Vector Identity with index notation

Question

I'm trying to proving the vector identity given by $$(\mathbf{A}\times \mathbf{B})\times (\mathbf{C}\times \mathbf{D})=(\mathbf{A}\cdot(\mathbf{B}\times \mathbf{D}))\mathbf{C}-(\mathbf{A}\cdot(\mathbf{B}\times \mathbf{C}))\mathbf{D}$$

I started from left hand side $$((\mathbf{A}\times \mathbf{B})\times (\mathbf{C}\times \mathbf{D}))_i=\epsilon_{ijk}(\mathbf{A}\times \mathbf{B})_j (\mathbf{C}\times \mathbf{D})_k$$ $$=\epsilon_{ijk}\epsilon_{jpq }\epsilon_{klm}A_pB_qC_lD_m$$ I don't idea what to do now. I tried using product property but that's just messed everything.

Please help me though this.

Answer 1

The first and third Levi-Civita symbols contract to $\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$, so the component is$$\color{blue}{\epsilon_{pqj}}(A_pB_qC_iD_j-A_pB_qC_jD_i)=(A\cdot B\times D)C_i-(A\cdot B\times C)D_i,$$where the blue factor uses $\epsilon_{jpq}=\epsilon_{pqj}$.

A proof with less calculation uses the fact that, since $C\times D$ is orthogonal to the plane $C,\,D$ span, $(A\times B)\times(C\times D)$ is a linear combination of $C,\,D$ that changes sign when they're exchanged, and also when $A,\,B$ are. This, together with multilinearity, gives the desired result up to a proportionality constant, which can be fixed with an example.

Answer 2

$$ \begin{aligned} \left[ \left( \mathbf{A}\times \mathbf{B} \right) \times \left( \mathbf{C}\times \mathbf{D} \right) \right] ^k&=\epsilon _{ij}^{k}\left( \epsilon _{mn}^{i}A^mB^n \right) \left( \epsilon _{pq}^{j}C^pD^q \right)\\ &=\epsilon ^{ijk}\left( \epsilon _{mni}A^mB^n \right) \left( \epsilon _{pqj}C^pD^q \right)\\ &=\delta _{pqj}^{ijk}\left( \epsilon _{mni}A^mB^n \right) \left( C^pD^q \right)\\ &=\delta _{pqj}^{kij}\left( \epsilon _{mni}A^mB^n \right) \left( C^pD^q \right)\\ &=\delta _{pq}^{ki}\left( \epsilon _{mni}A^mB^n \right) \left( C^pD^q \right)\\ &=2\left( \epsilon _{mni}A^mB^n \right) \left( C^{k}D^{i} \right)\\ &=\left( \epsilon _{mni}A^mB^n \right) \left( C^kD^i \right) -\left( \epsilon _{mni}A^mB^n \right) \left( C^iD^k \right)\\ &=\left( \epsilon _{mni}A^mB^nD^i \right) C^k-\left( \epsilon _{mni}A^mB^nC^i \right) D^k\\ &=A^m\left( \epsilon _{nim}B^nD^i \right) C^k-A^m\left( \epsilon _{nim}B^nC^i \right) D^k\\ &=\left( \mathbf{A}\cdot \left( \mathbf{B}\times \mathbf{D} \right) \right) C^k-\left( \mathbf{A}\cdot \left( \mathbf{B}\times \mathbf{C} \right) \right) D^k\\ \end{aligned} $$ As desired.

Answer 3

You can reduce your identity to the well-known vector identity given by

$$ \mathbf{X}\times(\mathbf{C}\times\mathbf{D})=(\mathbf{D}\cdot\mathbf{X})\,\mathbf{C}-(\mathbf{C}\cdot\mathbf{X})\,\mathbf{D}. $$

Now just set $\mathbf{X}=\mathbf{A}\times\mathbf{B}$ and use the following property of the triple product

$$ \mathbf{A}\cdot(\mathbf{B}\times\mathbf{C})=\det(\mathbf{A},\mathbf{B},\mathbf{C})=\det(\mathbf{C},\mathbf{A},\mathbf{B})=\mathbf{C}\cdot(\mathbf{A}\times\mathbf{B}), $$

that is, you can cyclically permute the vectors in the determinant.