Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/41

the ${\text{X-}},{\text{Y-}}$, and ${\text{Z-}}$components of $\lambda$ and adding. Hence the second equation may be regarded as the most general form of a scalar equation of the first degree in $a,b,c,d,$ etc., which can be derived from the original vector equation or its equivalent three scalar equations. If we wish to have two of the scalars, as $b$ and $c$, disappear, we have only to choose for $\lambda$ a vector perpendicular to $\beta$ and $\gamma$. Such a vector is $\beta \times \gamma$. We thus obtain

$a\alpha .\beta \times \gamma +d\delta .\beta \times \gamma +{\text{etc.}}=0.$

37. Relations of four vectors.—By this method of elimination we may find the values of the coefficients $a,b$, and $c$ in the equation

$\rho =a\alpha +b\beta +c\gamma ,$

(1)

by which any vector $\rho$ is expressed in terms of three others. (See No. 10.) If we multiply directly by $\beta \times \gamma ,\gamma \times \alpha$, and $\alpha \times \beta$, we obtain

$\rho .\beta \times \gamma =a\alpha .\beta \times \gamma ,$⁠$\rho .\gamma \times \alpha =b\beta .\gamma \times \alpha ,$⁠$\rho .\alpha \times \beta =c\gamma .\alpha \times \beta ;$

(2)

whence

$a={\frac {\rho .\beta \times \gamma }{\alpha .\beta \times \gamma }},$⁠$b={\frac {\rho .\gamma \times \alpha }{\alpha .\beta \times \gamma }},$⁠$c={\frac {\rho .\alpha \times \beta }{\alpha .\beta \times \gamma }}\cdot$

(3)

By substitution of these values, we obtain the identical equation,

$(\alpha .\beta \times \gamma )\rho =(\rho .\beta \times \gamma )\alpha +(\rho .\gamma \times \alpha )\beta +(\rho .\alpha \times \beta )\gamma .$

(4)

(Compare No. 31.) If we wish the four vectors to appear symmetrically in the equation we may write

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

(5)

If we wish to express $\rho$ as a sum of vectors having directions perpendicular to the planes of $\alpha$ and $\beta ,$ of $\beta$ and $\gamma$, and of $\gamma$ and $\alpha$, we may write

$\rho =e\beta \times \gamma +f\gamma \times \alpha +g\alpha \times \beta .$

(6)

To obtain the values of $e,f,g$, we multiply directly by $a$, by $\beta$, and by $\gamma$. This gives

$e={\frac {\rho .\alpha }{\beta .\gamma \times \alpha }},$⁠$f={\frac {\rho .\beta }{\gamma .\alpha \times \beta }},$⁠$g={\frac {\rho .\gamma }{\alpha .\beta \times \gamma }}\cdot$

(7)

Substituting these values we obtain the identical equation

$(\alpha .\beta \times \gamma )\rho =(\rho .\alpha )\beta \times \gamma +(\rho .\beta )\gamma \times \alpha +(\rho .\gamma )\alpha \times \beta .$

(8)

(Compare No. 32.)

38. Reciprocal systems of vectors.— The results of the preceding section may be more compactly expressed if we use the abbreviations

$\alpha '={\frac {\beta \times \gamma }{\alpha .\beta \times \gamma }},$⁠$\beta '={\frac {\gamma \times \alpha }{\beta .\gamma \times \alpha }},$⁠ $\gamma '={\frac {\alpha \times \beta }{\gamma .\alpha \times \beta }}\cdot$

(1)

The identical equations (4) and (8) of the preceding number thus become

$\rho =(\rho .\alpha ')\alpha +(\rho .\beta ')\beta +(\rho .\gamma ')\gamma ,$

(2)

$\rho =(\rho .\alpha )\alpha '+(\rho .\beta )\beta '+(\rho .\gamma )\gamma '.$

(3)