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Reconcile Geometric Form of Cross Product with Algebric Form

  1. Jul 18, 2012 #1
    Basically, we have to prove that A X B is equal to ABsinθ

    Now a crucial step towards the proof is proving that
    (AXB).(AXB) is equal to (AB)2 - (A.B)2

    After that it is fairly simple. But unfortunately, I can not prove the identity. I've tried expanding it into components but things are becoming messy. Any idea on how to prove this identity (namely : (AXB).(AXB)= (AB)2 - (A.B)2 ) ?

    Bold face indicates a Vector
     
  2. jcsd
  3. Jul 18, 2012 #2

    LCKurtz

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    I will assume you know the identities ##
    A \cdot B \times C = A \times B \cdot C## (interchange of dot and cross) and ##A\times (B\times C) = (A\cdot C)B-(A\cdot C)B##.

    $$\|A\times B\|^2=(A\times B)\cdot (A\times B) = A\cdot (B\times (A\times B))$$ $$
    =A\cdot ((B\cdot B)A - (B\cdot A)B)= \|B\|^2\|A\|^2-(B\cdot A)(B\cdot A)$$
     
  4. Jul 18, 2012 #3
    Hmm, thanks, :). That was short and sweet.

    By the way, I guess you made a small typo(in the second term of the vector triple product.), it will be AX(BXC) = B(A.C)-C(A.B).
     
  5. Jul 19, 2012 #4

    LCKurtz

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    Yes that was a typo. Unfortunately it's too late to correct it but that doesn't matter now anyway.
     
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