Proving AX(BXC) = B(A(dot)C) - C(A(dot)B)

[starbaj12]

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

 

[vincentchan]

expand the RHS, and see if they were equal

 

[matt grime]

Learn summation convention as that's how you ought to prove this result, otherwise it's just messy.

 

[dextercioby]

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.




Daniel.

 

[matt grime]

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).

 

[houserichichi]

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

 

[Dr Transport]

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

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

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.