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Prove Quadruple Product Identity from Triple Product Identities

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to prove the identity:

    (a×b)[itex]\cdot[/itex](c×d)= (a[itex]\cdot[/itex]c)(b[itex]\cdot[/itex]d)-(a[itex]\cdot[/itex]d)(b[itex]\cdot[/itex]c)

    using the properties of the vector and triple products:

    2. Relevant equations

    a×(b×c)=b(a[itex]\cdot[/itex]c)-c(a[itex]\cdot[/itex]b)
    a[itex]\cdot[/itex](b×c)=c[itex]\cdot[/itex](a×b)=b[itex]\cdot[/itex](c×a)

    3. The attempt at a solution
    I really don't know where to begin. I need to prove this identity simply so I can use it on a problem, and I know it CAN be proven using these identities from the triple products, but I'm lost on how to attempt such a proof.
     
  2. jcsd
  3. Sep 28, 2012 #2

    SammyS

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    What have you tried?

    Here's a hint:

    Let u = c×d. Then use the scalar triple product, then substitute c×d back in for u, and see where that's leading you.

    Added in Edit:

    Putting in u, then applying the scalar triple product will simply let you switch a sclar product and a vector product, but that will allow you to get the desired result.
     
    Last edited: Sep 28, 2012
  4. Sep 28, 2012 #3
    Okay, I did that, but I still am not understanding.

    Letting u=(c×d):

    (a×b)[itex]\cdot[/itex](c×d) = u[itex]\cdot[/itex](a×b) = b[itex]\cdot[/itex](u×a)
    =b[itex]\cdot[/itex]((c×da)
    =-b[itex]\cdot[/itex](a×(c×d))
    =b[itex]\cdot[/itex](c(a[itex]\cdot[/itex]d)-d(a[itex]\cdot[/itex]c))

    And I don't know what to do with this either...
     
  5. Sep 28, 2012 #4

    SammyS

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    Use the distributive law to distribute the b vector. Don't forget, ad and ac are just scalars.
     
  6. Sep 28, 2012 #5
    Right, I've got that:

    b[itex]\cdot[/itex]c(a[itex]\cdot[/itex]d)-b[itex]\cdot[/itex]d(a[itex]\cdot[/itex]c)

    Can I go from here to:

    (b[itex]\cdot[/itex]c)(a[itex]\cdot[/itex]d)-(b[itex]\cdot[/itex]d)(a[itex]\cdot[/itex]c) ?

    But this is backwards of the identity I'm after...
     
  7. Sep 28, 2012 #6

    SammyS

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    The vector product is not commutative.

    [itex]\textbf{r}\times\textbf{s}=-\textbf{s}\times\textbf{r}[/itex]
    Fix that & you'll be OK .
     
  8. Sep 28, 2012 #7
    Thanks! I had the negative sign there but forgot to carry it through the last step. Thanks for the help!!
     
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