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Dot/cross product proof

  1. Oct 6, 2008 #1

    rock.freak667

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    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. jcsd
  3. Oct 7, 2008 #2

    tiny-tim

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    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:
     
  4. Oct 7, 2008 #3

    rock.freak667

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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
     
  5. Oct 8, 2008 #4

    tiny-tim

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

    Big hint: what is (p x q) x (r x s) ? :wink:
     
  6. Oct 8, 2008 #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: Oct 8, 2008
  7. Oct 10, 2008 #6

    rock.freak667

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    :biggrin: thanks for this hint, finally got it out now!
     
  8. Oct 10, 2008 #7

    Dick

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    So what did you do?
     
  9. Oct 10, 2008 #8

    rock.freak667

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    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)
     
  10. Oct 11, 2008 #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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