Prove $(\vec a \times \vec b)\times(\vec c \times \vec d)=[\vec a,\vec b, \vec d]\vec c - [\vec a, \vec b, \vec c]\vec d$

Question

Let $\mathbf{a,b,c,d} \in \mathbb{R}^3$. Prove that $$ \newcommand{\bvec}[1]{\mathbf{#1}} (\bvec a \times \bvec b)\times(\bvec c \times \bvec d) = \begin{vmatrix} \vert & \vert & \vert \\ \bvec{a} & \bvec b & \bvec d \\ \vert & \vert& \vert \end{vmatrix} \bvec c - \begin{vmatrix} \vert & \vert & \vert \\ \bvec{a} & \bvec b & \bvec c \\ \vert & \vert& \vert \end{vmatrix}\bvec d $$

This is the progress I've made so far. I don't know what to do and it seems like a dead end, can someone please help me?

My friend also tried solving this, and this is her example

Answer 1

Can you use the identity $$ \mathbf{a}\times (\mathbf{b}\times \mathbf{c}) = (\mathbf{a}\cdot\mathbf{c})\mathbf{b} -(\mathbf{a}\cdot\mathbf{b}) \mathbf{c}? $$ A brief proof of this identity is given on the wikipedia article for the triple product.

With this result, your proof becomes straightforward: \begin{align} (\mathbf a \times \mathbf b)\times(\mathbf c \times \mathbf d) &= \left( (\mathbf a \times \mathbf b) \cdot \mathbf{d} \right) \mathbf{c} - \left( (\mathbf a \times \mathbf b) \cdot \mathbf{c} \right) \mathbf{d}\\ &= \det\begin{bmatrix} \vert & \vert & \vert \\ \mathbf{a} & \mathbf b & \mathbf d \\ \vert & \vert& \vert \end{bmatrix} \mathbf c - \det\begin{bmatrix} \vert & \vert & \vert \\ \mathbf{a} & \mathbf b & \mathbf c \\ \vert & \vert& \vert \end{bmatrix}\mathbf d \end{align}

Answer 2

Hint Use the triple product identities $$({\bf x} \times {\bf y}) \times {\bf z} = -({\bf y} \cdot {\bf z}) {\bf x} + ({\bf x} \cdot {\bf z}) {\bf y},$$ and $$\pmatrix{{\bf x}&{\bf y}&{\bf z}} = ({\bf x} \times {\bf y}) \cdot {\bf z}.$$