Cross Product

  • Thread starter starbaj12
  • Start date
  • #1
49
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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
 

Answers and Replies

  • #2
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expand the RHS, and see if they were equal
 
  • #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.
 
  • #4
dextercioby
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matt grime said:
Learn summation convention as that's how you ought to prove this result, otherwise it's just messy.
Yeah,use cartesian tensors.And Levi-Civita (cartesian) tensor.This is ugly.

:yuck:


Daniel.
 
  • #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
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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