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Double cross product

  1. Sep 22, 2015 #1
    1. The problem statement, all variables and given/known data
    If ##u,v,w\in\mathbb{R}^3##, show that ## u\times(v\times w) = (u.w) v - (u.v) w ##.

    2. Relevant equations

    3. The attempt at a solution
    Since ## u\times(v\times w)##, ##v## and ##w## are orthogonal to ##v\times w##, these vectors are coplanar. Therefore, there must be reals ## a,b## such that ## u\times(v\times w) = a v + b w ##.
    Then, ## 0 = u.(u\times(v\times w)) = a u.v + b u.w ## and ## a = u.w## and ## b = - u.v## work. But why not ## a = -u.w ## and ## b = u.v ## ? Thanks
     
  2. jcsd
  3. Sep 22, 2015 #2

    ehild

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    Write the vectors in Cartesian components and expand the double product.
     
  4. Sep 22, 2015 #3

    Ray Vickson

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    An alternative method is to write vectors as ##/vec{v} = (v_1,v_2,v_3)## for example, and use the so-called Levi-Civita symbol to express the cross-product:
    [tex] (\vec{u} \times \vec{v}) _i = \sum_{j,k} \epsilon_{ijk} u_j v_k [/tex]
    Here the Levi-Civita symbol ##\epsilon## is
    [tex] \epsilon_{ijk} = \begin{cases} +1 & \text{if} \:\; ijk \; \text{is an even permutation of} \: \; 123 \\
    -1 & \text{if} \:\; ijk \; \text{is an odd permutation of} \:\; 123 \\
    0 & \text{otherwise} .
    \end{cases}
    [/tex]
    There are identities available for sums of products like ##\sum_k \epsilon_{ijk} \epsilon_{klm}## which would be the thing needed in the triple product. See, eg., http://www.ucl.ac.uk/~ucappgu/seminars/levi-civita.pdf or https://en.wikipedia.org/wiki/Levi-Civita_symbol .
     
  5. Sep 23, 2015 #4
    Thank you for your answers, but is it possible to avoid lengthy calculations ?
    In the OP, I am to the point where ## (a,b) \in \{\pm (u.w, - u.v) \} ##.
    How should I decide for the sign ?
     
  6. Sep 23, 2015 #5

    Ray Vickson

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    Some detailed calculations are just unavoidable. However, if you pay attention to what I said in #3, that will lead to just about the shortest and most straightforward proof available. Besides, you have NOT reached the point where ## (a,b) \in \{\pm (u.w, - u.v) \} ##, because you had one relation between ##a## and ##b##, and you said that your particular choice "works". However, in principle, there are infinitely many other choices of ##a,b## that also work for that single condition. Somehow, you need more conditions in order to pin down ##a,b## convincingly and uniquely.
     
    Last edited: Sep 23, 2015
  7. Sep 23, 2015 #6
    Oops you're right, I'm just at ## \begin{pmatrix}a\\b\end{pmatrix} \in \mathbb{R} \begin{pmatrix}u.w \\ -u.v \end{pmatrix} ##.
    Thanks again, I will try your proof.
     
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