[If a, ß, y be coplanar, the simplified forms of the expression for p lead to the equation -1 -1 Saß.B-1a-1y = y-a-1Say+2ẞSa-11Say-B-1SBY, which, as before, we leave as an exercise to the student.] 157.] Example III. The solution of the equation Γερ = γ leads to the vanishing of some of the quantities m. Before, however, treating it by the general method, we shall deduce its solution from that of Γ.αβρ = γ already given. Our reason for so doing is that we thus have an opportunity of shewing the nature of some of the cases in which one or more of m, m1, m, vanish; and also of introducing an example of the use of vanishing fractions in quaternions. Far simpler solutions will be given in the following sections. The solution of the last-written equation is, § 154, a232 Saẞ.pa2ß2y-aẞ2 Say-Ba2 Sẞy +2ẞSaßSay. where e is a scalar, the solution of the first-written equation will evidently be derived from that of the second by making e gradually tend to zero. We have, for this purpose, the following necessary transformations: a2ß2 = aß K.aß = (e + €) (e − e) = e2 — e2, αβ aß2 Say+Ba2 Sẞy = aß.ẞSay + Ba.aSẞy, = (e + €) ẞSay + (e — €) aSßy, = e (BSay+aSẞy) + eV.yVaß, =e(BSay+aSẞy) + €Vye. Hence the solution becomes (e2 — €2) ep = (e2 — €2) y.—e (ẞSay + aSẞy) — eÏye + 2eẞSay, = = — (e2 — c2) y + eV.yVaß—eVye, (e2 — e2) y + eVy€ + ye2 — € Sy€, = e2y+eVye —ε Sye. Dividing by e, and then putting e = 0, we have —e2p =Vye—eRo (S¥€). e Now, by the form of the given equation, we see that Sye = 0. Hence the limit is indeterminate, and we may put for it x, where x Our solution is, therefore, is any scalar. The verification is obvious-for we have Ep = y +x. 158.] This suggests a very simple mode of solution. see that the given equation leaves Sep indeterminate. therefore, Sep=x For we Assume, 159.] To apply the general method, we may take e, y and ey Our warrant for putting xe, as the equivalent of 4-20 is this: The equation may be written V.ενεσ = 0 = σε2 — εδεσ. Hence, unless σ = 0, we have σ || € = XE. 160.] Example IV. As a final example let us take the most general form of 4, which, as will be soon proved, may be expressed as follows:: Here φρ == ppaSẞp+a1Sẞ1p + a2Sẞ2p = y. and, consequently, taking a, a1, a2, which are in this case noncoplanar vectors, for λ, μ, v, we have Hence the value of the determinant is -(SaаS. Vα1α2 Vα12+Sа1αS.VаVa1a2 + Sα,аS.Vaa, Va12) =S.a (Va2a,S.aa,a2) {by § 92 (3)} = —(§.aa1a2)2. The interpretation of this result in spherical trigonometry is very interesting. (See Ex. (6) p. 68.) 1 or, taking the terms by columns instead of by rows, 1 S.aa1a2 + + 1 [S.Vßß1 (Vaa1S.aa1α2) + S.aa1a2 S[aa1 (ẞSaa2+ẞ1Sа1a1⁄2 + ...) +α2a (ẞSaa1 + ...) +a ̧a1⁄2 (ẞSaa+ .....)], or, grouping as before, = = 1 S.aa, az 1 S.aa1a2 ·S [B(Vaa1Saa2+ Va1аSаa ̧+ Vα1α2Saa) + .....], S[B (aS.aa12) +......] (§ 92 (4)), · = 8(aẞ+a1ß1+a2ß2). And the solution is, therefore, 4-1yS.aa,a,S.ßß1ẞ2 = pS.aa,a,S.BB1B2 [It will be excellent practice for the student to work out in detail the blank portions of the above investigation, and also to prove directly that the value of p we have just found satisfies the given equation.] 161.] But it is not necessary to go through such a long process to get the solution-though it will be advantageous to the student to read it carefully-for if we operate on the equation by S.a,a,, S.aga, and S.aa, we get S.aaa1Sẞ1 = §.a,ay, S.aa.a2Sẞ2p =S.aa1y. From these, by § 92 (4), we have at once pS.ं¤‚S.ßß1ß2 = Vßß1S.àa ̧¥ + Vß1ß2S.à ̧a¿у + Vß2ẞS.a2ay. The student will find it a useful exercise to prove that this is equivalent to the solution in § 160: To verify the present solution we have (aSßp+a2Sß1p+a2Sẞ2p) S.aa‚a‚S.BB1B2 =aS.ẞB1ẞ2S.a,a2y+a1S.ẞ1 ẞ2ẞS.a2ay+a,S.ßßß1S.aa1y 162.] It is evident, from these examples, that for special cases we can usually find modes of solution of the linear and vector equation which are simpler in application than the general process of § 148. The real value of that process however consists partly in its enabling us to express inverse functions of 4, such as (+9)−1 for instance, in terms of direct operations, a property which will be of great use to us later; partly in its leading us to the fundamental cubic which is an immediate deduction from the equation of § 148, and whose interpretation is of the utmost importance with reference to the axes of surfaces of the second order, principal axes of inertia, the analysis of strains in a distorted solid, and various similar enquiries. 163.] When the function p is its own conjugate, that is, when δροσ = σφρ for all values of p and o, the vectors for which ρ (4-g) p = 0 form in general a real and definite rectangular system. This, of course, may in particular cases degrade into one definite vector, and any pair of others perpendicular to it; and cases may occur in which the equation is satisfied for every vector. Suppose the roots of m, 0 (§ 147) to be real and different, then P1=91P1) where P1, P2, P3 are three definite vectors determined by the constants involved in . PP2 = 92P2 PP3 = 93P3 91 92 which, as g1 and g2 are by hypothesis different, requires = If two roots be equal, as 92, 93, we still have, by the above proof, Sp1P20 and Sp1P30. But there is nothing farther to determine and P3 which are therefore any vectors perpendicular to P1. P2 If all three roots be equal, every real vector satisfies the equation (4-g)p = 0. 164.] Next, as to the reality of the three directions in this case. Suppose g2+h21 to be a root, and let p2+σ1⁄2√√=1 be the corresponding value of p, where 92 and h2 are real numbers, P2 and Մշ real vectors, and √-1 the old imaginary of algebra. and this divides itself, as in algebra, into the two equations Operating on these by S.σ2, S.p2 respectively, and subtracting the results, remembering our condition as to the nature of But, as σ and p2 are both real vectors, the sum of their squares cannot vanish. Hence h2 vanishes, and with it the impossible part of the root. 165.] When is self-conjugate, we have shewn that the equation g3 —m2g2+m1g—m = 0 has three real roots, in general different from one another. Hence the cubic in 4 may be written (p—91)(p—92) (P—93) = 0, |