of a spherical triangle, 145 Unit V., Definition of a. 8 Vector, Definition of a real and actual, 2 Vector, Definition of a null, 2 „ Reciprocal of a, Definition of the, 30 Vector-area, Definition of a, 51 „ -arcs, Circular, 86 „ „ Condition of equality of two, 86 Versor, Definition of a, 34 „ Right, Definition of a, 37 Spottiswoode & Co. Printers, N etc-street Square, LondonA CLASSIFIED CATALOGUE OF SCIENTIFIC WORKS PUBLISHED BY MESSRS. LONGMANS, GREEN, & CO. LONDON: 39 PATERNOSTER ROW, E.C. NEW YORK: 16 EA8T 16th STREET. CONTENTS. 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