MHTCET 18 Apr 2026 Shift 2 Paper

Concept:For a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$, the angle $\theta$ between them is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}$.Explanation:Step 1: Identify the direction vector of the line.The coefficient of $\lambda$ is $\vec{b} = \hat{i}+\hat{j}+\hat{k} = (1,1,1)$.Step 2: Identify the normal vector of the plane.From $\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=8$, the normal vector is $\vec{n} = 2\hat{i}-\hat{j}+\hat{k} = (2,-1,1)$.Step 3: Compute the dot product.$\vec{b} \cdot \vec{n} = (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2$.Step 4: Find 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{6}$.Step 5: Apply the formula for the angle.$\sin \theta = \frac{2}{\sqrt{3} \cdot \sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}$Therefore, $\theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)$.Answer:Option A: $\theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)$.