Q77 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
The angle between the line $\bar{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$ and the plane $\bar{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 8$ is $\ldots$
- Correct answerA$\theta = \sin^{-1}\left(\dfrac{\sqrt{2}}{3}\right)$
- B$\theta = \cos^{-1}\left(\dfrac{\sqrt{2}}{3}\right)$
- C$\theta = \cos^{-1}\left(\dfrac{2}{\sqrt{3}}\right)$
- D$\theta = \sin^{-1}\left(\dfrac{3}{\sqrt{2}}\right)$
Solution
The direction vector of the given line is $\vec{b} = \hat{i} + \hat{j} + \hat{k}$.
The normal vector to the given plane is $\vec{n} = 2\hat{i} - \hat{j} + \hat{k}$.
The angle $\theta$ between a line and a plane is given by the formula:
$\sin \theta = \dfrac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$
Calculating the dot product:
$\vec{b} \cdot \vec{n} = (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2$
Calculating the magnitudes of the vectors:
$|\vec{b}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$
$|\vec{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$
Substituting these values into the formula:
$\sin \theta = \dfrac{2}{\sqrt{3} \sqrt{6}} = \dfrac{2}{\sqrt{18}} = \dfrac{2}{3\sqrt{2}} = \dfrac{\sqrt{2}}{3}$
$\theta = \sin^{-1}\left(\dfrac{\sqrt{2}}{3}\right)$
Answer: $\theta = \sin^{-1}\left(\dfrac{\sqrt{2}}{3}\right)$
The normal vector to the given plane is $\vec{n} = 2\hat{i} - \hat{j} + \hat{k}$.
The angle $\theta$ between a line and a plane is given by the formula:
$\sin \theta = \dfrac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$
Calculating the dot product:
$\vec{b} \cdot \vec{n} = (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2$
Calculating the magnitudes of the vectors:
$|\vec{b}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$
$|\vec{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$
Substituting these values into the formula:
$\sin \theta = \dfrac{2}{\sqrt{3} \sqrt{6}} = \dfrac{2}{\sqrt{18}} = \dfrac{2}{3\sqrt{2}} = \dfrac{\sqrt{2}}{3}$
$\theta = \sin^{-1}\left(\dfrac{\sqrt{2}}{3}\right)$
Answer: $\theta = \sin^{-1}\left(\dfrac{\sqrt{2}}{3}\right)$
Hence, the correct answer is option: 'A'.