Line and Plane — Maths STD 12 Science — Question

$\sin \theta=\left|\frac{\bar{b} \cdot \bar{n}}{|\bar{b}||\bar{n}|}\right|$

$\ldots(1)$

Here, $\overline{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{n}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\therefore \bar{b} \cdot \bar{n}=(\hat{i}-\hat{j}+\hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k})$

= (2)(2) + (3)(-1) + (-6)(1) = 4 – 3 – 6 = -5

Also, $|\bar{b}|=\sqrt{1^2+1^2+(-1)^2}=\sqrt{2}=1$

$|\bar{n}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{4}$

$\therefore$ from (1), we have

$\sin \theta=\left|\frac{2 \sqrt{2}}{-3}\right|=\frac{2 \sqrt{2}}{3}$

$\therefore \theta=\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$