TITLE: Seeking an analytic proof of a vector identity
QUESTION [5 upvotes]: Show that for any vectors $\bf{u_1},\bf{u_2},\bf{v_1},\bf{v_2}\in\mathbb R^3$, we have $$(\bf{u_1}\times\bf{v_1})\cdot(\bf{u_2}\times\bf{v_2})=
\left|\begin{matrix}
\bf{u_1}\cdot\bf{u_2} & \bf{u_1}\cdot\bf{v_2}\\
\bf{v_1}\cdot\bf{u_2} & \bf{v_1}\cdot\bf{v_2}
\end{matrix} \right|.$$

I really don't want to "explode" it, is there a more analytic way to prove this identity? Thanks.

REPLY [4 votes]: An approach without too much calculation
The map $f:x\mapsto (\bf{u_1}\times\bf{v_1}$$)\cdot($$\bf{u_2}\times\bf{x}$$)$ is a linear form with kernel $\ker f= \langle\bf{u_2}\rangle$ and similary the map 
$$g: x\mapsto\left|\begin{matrix}
\bf{u_1}\cdot\bf{u_2} & \bf{u_1}\cdot\bf{x}\\
\bf{v_1}\cdot\bf{u_2} & \bf{v_1}\cdot\bf{x}
\end{matrix} \right|$$ is a linear form and $\ker f=\ker g$ hence we conclude that there's $\lambda\in\mathbb R$ such that $f=\lambda g$. Now take  $\bf{u_1}=\bf{u_2}$ and $\bf{v_1}=\bf{v_2}$ and $\bf{u_1}\perp \bf{v_1}$ and we find $\lambda=1$