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Cross Product

  1. Jan 11, 2005 #1
    AX(BXC) = B(A(dot)C) - C(A(dot)B)

    For the left hand side I got

    (AyBxCy-AyByCx+AzBxCz-AzBzCx)x(hat) + (-AxBxCy+AxByCx+AzByCz-AzBzCy)y(hat) + (-AxBxCz+AxBzCx-AyByCz+AyBzCy)z(hat)

    Is this right?

    Where would I go from here to prove the rest?

    Thanks for the help
  2. jcsd
  3. Jan 11, 2005 #2
    expand the RHS, and see if they were equal
  4. Jan 12, 2005 #3

    matt grime

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    Learn summation convention as that's how you ought to prove this result, otherwise it's just messy.
  5. Jan 12, 2005 #4


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    Yeah,use cartesian tensors.And Levi-Civita (cartesian) tensor.This is ugly.


  6. Jan 12, 2005 #5

    matt grime

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    Not sure about the neccesity to learn about tensors as such since I could do this question, with summation convention, well before I knew what a tensor was. Admittedly the things involved are tensors, but there's no need to know this (I mean, we aren't even transforming anything).
  7. Jan 12, 2005 #6
    Just in case the OP was interested...here's a cute little blurb on summation convention and the three dimensional cross product is expanded as such at the bottom. It's much prettier to use this method.
    Summation Convention
  8. Jan 12, 2005 #7

    Dr Transport

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    [tex] \epsilon_{ijk}\epsilon_{lmn}= \left[
    \delta_{il} & \delta_{im} & \delta_{in} \\
    \delta_{jl} & \delta_{jm} & \delta_{jn} \\
    \delta_{kl} & \delta_{km} & \delta_{kn} \\

    is the product of Levi-Civita tensors. Using this and
    [tex] \vec{a} \cross \vec{b} = \epsilon_{ijk}a_{i}b_{j} [/tex] and
    [tex] \vec{a} \cdot \vec{b} = a_{i}b_{i} [/tex]

    you should be able to prove any vector identity. I relearned this technique one week when it was really boring at work a couple of years ago.
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