# Help with vector cross product identity

You can use properties of the Levi-Civita tensor namely,

$$\epsilon_{kij}\epsilon_{kmn} = \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm}$$

so that

$$\vec a \times (\vec b \times \vec c) = \hat e_i \epsilon_{ijk} a_j (\epsilon_{kmn} b_m c_n )$$

$$= \hat e_i ( \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm} )a_j b_m c_n$$

$$= \hat e_i \delta_{im}\delta_{jn} a_j b_m c_n - \hat e_i \delta_{in}\delta_{jm} a_j b_m c_n$$

$$= \hat e_i a_j b_i c_j - \hat e_i a_j b_j c_i$$

Any steps in between and the rest should be straight forward - you may want to double check these indices are in the correct spot. See the link in the answer provided by Puk.

I always have trouble with this identity, so here's a fun way to derive it in three-dimensions. It may be argued that this method is a little convoluted, but I find it much easier to remember than the Levi-Civita contraction formula, and much less tedious than working out the components! Let's call the vector $$\mathbf{a \times (b \times c) = d}$$, and see what we can say about $$\mathbf{d}$$, using our intuition.

Now, $$\mathbf{d}$$ must be perpendicular to $$\mathbf{a}$$ by the definition of the cross-product. Furthermore, $$\mathbf{d}$$ must also be perpendicular to the vector $$\mathbf{(b\times c)}$$. From these two facts, you should be able to see that $$\mathbf{d}$$ must lie in the plane formed by the vectors $$\mathbf{b}$$ and $$\mathbf{c}$$! (If you're not convinced, try it out: the first cross-product between $$\mathbf{b}$$ and $$\mathbf{c}$$ takes you out of the $$\mathbf{bc}-$$plane, and the second cross product (with $$\mathbf{a}$$) has to bring you back onto it, because we're in three dimensions!)

As a result, since $$\mathbf{d}$$ lies in the plane of the vectors $$\mathbf{b}$$ and $$\mathbf{c}$$, it can therefore be written as a linear combination of them: $$\mathbf{d} = \alpha \, \mathbf{b} + \beta \, \mathbf{c},$$ where $$\alpha$$ and $$\beta$$ are scalars.

We now use the fact that $$\mathbf{d}$$ must be linear in $$\mathbf{a}$$,$$\mathbf{b}$$, and $$\mathbf{c}$$, and therefore all the terms on the right hand side must have only one power of each of these vectors respectively. Thus, $$\alpha$$ must be proportional to $$(\mathbf{a\cdot c})$$, since it has to be a scalar constructed from $$\mathbf{a}$$ and $$\mathbf{c}$$, and similarly $$\beta$$ must be proportional to $$(\mathbf{a\cdot b})$$. Thus, $$\mathbf{d} = A\, (\mathbf{a\cdot c}) \mathbf{b} + B\, (\mathbf{a\cdot b}) \mathbf{c},$$ where $$A$$ and $$B$$ are two absolute constants (dimensionless numbers) that are independent of the vectors.

Using the fact that $$\mathbf{d}$$ changes sign if $$\mathbf{b}$$ and $$\mathbf{c}$$ are interchanged, you should trivially be able to show that $$A=-B$$, and so $$\mathbf{d} = A\, \Big( (\mathbf{a\cdot c}) \mathbf{b} - (\mathbf{a\cdot b}) \mathbf{c}\Big).$$ All that's left to do now is to determine $$A$$, which is easily done by taking a special case (since the above equation is valid for all vectors), so we could set $$\mathbf{a} = \hat{\mathbf{x}}, \mathbf{b} = \hat{\mathbf{x}}, \mathbf{c} = \hat{\mathbf{y}},$$ for example, and we'd see in this case that $$A= 1$$, and so $$\mathbf{a \times (b\times c)} = (\mathbf{a\cdot c}) \mathbf{b} - (\mathbf{a\cdot b}) \mathbf{c}$$

A more down-to-earth approach is to prove this identity in three-dimensional space by writing it down in terms of vector components: $$\vec{a} = (a_x, a_y, a_z)$$, etc., and using the expression for the vector product $$\vec{a}\times\vec{b} = \left| \begin{matrix} \hat{e}_x & \hat{e}_y & \hat{e}_z \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{matrix} \right|$$ It may seem a bit tedious, but it is straightforward and foolproof.

Update
In some corners this identity is called Bee-Ay-Cee minus Cee-Ay-Bee, which is a simple mnemonic rule for memorizing it.