# Cross Product

1. Jan 11, 2005

### 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

2. Jan 11, 2005

### vincentchan

expand the RHS, and see if they were equal

3. Jan 12, 2005

### matt grime

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

4. Jan 12, 2005

### dextercioby

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

:yuck:

Daniel.

5. Jan 12, 2005

### 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).

6. Jan 12, 2005

### houserichichi

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

### Dr Transport

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

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.

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