# Cross Products

## Homework Statement

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

## Homework 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

## 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:

Related Calculus and Beyond Homework Help News on Phys.org
LCKurtz
Homework Helper
Gold Member

## Homework Statement

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

## Homework 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

## 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.
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)

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