Vectors — Maths STD 12 Science — Question

Then $\bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 1 & 0\end{array}\right|$

$\begin{aligned} & =(0-2) \hat{i}-(0-2) \hat{j}+(0-1) \hat{k} \\ & =-2 \hat{i}+2 \hat{j}-\hat{k}\end{aligned}$

$\therefore|\vec{a} \times \bar{b}|=\sqrt{(-2)^2+2^2+(-1)^2}=\sqrt{4+4+1}=\sqrt{9}=3$

Unit vector perpendicular to both $\bar{a}$ and $\bar{b}$

$= \pm \frac{\bar{a} \times \bar{b}}{|\bar{a} \times \bar{b}|}= \pm\left(\frac{-2 \hat{i}+2 \hat{j}-\hat{k}}{3}\right)$

$= \pm\left(-\frac{2}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}\right)$