# Help with vector cross product identity [closed]

In EM Physics we were given the problem to show that

$$\vec a \times (\vec b \times \vec c) = \vec b (\vec a\cdot \vec c) - \vec c (\vec a \cdot \vec b).$$

I know first

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

but don't know where to go from here. I don’t want to fully expand. If someone can give me pointer that would be great.

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

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

so that

\begin{align} \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 \\ &=\vec b (\vec a\cdot \vec c) - \vec c(\vec a\cdot\vec b) \end{align}

Any steps in between 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'm not sure if the Levi-Civita symbol is standard knowledge on undergraduate physics courses. Also, it's actually a tensor density, and not a tensor (it doesn't transform as a tensor).
– Tom
Commented Oct 14, 2020 at 16:08
• @Tom It probably depends on the university. It was part of my undergraduate course in Germany, first year even. Commented Oct 14, 2020 at 17:54
• OK yes it probably depends. Now that I think about it I probably did actually learn it on my electromagnetism course I took, so maybe depends on the university.
– Tom
Commented Oct 14, 2020 at 18:39
• The Levi-Civita symbol can be interpreted as an orientation tensor which does transform as a tensor, it just has negative components in “left-handed” choices of basis. I would just share that my mnemonic is basically to look at $\epsilon_{abc}\epsilon_{ak\ell}$ and say “well these are zero for non-permutations so $abc$ and $ak\ell$ are both permutations, either they are the same permutation giving +1, or they are opposite, giving -1. They are the same if $b=k$ and $c=\ell$ so that's $\delta_{bk}\delta_{c\ell},$ they are opposite if $b=\ell$ and $c=k$ so that's $-\delta_{b\ell}\delta_{ck}.$” Commented Oct 14, 2020 at 19:03

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 $$\mathbf{a \times (b\times c)} = (\mathbf{a\cdot c}) \mathbf{b} - (\mathbf{a\cdot b}) \mathbf{c}$$

• Given that this is a Physics EM question, I think this is much more likely what is being looked for since undergrad physics won't be covering Levi-Civita and this requires an intuitional understanding of vectors which is more valuable in Physics Commented Oct 14, 2020 at 15:03

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.

Use the contraction identities of the Levi-Civita symbol. I strongly encourage you to prove these identities yourself as well, I think you will find it worth the effort in the long run.

Rotation of the axes does not affect the cross and dot products. Since the equation is obvious if $$\mathbf{b} \times \mathbf{c}$$ is zero, we can assume that $$\mathbf{b} \times \mathbf{c}$$ is a nonzero multiple of $$\mathbf{i}.$$ Now, we can assume that $$\mathbf{b}=(0,r,0)$$ and that $$\mathbf{c}=(0,s,t),$$ with both $$r$$ and $$t$$ nonzero. Thus, $$\mathbf{a} \times (\mathbf{b} \times \mathbf{c})=rt(0,a_3,-a_2).$$ As observed in another answer, $$\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$$ is linear combination of $$\mathbf{b}$$ and $$\mathbf{c}.$$ Thus, $$\mathbf{a} \times (\mathbf{b} \times \mathbf{c})=$$ $$rt(0,a_3,-a_2)=k_1\mathbf{b} +k_2\mathbf{c} =k_1(0,r,0)+k_2(0,s,t).$$ Clearly, $$k_2=-ra_2=-\mathbf{a}\cdot \mathbf{b}.$$ From $$rta_3=k_1r+k_2t,$$ we conclude that $$k_1=sa_2+ta_3=\mathbf{a}\cdot \mathbf{c}.$$ Observe that the cross-product calculations above were extremely easy.

You want to know when $$\epsilon_{ijk} \epsilon_{kmn}$$ is different from zero. A basic property is that $$\epsilon_{ijk} = \epsilon_{jki} = \epsilon_{kij} = -\epsilon_{ikj}$$ (cyclical and anti-cyclical permutations of the indices). So to make reasoning simpler, let's rewrite your product as $$\epsilon_{kij} \epsilon_{kmn}$$. Note that this can only be non-zero if all indices $$1,2,3$$ appear exactly once in each term, and since $$k$$ appears in both this means that in all non-zero terms either $$i=m$$ and $$j=n$$ or $$i=n$$ and $$j=m$$. In the first case both $$\epsilon$$s have the same indices and thus evaluate to the same value giving a product equal $$+1$$, in the second case they evaluate to the opposite value, and thus their product is $$+1\times -1=-1$$. This can be summed up in the identity given in the accepted answer:$$\epsilon_{ijk}\epsilon_{kmn} = \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm}.$$ The remainder is identifying the products that remain after simplifying via $$a_i \equiv \delta_{ij} a_j$$.