Question:
If either vector $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}$, then $\vec{a} \cdot \vec{b}=0$. But the converse need not be true. Justify your answer with an example.
Solution:
Consider $\vec{a}=2 \hat{i}+4 \hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}+3 \hat{j}-6 \hat{k}$.
Then,
$\vec{a} \cdot \vec{b}=2.3+4.3+3(-6)=6+12-18=0$
We now observe that:
$|\vec{a}|=\sqrt{2^{2}+4^{2}+3^{2}}=\sqrt{29}$
$\therefore \vec{a} \neq \overrightarrow{0}$
$|\vec{b}|=\sqrt{3^{2}+3^{2}+(-6)^{2}}=\sqrt{54}$
$\therefore \vec{b} \neq \overrightarrow{0}$
Hence, the converse of the given statement need not be true.
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