146.] It remains to find the value of the constant m, and to express the vector

as a function of Vλμ.

Γφλφ μ

Operate on (4) by S.d'v, where v is any vector not coplanar with

λ and μ, and we get

mS.p'vp-1Vλμ = mS.vpp-1Vλμ (by (3) of § 144)

= mδ.λμν = δ.φ'λφμόν, οι

m =

δ. φλφό μφ'ν
δ.λμν

(5)

[That this quantity is independent of the particular vectors A, μ, v is evident from the fact that if

X'=pλ+qμ+rv, μí2 = P1λ+q1m+r1v, and = p2λ+Q2M+r2v be any other three vectors (which is possible since λ, μ, v are not coplanar), we have

p'x'= pp'λ+qo'μ+ro'v, &c., &c.;

so that the numerator and denominator of the fraction which expresses m are altered in the same ratio. Each of these quantities is in fact an Invariant, and the numerical multiplier is the same for both when we pass from any one set of three vectors to another. A still simpler proof is obtained at once by writing λ+pu for a in (5), and noticing that neither numerator nor denominator is altered.]

147.] Let us now change to +g, where g is any scalar. It is evident that ' becomes d'+g, and our equation (4) becomes

mg (Þ+g)−1Vλμ = V (p′+g)λ (p′+9) μ ;

= φλόμ+9 (φλμ +λφμ)+g Γλμ,

is what m becomes when is changed into +g; m, and m, being two new scalar constants whose values are

in

p

If, in these expressions, we put X+pu for λ, we find that the terms vanish identically; so that they also are invariants. Substituting for m,, and equating the coefficients of the various powers of g after operating on both sides by 4+g, we have two identities and the following two equations,

[The first determines x, and shews that we were justified in treating (p'λμ+λp'u) as a linear and vector function of V.A. The result might have been also obtained thus,

δ.λχλμ = δ.λφ'λμ = - δ.λμφ' λ = - δ.λφ Γλμ,

δ.μχ Γλμ = 5.μλφόμ = - δ.μφ Γλμ,

δ.νχλμ = 8.(υφ' λμ + υλφ'μ)

= m2 βλμν- δ.λμφ'ν

= S.v (mgΓλμ-φίλμ);

and all three (the utmost generality) are satisfied by

which contains the complete solution of linear and vector equations. 149.] More to satisfy the student of the validity of the above investigation, about whose logic he may at first feel some difficulties, than to obtain easy solutions, we take a few very simple examples to begin with: we treat them with all desirable prolixity, and we append for comparison easy solutions obtained by methods specially adapted to each case.

Now λ, μ, v are any three non-coplanar vectors; and we may therefore put for them a, ß, y if the latter be non-coplanar.

a232 Saß.4-1y=a232 Saß.pa2ß2y + SaẞV.ayẞ+V.a (V.ayẞ) B, which is one form of solution.

By expanding the vectors of products we may easily reduce it to the form a232 Saß.pa2ß2y+aẞ2 Say + Ba2Spy,

1

V.apẞ= (B Say+a SBy-V.ayẞ) = y,
Saß

which is the given equation.

152.] An easier mode of arriving at the same solution, in this simple case, is as follows:

Operating by S.a and S.ẞ on the given equation

But the given equation may be written

aSẞp-pSaß+ẞSap = y.

Substituting and transposing we get

pSaß = a 1Say+B-1SBY-Y,

ρδαβ

which agrees with the result of § 150.

153.] If a, ß, y be coplanar, the above mode of solution is applicable, but the result may be deduced much more simply.

=

For (§ 101) S.aßy 0, and the equation then gives S.aßp that p is also coplanar with a, ẞ, y.

= 0, so

and this, being a vector, may be written

-1
= a ̄1SB-1y+B-1Sa-1y-ySa-1ẞ-1.

This formula is equivalent to that just given, but not equal to it term by term. [The student will find it a good exercise to prove directly that, if a, ß, y are coplanar, we have

by comparison of which with the given equation we find

Sap and SBp.

The value of p remains therefore with one indeterminate scalar.

Suppose a, ẞ, y not to be coplanar, and employ them as λ, μ, v to calculate the coefficients in the equation for 4-1. We have

Hence

a232 Saß.p-1y = a2ß2Saß.p

=

(2 (Saß)2 + a2ß2) y-3 SaẞF.aẞy+V.aẞV.aẞy,

which, by expanding the vectors of products, takes easily the simpler form a2 32 Saẞ.p=a2ẞ2y—aß2 Say + 2ß Saß Say-Ba2 Spy.

155.] To verify this, operate by V.aẞ on both sides, and we have a2ß2 SaẞV.aßp = a2ß2 V.aßy — V.aßaß2 Say+2a32 Saẞ Say-aa22 Spy = a2ß2 (aSẞy-BẞSay+ySaß) — (2aSaß-Ba2) ẞ2 Say

=a232 Saß.y,

or

+2 aß2 Saß Say-aa2ß2 SBY

V.aẞp=y.

156.] To solve the same equation without employing the general method, we may proceed as follows :—

y=V.aẞppSaß + V.V (aß) p.

Operating by S. Vaß we have

S.aßy S.aßpSaß.

Divide this by Saß, and add it to the given equation. We thus

a form of solution somewhat simpler than that before obtained. To shew that they agree, however, let us multiply by a2ß2Saß, and we get

a232 Saẞ.p = BaySaß + BaS.aẞy.

In this form we see at once that the right-hand side is a vector, since its scalar is evidently zero (§ 89). Hence we may write

αβ δαβ.ρ = Γ.βαγδαβ - Γαβδ.αβγ.

But by (3) of § 91,

-YS.aẞVaẞ+aS.ẞ (Vaß) y+BS.V (aß) ay + VaẞS.aßy = 0. Add this to the right-hand side, and we have

a232 Saß.py ((Saß)2-S.aßaß)

But

= ((Saß)2 - S.aẞVaß) — a (SaẞSẞy—S.ẞ (Vaß) y)
+B (SaẞSay + S.V (aß) ay).

(Saẞ)2-S.aẞTaẞ = (Saẞ)2- (Vaß)2 = a232,
SaẞSBy-S.ẞ(Vaẞ) y = SaẞSẞy-SẞaSẞy + B2 Say = B2 Say
SaẞSay+S.V (aß) ay = SaẞSay+SaßSay-a Spy

=

2 Saß Say-a2SBY;

and the substitution of these values renders our equation identical with that of § 154.