Prove that vec a×(vec b+λvec a)=vec a×vec b for every scalar λ

Prove that $\vec a\times(\vec b+\lambda\vec a)=\vec a\times\vec b$ for every scalar $\lambda$.

Explanation of Method

Proving that a×(b+λa) equals a×b for every λ.

Step 1: Proof

\[ \vec a\times(\vec b+\lambda\vec a)=\vec a\times\vec b+\lambda\,(\vec a\times\vec a)=\vec a\times\vec b+\lambda\,\vec 0=\vec a\times\vec b.\qquad\blacksquare \]

\[\vec a\times(\vec b+\lambda\vec a)=\vec a\times\vec b\quad\forall\lambda\]

Step 2: Geometric meaning

Adding $\lambda\vec a$ moves the tip of $\vec b$ parallel to $\vec a$ (a shear). The parallelogram keeps its base $|\vec a|$ and its height, so the area is unchanged. It also stays in the same plane with the same orientation, so the vector $\vec a\times\vec b$ itself is unchanged, not only its length.

$O$ $\vec a$ $\vec b$ $\vec b+0.8\vec a$ $\vec b-\vec a$

Adding $\lambda\vec a$ slides the tip of $\vec b$ parallel to $\vec a$: same base, same height, so the same area and the same $\vec a\times\vec b$.

Result

\[\boxed{\,\vec a\times(\vec b+\lambda\vec a)=\vec a\times\vec b\quad\forall\lambda\,}\]