Dot/cross product proof

  • #1
rock.freak667
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Homework Statement



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.

Homework Equations





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.
 

Answers and Replies

  • #2
tiny-tim
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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.

Hi rock.freak667! :smile:

(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) :wink:
 
  • #3
rock.freak667
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Use the only other relation you know for the cross-product …

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

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
tiny-tim
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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

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

Big hint: what is (p x q) x (r x s) ? :wink:
 
  • #5
Dick
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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:
  • #6
rock.freak667
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Big hint: what is (p x q) x (r x s) ? :wink:

:biggrin: thanks for this hint, finally got it out now!
 
  • #7
Dick
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  • #8
rock.freak667
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So what did you do?

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
Dick
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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.
 

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