If either vector

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.