Question

If the angle $\theta$ between the line $\frac{x+1}{1} = \frac{y-1}{2} = \frac{z-2}{2}$ and the plane $2x - y + \sqrt{\lambda}z + 4 = 0$ is such that $\sin \theta = \frac{1}{3}$, then $\lambda + 1 =$

• A. $\frac{5}{3}$
• B. $\frac{-5}{3}$
• C. $\frac{8}{3}$
• D. $\frac{-8}{3}$

Hint:

Line–plane angle uses sine with dot product of direction and normal vectors.

Answer 1

The Correct Option is C

Concept:
Angle between a line and a plane: \[ \sin \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| \, |\vec{n}|} \] Step 1: Direction vector of line. \[ \vec{d} = (1, 2, 2) \]
Step 2: Normal vector of plane. \[ \vec{n} = (2, -1, \sqrt{\lambda}) \]
Step 3: Apply formula. \[ \sin \theta = \frac{|2 - 2 + 2\sqrt{\lambda}|}{\sqrt{1^2+2^2+2^2} \cdot \sqrt{4+1+\lambda}} \] \[ = \frac{|2\sqrt{\lambda}|}{3\sqrt{5+\lambda}} \] Given: \[ \frac{2\sqrt{\lambda}}{3\sqrt{5+\lambda}} = \frac{1}{3} \]
Step 4: Solve. \[ 2\sqrt{\lambda} = \sqrt{5+\lambda} \] \[ 4\lambda = 5 + \lambda \Rightarrow 3\lambda = 5 \Rightarrow \lambda = \frac{5}{3} \]
Step 5: Final answer. \[ \lambda + 1 = \frac{5}{3} + 1 = \frac{8}{3} \]