Question

The angle between plane

$x-y+3z=5$ and the line

$\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})$ is

• A. ${\sin }^{-1}\left(\frac{5}{\sqrt{33}}\right)$
• B. ${\cos }^{-1}\left(\frac{5}{\sqrt{33}}\right)$
• C. ${\sin }^{-1}\left(\frac{1}{\sqrt{33}}\right)$
• D. ${\cos }^{-1}\left(\frac{3}{\sqrt{33}}\right)$

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{5}{\sqrt{33}}\right)$

Given plane equation is $x-y+3z=5$
D.R's of Normal to the plane $\overrightarrow{n}=(n_1,n_2,n_3)=(1,-1,3)$

Line equation is: $r=(\overline{i}+\hat{j}-\overline{k})+\lambda (\hat{i}-\hat{j}+\hat{k})$

$r=\overrightarrow{a}+\lambda \overrightarrow{b}$

$\overrightarrow{a}=(a_1,a_7,a_3)=(1,1,-1)$

$\overrightarrow{b}=(b_1,b_2,b_3)=(1,-1,1)$
Angle between the line and plane $\theta$ is given by

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

$\theta ={\sin }^{-1}\left|\frac{n_1{b}_1+n_2{b}_2+n_3{b}_3}{\sqrt{n_1^2+n_2^2+n_3{}^2}\sqrt{b_1{}^2+b_2{}^2+b_3^2}}\right|$

$={\sin }^{-1}\left|\frac{1+1+3}{\sqrt{1+1+9}\sqrt{1+1+1}}\right|$

$={\sin }^{-1}\frac{5}{\sqrt{11}\sqrt{3}}$

$={\sin }^{-1}\left(\frac{5}{\sqrt{33}}\right)$