# Vector proof (cross product)

1. May 5, 2013

### Jbreezy

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

If x + v = u
Prove x x v = u x v = x x u

3. The attempt at a solution

I don't even know where to start with this. I thought that magnitude of the resultant vector would have to be equal. So I started messing with each to see if I could find a pattern.

x x v = | x|| v| sin θ

This is a crap approach I can't find anything. Please give me a hint not the answer. I just can't seem to draw the information in my text together to give myself enough to show this. Thank you.

2. May 5, 2013

### Staff: Mentor

Just use your definition of u in the equations, and simplify with known properties of the cross product.
It does not help to consider the magnitude of the vectors, as their "direction" has to fit, too.

3. May 5, 2013

### tiny-tim

Hi Jbreezy!

Are you allowed to use the associative rule, a x (b + c) = a x b + a x c ?

Or do you have to use coordinates?

4. May 5, 2013

### Jbreezy

I guess you are allowed to use whatever. I don't understand this problem at all.

5. May 5, 2013

### tiny-tim

ok, then what is x x (u - v) ?

6. May 5, 2013

### Jbreezy

0 vector? This is not anywhere in my eq. though. In terms of the original.

7. May 5, 2013

### Jbreezy

This was a good problem to make me feel like a monkey with a stick.

so, If x + v = u
Prove x x v = u x v = x x u

Sub in u.

x x v = (x + v ) x v = x x(x + v)

So when you distribute.

X x V = X x V + V x V = X x X + X x v

So, V x V = 0 , X x X = 0
So
X x V = X x V = X x V
Thanks

8. May 6, 2013

### tiny-tim

Hi Jbreezy!

(just got up :zzz:)

Yes, that's correct.

But it's a bit long-winded …

you could have done x x v = (x + v ) x v (because v x v = 0)

= u x v,​

or x x v - u x v = … (you finish it )