# Cross Product

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

expand the RHS, and see if they were equal

matt grime
Science Advisor
Homework Helper
Learn summation convention as that's how you ought to prove this result, otherwise it's just messy.

dextercioby
Science Advisor
Homework Helper
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.

matt grime
Science Advisor
Homework Helper
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).

Dr Transport
Science Advisor
Gold Member
$$\epsilon_{ijk}\epsilon_{lmn}= \left[ \begin{array}{ccc} \delta_{il} & \delta_{im} & \delta_{in} \\ \delta_{jl} & \delta_{jm} & \delta_{jn} \\ \delta_{kl} & \delta_{km} & \delta_{kn} \\ \end{array} \right]$$

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.