# The identity $(u\times v)\cdot(x\times y)=\begin{vmatrix}u\cdot x&v\cdot x\\ u\cdot y&v\cdot y\\\end{vmatrix}$

I know by brutal calculation this identity holds always:

$$(u × v) \cdot (x × y) = \begin{vmatrix} u \cdot x & v \cdot x \\ u \cdot y & v \cdot y \\ \end{vmatrix}$$

for arbitrary vectors $$u$$, $$v$$, $$x$$, $$y$$.

I'd like to know where it come from naturally.

Thank you.

• To answer your question on where it comes from naturally, the point is that in $\Bbb R^3$ the cross product of vectors can be interpreted as the wedge (exterior) product of the vectors. This formula actually generalizes to give a natural formula for the dot product of two wedge (exterior) products of $k$ vectors for any positive integer $k$. If $e_i$ are an orthonormal basis for $\Bbb R^n$, you declare $e_i\wedge e_j$ (for $1\le i<j\le n$) to be an orthonormal basis for $\Lambda^2\Bbb R^n$, and you end up with this formula in $\Bbb R^3$. Similarly for $k\ge 3$ ... – Ted Shifrin Sep 30 at 23:22

I suppose you are talking about dot products and cross products in $$\mathbb R^3$$. The identity can be seen as a special case of Cauchy-Binet formula. Alternatively, note that $$\pmatrix{x^T\\ y^T\\ (u\times v)^T}\pmatrix{u&v&x\times y} =\pmatrix{u\cdot x&v\cdot x&0\\ u\cdot y&v\cdot y&0\\ 0&0&(u\times v)\cdot(x\times y)}.\tag{1}$$ Therefore the RHS has determinant $$\det\pmatrix{u\cdot x&v\cdot x\\ u\cdot y&v\cdot y}[(u\times v)\cdot(x\times y)].\tag{2}$$ On the other hand, since $$\det\pmatrix{p&q&r}\equiv(p\times q)\cdot r$$ (this actually is the definition of cross product), the LHS of $$(1)$$ has determinant $$[(u\times v)\cdot(x\times y)]^2.\tag{3}$$ The result now follows by equating $$(2)$$ and $$(3)$$ with the use of a continuity argument.
Since $$\epsilon_{ijk}\epsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$$, the dot product is$$\epsilon_{ijk}u_jv_k\epsilon_{ilm}x_ly_m=(\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl})u_jv_kx_ly_m=(u\cdot x)(v\cdot y)-(u\cdot y)(v\cdot x)=\begin{vmatrix} u\cdot x & v\cdot x \\ u\cdot y & v\cdot y \\ \end{vmatrix}.$$This is also the only linear combination of the four vectors, with coefficient $$1$$ for $$u_1v_2x_1y_2$$, that's symmetric under simultaneously exchanging $$u,\,x$$ and $$v,\,y$$, but antisymmetric under exchanging $$u$$ with $$v$$, or $$x$$ with $$y$$.