I think I have done this question but is there a better way? Prove that if the numbers p, q, r and s satisfy ps=qr, then (pa+qb) x (ra+sb) = 0. a and b are vectors.
It looks good to me though a little more complicated than necessary.
$\displaystyle (pa+ qb)\times(ra+ sb)= pa\times ra+ pa\times sb+ qb\times ra+ qb\times sb$
Of course, you can take out the numbers: pr(a\times a)+ ps(a\times b)+ qr(b\times a)+ qs(b\times b)[tex]
$\displaystyle a\times a= b\times b= 0$ so that reduces to $\displaystyle ps(a\times b)+ ar(b\times a)= ps(a\times b)- qr(a\times b)= (ps- qr)(a\times b)$.