Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/37

Again, if $\alpha$ be directed toward the east, and $\beta$ lie in the same horizontal plane and on the north side of $\alpha$, $\alpha \times \beta$ will be directed upward.

15. It is evident from the preceding definitions that

$\alpha .\beta =\beta .\alpha ,$and$\alpha \times \beta =\beta \times \alpha .$

16. Moreover,$[n\alpha ].\beta =\alpha .[n\beta ],$

and$[n\alpha ]\times \beta =\alpha \times [n\beta ].$

The brackets may therefore be omitted in such expressions.

17. From the definitions of No. 11 it appears that

$i.i=j.j=k.k=1,$
$i.j=j.i=i.k=k.i=j.k=k.j=0,$
$i\times i=0,$	$j\times j=0,$	$k\times k=0,$
$i\times j=k,$	$j\times k=i,$	$k\times i=j,$
$j\times i=-k,$	$k\times j=-i,$	$i\times k=-j.$

18. If we resolve $\beta$ into two components $\beta '$ and $\beta ''$, of which the first is parallel and the second perpendicular to $\alpha$, we shall have

$\alpha .\beta =\alpha .\beta '$and$\alpha \times \beta =\alpha \times \beta ''.$

19. $\alpha .[\beta +\gamma ]=\alpha .\beta +\alpha .\gamma$ and $\alpha \times [\beta +\gamma ]=\alpha \times \beta +\alpha \times \gamma .$

To prove this, let $\sigma =\beta +\gamma$, and resolve each of the vectors $\alpha ,\beta ,\sigma$ into two components, one parallel and the other perpendicular to $\alpha$. Let these be $\beta ',\beta '',\gamma ',\gamma '',\sigma ',\sigma ''.$ Then the equations to be proved will reduce by the last section to

$\alpha .\sigma '=\alpha .\beta '+\alpha .\gamma '$and$\alpha \times \sigma ''=\alpha \times \beta ''+\alpha \times \gamma ''.$

Now since $\sigma =\beta +\gamma$ we may form a triangle in space, the sides of which shall be $\beta ,\gamma$, and $\sigma .$ Projecting this on a plane perpendicular to $\alpha$, we obtain a triangle having the sides $\beta '',\gamma '',$ and $\sigma '',$ which affords the relation $\sigma ''=\beta ''+\gamma ''.$ If we pass planes perpendicular to a through the vertices of the first triangle, they will give on a line parallel to a segments equal to $\beta ',\gamma ',\sigma '.$ Thus we obtain the relation $\sigma '=\beta '+\gamma '.$ Therefore $\alpha .\sigma '=\alpha .\beta '+\alpha .\gamma ',$ since all the cosines involved in these products are equal to unity. Moreover, if $\alpha$ is a unit vector, we shall evidently have $\alpha \times \sigma ''=\alpha \times \beta ''+\alpha \times \gamma '',$ since the effect of the skew multiplication by a upon vectors in a plane perpendicular to a is simply to rotate them all 90° in that plane. But any case may be reduced to this by dividing both sides of the equation to be proved by the magnitude of $\alpha .$ The propositions are therefore proved.

20. Hence,

$[\alpha +\beta ].\gamma =\alpha .\gamma +\beta .\gamma ,$$[\alpha +\beta ]\times \gamma =\alpha \times \gamma +\beta \times \gamma ,$ $[\alpha +\beta ].[\gamma +\delta ]=\alpha .\gamma +\alpha .\delta +\beta .\gamma +\beta .\delta ,$ $[\alpha +\beta ]\times [\gamma +\delta ]=\alpha \times \gamma +\alpha \times \delta +\beta \times \gamma +\beta \times \delta ;$