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Homework Help: Vector proof

  1. Jun 19, 2010 #1
    i) Show that: a x ( b x c) + b x ( c x a) + c x (a x b ) =0
    I managed to this, by expanding each term using the definition of the triple vector product i.e. a x ( b x c) = (a.c)b-(a.b)c and adding the results.

    ii) and deduce that

    a x { b x ( c x d ) } + b x { c x ( d x a ) } + c x { d x ( a x b) } + d x { a x ( b x c ) } =
    ( a x c ) x ( b x d)

    I expanded each term like i did in the first an added the results an obtained:
    -(d x b)(b.a) - (d.a)(c x b) - (b x a)(c.d) - (b.c)(a x d )

    Clearly the result does not agree, and I can't find any means how to simplify it. Some help anyone? Thanks
     
  2. jcsd
  3. Jun 20, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi cordines! Welcome to PF! :smile:
    Noooo :redface:

    you've ignored the hint

    it says "deduce", which means that you should use i) to do it. :wink:
     
  4. Jun 20, 2010 #3
    Actually at first that's what I did by substituting a x ( b x c) = - b x ( c x a) - c x (a x b ) into d x { a x ( b x c ) } and using the triple vector definition for the other three, however it was all but in vain.

    Any other suggestions on how I can substitute i) in ii) ?
     
  5. Jun 20, 2010 #4

    tiny-tim

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    Yes … replace c in i) by c x d :wink:
     
  6. Jun 20, 2010 #5
    As in...

    To prove: a x { b x ( c x d ) } + b x { c x ( d x a ) } + c x { d x ( a x b) } + d x { a x ( b x c ) } = ( a x c ) x ( b x d)

    Proof:

    a x ( b x c ) = ( a . c )b - (a . b)c
    a x { b x ( c x d ) } = { a . (c x d) }b - (a.b)(c x d) ..... [1]

    b x ( c x a ) = (b . a)c - ( b . c)a
    b x { c x ( d x a ) } = (b . a)(c x d) - { b. (c x d) }a ..... [2]

    c x ( a x b ) = ( c . b )a - (c . a)b
    c x { d x ( a x b) } = { (c x d) . b }a - { (c x d) . a }b ..... [3]

    Adding [1], [2] and [3] we obtain 0 which verifies with the proof in i).

    For the last term,

    d x { a x ( b x c ) } = d x { (a . c)b - (a . b) c }
    = (a . c) ( d x b) - (a . b) (d x c)

    which does not agree. I think I'm missing something. Did I expand it correctly? Thanks for your patience.
     
  7. Jun 20, 2010 #6

    tiny-tim

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    Your [2] and [3] are completely wrong. :redface:

    Try again. :smile:
     
  8. Jun 20, 2010 #7
    b x ( c x d ) = (b . d)c - (b . c)d
    a x { b x ( c x d ) } = (a x c)(b . d) - (a x d)(b .c) ...[1]

    or should [1] stay as: a x { b x ( c x d ) } = { a . (c x d) }b - (a.b)(c x d)

    c x ( d x a ) = (c . a)d - (c . d)a
    b x { c x ( d x a ) } = (b x d)(c . a) - (b x a)(c . d) ....[2]

    d x (a x b ) = (d . b)a - (d . a)b
    c x { d x (a x b) } = ( c x a)(d . b) - (c x b)(d . a) ...[3]

    a x ( b x c ) = (a . c)b - (a . b) c
    d x { a x ( b x c ) } = (a . c) ( d x b) - (a . b) (d x c) ...[4]

    Is it correct?
     
  9. Jun 21, 2010 #8

    tiny-tim

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    hi cordines! :smile:

    (just got up :zzz: …)
    yes!! :rolleyes:

    (and [2] [3] and [4] should look similar)

    get some sleep, then try again :smile:
     
  10. Jun 21, 2010 #9
    The proof you are seaching are derived from the axioms of the vector space and can be found in any of Schaums compendiums!
     
  11. Jun 22, 2010 #10
    To prove: a x { b x ( c x d ) } + b x { c x ( d x a ) } + c x { d x ( a x b) } + d x { a x ( b x c ) } = ( a x c ) x ( b x d)

    Proof:

    a x { b x ( c x a ) } = {a . (c x d) }b - (a.b)(c x d)

    b x { c x ( d x a ) } = b x { (c.a)d - (c.d)a }
    = (c.a)(b x d) - (c.d)(b x a)

    c x { d x ( a x b ) } = {c . (a x b) }d - (c.d)(a x b)

    d x ( a x ( b x c ) } = d x { (a . c)b - (a . b)c }
    = (a . c)(d x b) - (a . b)(d x c)

    Adding gives:

    {a . (c x d) }b + (c . (a x b) }d = { a x c . d }b - { a x c . b }d
    = ( a x c ) x ( b x d)

    Proved! Thanks
     
  12. Jun 22, 2010 #11
    Looks good, well done! I just spotted one typo in the first line of your proof: you wrote a instead of d on the left hand side.
     
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