Proving a triple vector product identity In a course on skew algebra the following relationship was used in a proof.
$\omega\wedge(r\wedge\omega)=r\left\|\omega\right\|^2$
In this $\omega$ is the angular velocity and $r$ a position vector which are both in $\mathbb{R}^3$. When I try some examples I find that this indeed is the case but I would like to know a proper proof.

 A: I assume here that $\wedge$ is the three-dimensional cross product: in this case the identity you have written is true only when $r$ and $\omega$ are orthogonal (note for instance the identity can't possible be true when $r=\omega$, since the LHS vanishes but the RHS does not).
In the orthogonal case it should be easy to work out the identity in coordinates, but here's a coordinate-free argument: $r$, $\omega$, and $n = r\wedge \omega$ form an orthogonal basis for $\mathbb{R}^3$. Since $\omega \wedge n$ must be orthogonal to both $\omega$ and $n$, $\omega \wedge (r\wedge \omega)$ must be parallel to $r$. Since all vectors are orthogonal the magnitude is $\|\omega\|\|r\wedge \omega\| = \|r\|\|\omega\|^2,$ and thus
$$\omega \wedge (r\wedge \omega) = \pm r\|\omega\|^2.$$
Finally apply the right-hand rule to see the sign must be positive.
A: Using the vector triple product identity $\,a \times (b \times c) = \langle a,c\rangle \,b\, - \langle a,b\rangle \,c\,$ with $\,a=c=\omega\,$, $\,b=r\,$:
$$
\omega\wedge(r\wedge\omega) = \langle \omega,\omega\rangle \,r\, - \langle \omega,r\rangle \,\omega = \left\|\omega\right\|^2\,r - \langle \omega,r\rangle \,\omega
$$
If $\,\omega \perp r \iff \langle \omega,r\rangle=0\,$ the above does indeed reduce to $\,\omega\wedge(r\wedge\omega)=\left\| \omega \right\| ^2\,r\,$.
