VECTOR ALGEBRA — Maths STD 12 Science — Question

Then, $\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=0$ ($\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$)

Now, let us assume that $\vec{\text{a}}.\vec{\text{b}}=0$

$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=0$

But here we cannot say that either $|\vec{\text{a}}|=0$ or $\big|\vec{\text{b}}\big|=0$. (Because even $\cos\theta$ can be zero)

For example, let

$\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{b}}=-3\hat{\text{i}}+2\hat{\text{k}}$

Here, $|\vec{\text{a}}|=\sqrt{4+1+9}=\sqrt{14}\neq0$

$\big|\vec{\text{b}}\big|=\sqrt{9+4}=\sqrt{13}\neq0$

But $\vec{\text{a}}.\vec{\text{b}}=\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big).\big(-3\hat{\text{i}}+2\hat{\text{k}}\big)=-6+0+6=0$