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.
(B) The converse of the statement is: If $\vec{a} \times \vec{b}=\overrightarrow{0},$ then either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}.$
This converse is not necessarily true.
Consider two non-zero parallel vectors $\vec{a}$ and $\vec{b}.$
Let $\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{b}=4 \hat{i}+6 \hat{j}+8 \hat{k}.$
Calculating the cross product:
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & 6 & 8\end{array}\right| = \hat{i}(24-24) - \hat{j}(16-16) + \hat{k}(12-12) = \overrightarrow{0}.$
Here,$|\vec{a}| = \sqrt{2^2+3^2+4^2} = \sqrt{29} \neq 0$ and $|\vec{b}| = \sqrt{4^2+6^2+8^2} = \sqrt{116} \neq 0.$
Since $\vec{a} \times \vec{b} = \overrightarrow{0}$ even though $\vec{a} \neq \overrightarrow{0}$ and $\vec{b} \neq \overrightarrow{0},$ the converse is not true.
This converse is not necessarily true.
Consider two non-zero parallel vectors $\vec{a}$ and $\vec{b}.$
Let $\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{b}=4 \hat{i}+6 \hat{j}+8 \hat{k}.$
Calculating the cross product:
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & 6 & 8\end{array}\right| = \hat{i}(24-24) - \hat{j}(16-16) + \hat{k}(12-12) = \overrightarrow{0}.$
Here,$|\vec{a}| = \sqrt{2^2+3^2+4^2} = \sqrt{29} \neq 0$ and $|\vec{b}| = \sqrt{4^2+6^2+8^2} = \sqrt{116} \neq 0.$
Since $\vec{a} \times \vec{b} = \overrightarrow{0}$ even though $\vec{a} \neq \overrightarrow{0}$ and $\vec{b} \neq \overrightarrow{0},$ the converse is not true.
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