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Vectors mixed product proof

  1. Mar 5, 2014 #1
    1. The problem statement, all variables and given/known data

    Prove that (A×B) . [(B×C)×(C×A)]=(A,B,C)^2
    where A, B, C are vectors in R3.

    2. Relevant equations

    W×(U×V)=(W . V) U - (W × U) V

    3. The attempt at a solution

    Assuming K=(A×B), M=(B×C):
    K . [M×(C×A)]
    K . [(M . A) C - (M . C) A]
    [(M . A)(K . C) - (M . C)(K . A)]

    Then:
    [(B×C) . A] [(A×B) . C] - [(B×C) . C] [(A×B) . A]
     
  2. jcsd
  3. Mar 5, 2014 #2

    SammyS

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    You're almost there!

    What is ##\displaystyle \left(\vec{B}\times\vec{C}\right)\cdot\vec{C} \ ?##
     
  4. Mar 5, 2014 #3
    (b×c) . C = (c×b) . C = (c×c) . B = (0) . B = 0
     
  5. Mar 5, 2014 #4

    SammyS

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    Yes.

    Does that get you to the result?
     
  6. Mar 5, 2014 #5
    (A×B) . A will vanish as well, and I'll end up with [(B×C) . A] and [(A×B) . C] which are equal.

    But how - by algebra - can they be equal to (A,B,C)^2?
     
  7. Mar 5, 2014 #6

    SammyS

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    What is (A,B,C) ?

    Isn't it the triple scalar product ?
     
  8. Mar 5, 2014 #7
    I understand now.

    Thank you very much.
     
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