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 $yo1\pi$ answer with an example.

Let $a = \hat i - 2\hat j + \hat k$ and $\vec b = \hat i + 3\hat j + 5\hat k$

Here, $|\vec a| = \sqrt {{1^2} + {{( - 2)}^2} + {1^2}} = \sqrt {1 + 4 + 1} = \sqrt 6$
and $|\vec b| = \sqrt {{1^2} + {3^2} + {5^2}} = \sqrt {1 + 9 + 25} = \sqrt {35}$

Clearly, $|\vec a| \ne 0,\,\,|\vec b| \ne 0$ but,
$\vec a \cdot \vec b = (\hat i - 2\hat j + \hat k) \cdot (\hat i + 3\hat j + 5\hat k) = (1)(1) + ( - 2)(3) + (1)(5)$

$= 1 - 6 + 5 = 0$
Hence, $\vec a \cdot \vec b = 0$ though $\vec a \ne \vec 0,\vec b \ne 0$.