If either vector $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0},$ then $\vec{a} \cdot \vec{b}=0 .$ But the converse need not be true. Justify your answer with an example.

(N/A) 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,the dot product is:
$\vec{a} \cdot \vec{b}=(2)(3)+(4)(3)+(3)(-6)=6+12-18=0.$
We now observe the magnitudes:
$|\vec{a}|=\sqrt{2^{2}+4^{2}+3^{2}}=\sqrt{4+16+9}=\sqrt{29} \neq 0.$
$|\vec{b}|=\sqrt{3^{2}+3^{2}+(-6)^{2}}=\sqrt{9+9+36}=\sqrt{54} \neq 0.$
Since $\vec{a} \cdot \vec{b}=0$ but both $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0},$ the converse of the statement is not necessarily true.