The cross product between vectors in $\mathbb{R}^3$ obeys two pleasant identities (sometimes named after Lagrange), namely

  • $a\times(b\times c)=b(a\cdot c)-c(a\cdot b)$
  • $(a\times b)\cdot(c\times d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)$.

The first one can be proved by invoking multilinearity and thus reducing oneself to a tedious check when $\{a,b,c\}\subseteq\{e_1,e_2,e_3\}$. The second one can be deduced from the first using properties of the triple product.

I am looking for some slick/conceptual proof of the two identities (or one of them); maybe exterior algebra has something to tell us?

  • $\begingroup$ You know levi-cevita symbol? $\endgroup$ – SchrodingersCat Dec 27 '15 at 15:53
  • 2
    $\begingroup$ Yes, I forgot mentioning that alternative. But using it means invoking the identity $\epsilon _{ijk}\epsilon_{imn}=\delta_{jm}\delta_{kn}-\delta_{jn}\delta_{km}$, which looks too magical to me, so I'd like to avoid it. $\endgroup$ – Mizar Dec 27 '15 at 15:57

You can definitely approach this from the perspective of exterior algebra. Indeed, the part of your question relating to your first identity has been asked before, so for the coordinate-free, exterior algebra proof of that, please take a look at my answer to the earlier question. So, let me know turn to your second identity. Before continuing, let me recycle some background and notation from my old answer.

Recall that if $V$ is an $n$-dimensional inner product space, then the Hodge star is a linear isomorphism $\ast : \bigwedge^k V \to \bigwedge^{n-k} V$ for each $0 \leq k \leq n$, satisfying the following:

  1. for $v$, $w \in V$, $\langle v,w\rangle \omega = v \wedge \ast w$ for $\omega = \ast 1$ the generator of $\bigwedge^n V$ satisfying $\omega = e_1 \wedge \cdots \wedge e_n$ for any orthonormal basis $\{e_k\}$ of $V$ with the appropriate orientation (e.g., the volume form in $\bigwedge^n (\mathbb{R}^n)^\ast$);

  2. in particular, $\ast \ast = \operatorname{Id}$ when $n$ is odd.

Also, recall that the inner product on $\bigwedge^k V$ is given by $$\langle v_1 \wedge \cdots \wedge v_k, w_1 \wedge \cdots \wedge w_k \rangle = \det(\langle v_i,w_j \rangle). $$

So, suppose that $V$ is $3$-dimensional, in which case we can define the cross product of $a$, $b \in V$ by $$a \times b := \ast (a\wedge b).$$

Now, let $a$, $b$, $c$, $d \in \mathbb{R}^3$. Then $$ \langle a \times b, c \times d \rangle \omega = \langle \ast (a \wedge b), \ast (c \wedge d) \rangle \omega \\ = \ast (a \wedge b) \wedge \ast \ast (c \wedge d) \\ = \ast (a \wedge b) \wedge (c \wedge d)\\ = (-1)^{1 \cdot 2} (c \wedge d) \wedge \ast (a \wedge b)\\ = \langle c \wedge d, a \wedge b \rangle\omega \\ = \langle a \wedge b, c \wedge d \rangle\omega\\ = \begin{vmatrix} \langle a,c \rangle & \langle a,d \rangle \\ \langle b,c \rangle & \langle b,d \rangle \end{vmatrix}\omega\\ = (\langle a,c \rangle \langle b,d \rangle - \langle a,d \rangle \langle b,c \rangle)\omega, $$ as was required. Observe, in particular, that the right-hand side of your second identity looks like a $2 \times 2$ determinant because once you've unpacked all relevant definitions, it really is a $2 \times 2$ determinant, namely the inner product $$ \langle a \wedge b, c \wedge d \rangle := \begin{vmatrix} \langle a,c \rangle & \langle a,d \rangle \\ \langle b,c \rangle & \langle b,d \rangle \end{vmatrix} $$ of the bivectors $a \wedge b$, $c \wedge d \in \bigwedge^2 \mathbb{R}^3$.


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