- #1

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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

- Thread starter starbaj12
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- #1

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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

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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.

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Yeah,use cartesian tensors.And Levi-Civita (cartesian) tensor.This is ugly.matt grime said:Learn summation convention as that's how you ought to prove this result, otherwise it's just messy.

:yuck:

Daniel.

- #5

matt grime

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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.

http://mit.fnal.gov/~paus/8.21-IAP2001/notes/notes/node5.html

http://mit.fnal.gov/~paus/8.21-IAP2001/notes/notes/node5.html

Last edited by a moderator:

- #7

Dr Transport

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\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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