Prove that if A x B = O and A . B = 0 then A or B is O.

Prove that if $A \times B = O$ and $A \cdot B = 0$ , then at least one of $A,B$ is the zero vector. Give a geometric interpretation of this result.
Prove that if $A$ is a nonzero vector, and if $A \times B = A \times C$ and $A \cdot B = A \cdot C$ then $B = C$ .

Proof. If we have $A \times B = O$ then
$\begin{align*} A \times B = O && \implies && \lVert A \times B \rVert &= 0 \\ && \implies && \lVert A \rVert \lVert B \rVert \sin \theta &= 0. \end{align*}$

But, since $A \cdot B = 0$ we know $A$ and $B$ are orthogonal, so $\sin \theta = 1$ . Therefore, we must have either $\lVert A \rVert = 0$ or $\lVert B \rVert = 0$ . Hence, either $A$ or $B$ is the zero vector $. \qquad \blacksquare$
Proof. First,
$A \times B = A \times C \quad \implies \quad A \times B - A \times C = O \quad \implies \quad A \times (B -C) = O.$

Then,

$A \cdot B = A \cdot C \quad \implies \quad A \cdot B - A \cdot C = 0 \quad \implies \quad A \cdot (B-C) = 0.$

But, $A \times (B-C) = O$ and $A \cdot (B-C) = 0$ implies $A = O$ or $B-C = O$ (by part (a)). Since $A \neq O$ by hypothesis, we must have $B-C = O$ . Hence, $B = C. \qquad \blacksquare$