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Proved A x [B x (Cx D) ] = 0

  1. Nov 21, 2011 #1
    Hello I hope you can help me in solving this proof

    proved A x [B x (Cx D) ] = 0


    Thank you
     
  2. jcsd
  3. Nov 21, 2011 #2

    LCKurtz

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    Hello Samia, welcome to PF. Since this is your first post you may not have read the rules, one of which is you must show what you have tried. Nevertheless I will give you a Hint: Try the calculation with 4 different vectors.
     
  4. Nov 21, 2011 #3

    Deveno

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    that's just...cruel.

    presumably A,B,C and D lie in R3?

    if so, A must be a linear combination of B,C, and D.

    the cross-product is distributive, is it not?

    from here it get easy....because of a certain elementary property of cross-products.
     
  5. Nov 21, 2011 #4

    LCKurtz

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    How does it get easy when it is false? And what is cruel about suggesting that most anything will give a counterexample?
     
  6. Nov 21, 2011 #5

    Deveno

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    except, obviously, for the ones i tried. please excuse me while i shave my facial omelette.
     
  7. Nov 21, 2011 #6
    This theorem can't be proved, because (as noted by LCKurtz) it is simply false. A simple counter can be shown by letting A=B=C=i, D=j where i,j, and k are the unit vectors in the x,y, and z directions (respectively). Then

    A x (B x (C x D))=i x (i x (i x j))
    =i x (i x (k))
    =i x (-j)
    A x (B x (C x D))=-k0
     
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