Question

The angle between the straight line $r=(\hat{i}+2\hat{j}+\hat{k})+(\hat{i}-\hat{j}+\hat{k})$ and the plane $r\cdot (2\hat{i}-\hat{j}+\hat{k})=4$ is

• A. ${\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$
• B. ${\sin }^{-1}\left(\frac{\sqrt{2}}{6}\right)$
• C. ${\sin }^{-1}\left(\frac{\sqrt{2}}{3}\right)$
• D. ${\sin }^{-1}\left(\frac{2}{\sqrt{3}}\right)$
• E. ${\sin }^{-1}\left(\frac{2}{3}\right)$

Answer 1

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

Let $\theta$ be the angle between given line and the plane.

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