# Cross Products

1. Dec 31, 2009

### Neen87

1. The problem statement, all variables and given/known data

Consider the vector equation a × x = b in R3, where a doesn't = 0. Show that:
(a) a · b = 0
(b) x = (b × a / ||a||^2) + ka is a solution to the equation, for any scalar k

2. Relevant equations

I'm not really sure, but I've been messing with these:

1. u x (v x w) = (u · w)v - (u · v)w
2. v x w = v2w3 - v3w2, v3w1 - v1w3, v1w2 - v2w1

3. The attempt at a solution

(a) for a · b = 0, b must = 0 since a can't = 0. for b = 0, x must = 0 since a can't = 0.

(b) i'm really lost!

x = ((a x X) x a / (a · a)) + ka

I have no idea if I'm headed in the right direction or where to go with this. Even a few hints would be greatly appreciated!

Thanks!
Tina

Last edited: Dec 31, 2009
2. Dec 31, 2009

### LCKurtz

Remember that a · b = 0 means a or b is the zero vector or a and b are perpendicular to each other. Dot a into both sides of the equation to prove (a).

For (b) use your equation (1) above and the result of (a)

3. Dec 31, 2009

### diazona

But remember that you don't know that a · b = 0. You have to prove it.

If the vector identities (the ones you listed under "relevant equations") are what you have to work with, I'd recommend trying to get the original equation (a × x = b) into some form that appears in one of the identities. For instance, try multiplying both sides of a × x = b by something.