Page:Linear Algebra (1882) Tevfik.djvu/24

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And again if $\alpha$ and $\beta$ being in diifferent directions, we have

$a\alpha +b\beta =k\alpha +l\beta$

we must also have

$(a-k)\alpha +(b-l)\beta =0$;

$\therefore$

$a-k=0$ and $b-l=0$.

19. If $\alpha ,\beta ,\gamma$ are non parallel lines in the same plane, it is always possible to find the numerical values of $a,b,c,$ so that,

$a\alpha +b\beta +c\gamma$ shall $=0$.

For as these $\alpha ,\beta ,$ and $\gamma$ are on the same plane, a triangle can be constructed the sides of which shall be parallel respectively to $\alpha ,\beta ,\gamma$. Now if the sides of this triangle taken in order be

$a\alpha ,b\beta ,c\gamma$

we shall have, by going around the triangle,

$a\alpha +b\beta +c\gamma =0$.

20. If $\alpha ,\beta ,\gamma$ are three lines neither parallel, nor in the same plane, it is impossible to find numerical values of $a,b,c$, not equad to zero, which shall render $a\alpha +b\beta +c\gamma =0$, for $a\alpha +b\beta$ can be represented by a line in the plane parallel to $\alpha ,\beta$. Now $c\gamma$ is not in that plane, therefore the sum of $a\alpha +b\beta$ and $c\gamma$ cannot equal $0$. It follows that, if $a\alpha +b\beta +c\gamma =0$ and $\alpha ,\beta ,\gamma$ are not parallel to each other, they are in the same plane.

21 . There is but one way of making the sum of the multiples of $\alpha ,\beta ,\gamma$ equal to $0$.

Let

$a\alpha +b\beta +c\gamma =0$

and also

$a,\alpha +b,\beta +c,\gamma =0$.

By eliminating $\gamma$ we get

$(ac,-ca,)\alpha +(bc,-cb,)\beta =0$;

but as $\alpha ,\beta$ are in different directions,

$ac,-ca,=0$ and $bc,-cb,=0$;

$\therefore$

$ac,=ca$ and $bc,=cb,$

or

$a:b:c::a,:b,:c,$,

so that the second equation is simply a multiple of the first. If we observe that the triangles which give the different values of $a,b,c,$ are similar the last proposition will be accepted a priori.

22. If $\alpha ,\beta ,\gamma$ are coinitial coplanar lines, terminating in a straight line, then the