# 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.