Prove that (vec a×vec b)×(vec b×vec c)=[(vec a×vec b)·vec c]…

Prove that $(\vec a\times\vec b)\times(\vec b\times\vec c)=[(\vec a\times\vec b)\cdot\vec c]\,\vec b$.

Step 1: Proof

\[ (\vec a\times\vec b)\times(\vec b\times\vec c)=\vec b\,\big[(\vec a\times\vec b)\cdot\vec c\big]-\vec c\,\big[(\vec a\times\vec b)\cdot\vec b\big]. \]

The second bracket is $0$, since $\vec a\times\vec b\perp\vec b$.

\[(\vec a\times\vec b)\times(\vec b\times\vec c)=[\vec a,\vec b,\vec c]\,\vec b\qquad\blacksquare\]

This is also Problem 144 with $(\vec c,\vec d)\to(\vec b,\vec c)$.

Step 2: Meaning

Both normals $\vec a\times\vec b$ and $\vec b\times\vec c$ are perpendicular to $\vec b$. Their vector product therefore points along the common line $\vec b$ of the two planes. Check with $\vec a=\vec i$, $\vec b=\vec j$, $\vec c=\vec k$: $\vec k\times\vec i=\vec j=1\cdot\vec b$ $\checkmark$