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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

    dextercioby

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

    :yuck:


    Daniel.
     
  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
    Last edited by a moderator: Apr 21, 2017
  8. Jan 12, 2005 #7

    Dr Transport

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    [tex] \epsilon_{ijk}\epsilon_{lmn}= \left[
    \begin{array}{ccc}
    \delta_{il} & \delta_{im} & \delta_{in} \\
    \delta_{jl} & \delta_{jm} & \delta_{jn} \\
    \delta_{kl} & \delta_{km} & \delta_{kn} \\
    \end{array}
    \right]
    [/tex]

    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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