Question

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

Answer 1

It is given that when ${\displaystyle \overrightarrow{a}=0}$ or ${\displaystyle \overrightarrow{b}=0}$, then ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=0}$. But if ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=0}$, then it is not true that ${\displaystyle \overrightarrow{a},\overrightarrow{b}}$ is zero.

Let ${\displaystyle \overrightarrow{a}=\hat{i}-2\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+3\hat{j}+5\hat{k}}$

${\displaystyle \left|\overrightarrow{a}\right|=\sqrt{1^2+{(-2)}^2+1^2}}$

${\displaystyle =\sqrt{1+4+1}}$

${\displaystyle =\sqrt{6}}$

and ${\displaystyle \left|\overrightarrow{b}\right|=\sqrt{1^2+3^2+5^2}}$

${\displaystyle =\sqrt{1+9+25}}$

${\displaystyle =\sqrt{35}}$

${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=(\hat{i}-2\hat{j}+\hat{k})(1+3\hat{j}+5\hat{k})}$

= 1.1 - 2.3 + 1.5

= 0

Hence, ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=0}$ thought ${\displaystyle \overrightarrow{a}\ne 0,\overrightarrow{b}\ne 0.}$