If either
Question:

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

Solution:

Take any parallel non-zero vectors so that $\vec{a} \times \vec{b}=\overrightarrow{0}$.

Let $\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{b}=4 \hat{i}+6 \hat{j}+8 \hat{k}$

Then,

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & 6 & 8\end{array}\right|=\hat{i}(24-24)-\hat{j}(16-16)+\hat{k}(12-12)=0 \hat{i}+0 \hat{j}+0 \hat{k}=\overrightarrow{0}$

It can now be observed that:

$|\vec{a}|=\sqrt{2^{2}+3^{2}+4^{2}}=\sqrt{29}$

$\therefore \vec{a} \neq \overrightarrow{0}$

$|\vec{b}|=\sqrt{4^{2}+6^{2}+8^{2}}=\sqrt{116}$

$\therefore \vec{b} \neq \overrightarrow{0}$

Hence, the converse of the given statement need not be true.

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