# Homework Help: Dot/cross product proof

1. Oct 6, 2008

### rock.freak667

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

Prove that p(q.rXs)-q(r.sXp)+r(s.pXq)-s(p.qXr)=0

Where p,q,r,s are vectors and X denotes the cross product.
2. Relevant equations

3. The attempt at a solution

So far all I have is this ( I can't get any further after)

q.rXs=r.sXq=s.qXr

r.sXp=p.sXr=s.rXp

s.pXq=q.sXp=p.qXs

p.qXr=r.pXq=q.rXp.

2. Oct 7, 2008

### tiny-tim

Hi rock.freak667!

(please either use more brackets, or write the triple product as [q,r,s])

Use the only other relation you know for the cross-product …

a x (b x c) = b(a.c) - c(b.a)

3. Oct 7, 2008

### rock.freak667

But I do not have anything like b(a.c)-c(b.a) to convert back to a x (b x c) or vice versa

4. Oct 8, 2008

### tiny-tim

D'oh! forget reality … this is maths … use your imagination!

Big hint: what is (p x q) x (r x s) ?

5. Oct 8, 2008

### Dick

Try this. Call your expression F(p,q,r,s). Can you show F(p,q,r,s)=-F(q,p,r,s)=-F(p,q,s,r)=-F(p,r,q,s)? It's not very hard if you rearrange the triple products. I.e. interchanging any two arguments flips the sign of F. That means, for example, F(p,p,r,s)=0. If any argument is repeated F is zero. It's also clear F is linear in each of it's arguments. That means we can figure out what F is by evaluating it on all combinations of the basis elements i,j,k. But in three dimensions we only have three basis elements. So F must vanish on all combinations. So F is zero. tiny-tim's suggestion sort of looks along the right lines, but (ixj)x(ixk) isn't zero. So (pxq)x(rxs) can't be the same as what you are looking at. It's not 'antisymmetric' enough.

Last edited: Oct 8, 2008
6. Oct 10, 2008

### rock.freak667

thanks for this hint, finally got it out now!

7. Oct 10, 2008

### Dick

So what did you do?

8. Oct 10, 2008

### rock.freak667

Well if I did it correctly, I believe (s.pXq)-s(p.qXr) works out as (p x q) x (r x s).

And p(q.rXs)-q(r.sXp) works out as (r x s) x (p x q)

9. Oct 11, 2008

### Dick

Yep, I think you did it correctly. I was trying to find a trick where I didn't have to do the general antisymmetric thing, and couldn't find one. You and tiny-tim did.