Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/43

is also an identical equation. (The reader will observe that in each of these equations the second member may be expressed as a determinant.)

From these transformations, with those already given, it follows that a product formed of any number of letters (representing vectors and scalars), combined in any possible way by scalar, direct, and skew multiplications, may be reduced to a sum of products, containing each the sign $\times$ once and only once, when the original product contains it an odd number of times, or entirely free from the sign, when the original product contains it an even number of times.

39. Scalar equations of the first degree with respect to an unknown vector.—It is easily shown that any scalar equation of the first degree with respect to an unknown vector $\rho$, in which all the other quantities are known, may be reduced to the form

$\rho .\alpha =a,$

in which $\alpha$ and $a$ are known. (See Na 35.) Three such equations will afford the value of $\rho$ (by equation (8) of No. 37, or equation (3) of No. 38), which may be used to eliminate $\rho$ from any other equation either scalar or vector.

When we have four scalar equations of the first degree with respect to $\rho$, the elimination may be performed most symmetrically by substituting the values of $\rho .\alpha$, etc., in the equation

$(\rho .\alpha )(\beta .\gamma \times \delta )-(\rho .\beta )(\gamma .\delta \times \alpha )+(\rho .\gamma )(\delta .\alpha \times \beta )-(\rho .\delta )(\alpha .\beta \times \gamma )=0,$

which is obtained from equation (8) of No. 37 by multiplying directly by $\delta$, It may also be obtained from equation (5) of No. 37 by writing $\delta$ for $\rho$, and then multiplying directly by $\rho$.

40. Solution of a vector equation of the first degree with respect to the unknown vector.—It is now easy to solve an equation of the form

$\delta =\alpha (\lambda .\rho )+\beta (\mu .\rho )+\gamma (\nu .\rho ),$

(1)

where $\alpha ,\beta ,\gamma ,\delta ,\lambda ,\mu$, and $\nu$ represent known vectors. Multiplying directly by $\beta \times \gamma$, by $\gamma \times \alpha$, and by $\alpha \times \beta$, we obtain

$\beta .\gamma \times \delta =(\beta .\gamma \times \alpha )(\lambda .\rho ),$$\gamma .\alpha \times \delta =(\gamma .\alpha \times \beta )(\mu .\rho ),$

$\alpha .\beta \times \delta =(\alpha .\beta \times \delta )(\nu .\rho );$

or	$\alpha '.\delta =\lambda .\rho ,$$\beta '.\delta =\mu .\rho ,$$\gamma '.\delta =\nu .\rho ,$

where $\alpha ',\beta ',\gamma '$ are the reciprocals of $\alpha ,\beta ,\gamma$. Substituting these values in the identical equation

$\rho =\lambda '(\lambda .\rho )+\mu '(\mu .\rho )+\nu '(\nu .\rho ),$

in which $\lambda ',\mu ',\nu '$ are the reciprocals of $\lambda ,\mu ,\nu$ (see No. 88), we have

$\rho =\lambda '(\alpha '.\delta )+\mu '(\beta '.\delta )+\nu '(\gamma '.\delta ),$

(2)

which is the solution required.