Question

If either $\vec{a}=0$ or $\vec{b}=0$,then $\vec{a}\times\vec{b}=0$. Is the converse true? Justify your answer with an example.

Answer 1

Take any parallel non-zero vectors so that $\vec{a}\times\vec{b}=0$
Let $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k},\vec{b}=4\hat{i}+6\hat{j}+8\hat{k}$.
Then,
\vec{a}\times\vec{b}=$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\\ 2 & 3 & 4 \\\4&6&8\end{vmatrix}$$=\hat{i}(24-24)-\hat{j}(16-16)+\hat{k}(12-12)=0\hat{i}+0\hat{j}+0\hat{k}=\vec{0}
It can be observed that:
$|\vec{a}|=\sqrt{22+32+42}=\sqrt{29}$
$∴\vec{a}\ne\vec{0}$
$|\vec{b}|=\sqrt{42+62+82}=\sqrt{116}$
$∴\vec{b}\ne\vec{0}$
Hence,the converse of the given statement need not to be true.