Question

If either a = 0 or b = 0, then a.b = 0. But the converse need not to be true. Justify your answer with an example.

Answer 1

If $a=0=0\hat{i}+0\hat{j}+0\hat{k}$ and b is non-zero i.e., $b=x\hat{i}+y\hat{j}+z\hat{k}$
$\therefore a.b=(0\hat{i}+0\hat{j}+0\hat{k})(x\hat{i}+y\hat{j}+z\hat{k})=(0\times x)+(0\times y)+(0\times z)=0$
So, if a = 0 or b = 0, then for same a.b = 0
To prove that converse need not be we have to prove that for two non-zero vectors a and b, a.b can be zero.
Let $a=2\hat{i}+4\hat{j}+3\hat{k}$ and $b=3\hat{i}+3\hat{j}-6\hat{k}$
Then, $a.b=2.3+4.3+3.(-6)=6+12-18=0$
We now observe that $|a|=\sqrt{2^2+4^2+3^2}=\sqrt{29}\therefore a\ne 0$
$|b|=\sqrt{3^2+3^2+(-6{)}^2}=\sqrt{54}$
$\therefore b\ne 0$
Hence, the converse of the given statement need not be true.