Question

The acute angle between the line $\overline{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda (\hat{i}+\hat{j}+\hat{k})$ and the plane $\overline{r}.(2\hat{i}-\hat{j}+\hat{k})=5$

• A. ${\cos }^{-1}\left(\frac{\sqrt{2}}{3}\right)$
• B. ${\sin }^{-1}\left(\frac{\sqrt{2}}{3}\right)$
• C. ${\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)$
• D. None of these

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{\sqrt{2}}{3}\right)$

Now acute angle $\theta$ between a line $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ and plane $\overrightarrow{r}.\overrightarrow{n}=d$ is

$\sin \theta =\frac{\overrightarrow{b}.\overrightarrow{n}}{|\overrightarrow{b}.\overrightarrow{n}|}$

So here,

$\sin \theta =\frac{2-1+1}{\sqrt{3}\times \sqrt{6}}\sin \theta =\frac{2}{\sqrt{18}}\theta ={\sin }^{-1}\left(\frac{2}{\sqrt{18}}\right)$

$\therefore$ angle between the line a vector perpendicular to the plane is ${\sin }^{-1}\left(\frac{\sqrt{2}}{3}\right).$

$\therefore$ the angle between plane and line is $\frac{\pi }{2}-{\sin }^{-1}\left(\frac{\sqrt{2}}{3}\right)={\cos }^{-1}\left(\frac{\sqrt{2}}{3}\right)$