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.

  • $\begingroup$ 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 $k\ge 3$ ... $\endgroup$ – 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.

  • $\begingroup$ Wow! the equation (1) is what I couldn't come up with. thx :) $\endgroup$ – glimpser Sep 28 at 9:54
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    $\begingroup$ What a beautiful proof! $\endgroup$ – Omnomnomnom Sep 28 at 9:58

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 $y$.


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