Finding position vector of cross product

869 Views Asked by At

I have been stuck trying to find the solution doe this following question:

If a = (-1, 5, r ) b = (-5, 1, 1) and c = (-1, 1, -3) find the values of for which a x b is perpendicular to a x c. (state lowest first)

I first calculated a x b, which led me to (5-r)i-(-1+5r)j+24k. And b x c = -4i-16j+4k. I then counted the dot product of a x b and b x c

a x b.b x c = -20+4r-16+80r+96

This equals to 0 because they are perpendicular to each other and I ended up with r = 5/7.

I am supposed to submit two values of r. So I don't know what I did wrong and wondering if any of you can help me. It is much appreciated.

Thanks!

1

There are 1 best solutions below

0
On

Hint (Lagrange): $$\langle a\times b,c\times d\rangle=\langle a,c\rangle\langle b,d\rangle-\langle b,c\rangle\langle a,d\rangle. $$

Now happily solve $\langle a\times b,a\times c\rangle=0$.