Question

Angle between the line $\mathbf{r} = (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) + \lambda ⥂ (\mathbf{i} - \mathbf{j} + \mathbf{k})$ and the normal to the plane $\mathbf{r}.(2\mathbf{i} - \mathbf{j} + \mathbf{k}) = 4$ is

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

Answer 1

Answer: (A)

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

Answer 2

The plane is $2x - y + z = 4$ and the line is

$\frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z + 1}{1}$

$\therefore\sin\theta = \frac{2 + 1 + 1}{\sqrt{6}\sqrt{3}} = \frac{4}{\sqrt{18}} = \frac{2\sqrt{2}}{3} \Rightarrow \theta = \sin^{- 1}\left( \frac{2\sqrt{2}}{3} \right)$.