Line and Plane — Maths STD 12 Science — Question

$\cos \theta=\left|\frac{n_1 \cdot n_2}{\left|n_1 \|\right| n_2 \mid}\right|$

$\ldots(1)$

Here, $\bar{n}_1=\hat{i}+\hat{j}+2 \hat{k}, \bar{n}_2=2 \vec{i}-\hat{j}+\vec{k}$

$\begin{aligned} \therefore \bar{n}_1 \cdot \bar{n}_2 & =(\hat{i}+\hat{j}+2 \hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k}) \\ & =(1)(2)+(1)(-1)+(2)(1)\end{aligned}$

$=2-1+2=3$

Also, $\left|\bar{n}_1\right|=\sqrt{1^2+1^2+2^2}=\sqrt{6}$

$\left|\bar{n}_2\right|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}$

$\therefore$ from (1), we have

$\cos \theta=\left|\frac{3}{\sqrt{6} \cdot \sqrt{6}}\right|=\frac{3}{6}=\frac{1}{2} \cos 60^{\circ}$

$\therefore \theta=60^{\circ}$.