Help deriving Lagrange's Formula with the levi-civita symbol

[radonballoon]

Ok, so I'm really at a loss as to how to do this. I can prove the formula by just using determinants, but I don't really know how to do manipulations with the levi-civita symbol.
Here's what I have so far:
$$(\vec{B} \times \vec{C})_{i} = \epsilon_{ijk}(B_{j}C_{k})\vec{e_{i}}$$

And I'm trying to get to:
$$\vec{A} \times (\vec{B} \times \vec{C}) = B(A \bullet C) - C(A \bullet B)$$

Does anyone have any suggestions?
Thanks

 

[radonballoon]

Ok So I figured it out, I'll just post the answer for the sake of completeness:

$$(\vec{B} \times \vec{C})_{k} = \epsilon_{kmn} (B_{m} C_{n})$$

$$let (\vec{B} \times \vec{C}) = \vec{N}$$

$$\vec{A} \times (\vec{B} \times \vec{C}) = \vec{A} \times \vec{N}$$
$$(\vec{A} \times \vec{N})_{i} = \epsilon_{ijk} A_{j} N_{k}$$
$$= \epsilon_{ijk} A_{j} (\epsilon_{kmn} B_{m} C_{n})$$
$$= \epsilon_{ijk} \epsilon_{kmn} (A_{j} B_{m} C_{n})$$

$$\epsilon_{ijk} \epsilon_{mnk} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}$$

$$\vec{A} \times (\vec{B} \times \vec{C}) = (\delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}) A_{j} B_{m} C_{n}$$
$$= B_{i} A_{j} C_{j} - A_{j} B_{j} C_{i}$$
$$= \vec{B}(\vec{A} \bullet \vec{C}) - \vec{C}(\vec{A} \bullet \vec{B})$$

 

[Manchot]

It's threads like these that seem to be causing PF to accumulate helpful Google searches. :)

 

[malawi_glenn]

$$<br /> (\vec{B} \times \vec{C})_{i} = \epsilon_{ijk}(B_{j}C_{k})\vec{e_{i}}<br />$$

Should be:

$$<br /> (\vec{B} \times \vec{C})_{i} = \epsilon_{ijk}(B_{j}C_{k})<br />$$

And start using \cdot instead of that big black ball :-D

 

[radonballoon]

Haha wow this seems like so long ago. I couldn't find the dot for dot product, so thanks for that :D