Prove that vec a×(vec b×vec c) , vec b and vec c are coplanar

Prove that $\vec a\times(\vec b\times\vec c)$, $\vec b$ and $\vec c$ are coplanar.

Step 1: Algebraic ("BAC–CAB")

\[ \vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\,\vec b-(\vec a\cdot\vec b)\,\vec c, \] a linear combination of $\vec b$ and $\vec c$. Hence the three vectors are linearly dependent, i.e. coplanar.

Step 2: Triple product

\[ [\vec a\times(\vec b\times\vec c),\ \vec b,\ \vec c]=\big(\vec a\times(\vec b\times\vec c)\big)\cdot(\vec b\times\vec c)=0, \] because a vector product is perpendicular to each of its factors, in particular to $\vec b\times\vec c$.

Step 3: Geometric

$\vec b\times\vec c$ is normal to the plane of $\vec b,\vec c$. Any vector perpendicular to that normal lies in the plane. $\blacksquare$

\[\vec a\times(\vec b\times\vec c)\in\operatorname{span}\{\vec b,\vec c\}\]

(If $\vec b\parallel\vec c$, then $\vec b\times\vec c=\vec 0$, and the statement is trivial.)