Question:medium

If the angle between the line $x = \frac{y-1}{2} = \frac{z-3}{\lambda}$ and the plane $x + 2y + 3z = 4$ is $\cos^{-1} \sqrt{\frac{5}{14}}$, then the value of $\lambda$ is

Show Hint

Angle between line and plane uses $\sin \theta$, not $\cos \theta$.
Updated On: May 7, 2026
  • $\frac{1}{3}$
  • $\frac{4}{5}$
  • $\frac{2}{3}$
  • $\frac{2}{5}$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The angle $\theta$ between a line and a plane is the complement of the angle between the line's direction and the plane's normal.
Step 2: Key Formula or Approach:
If the angle between line and plane is $\theta$, then $\sin \theta = \frac{|\bar{\text{b}} \cdot \bar{\text{n}}|}{|\bar{\text{b}}||\bar{\text{n}}|}$.
Given $\theta = \cos^{-1} \sqrt{\frac{5}{14}}$, so $\cos^2 \theta = \frac{5}{14}$ and $\sin^2 \theta = 1 - \frac{5}{14} = \frac{9}{14}$.
Step 3: Detailed Explanation:
Line direction $\bar{\text{b}} = (1, 2, \lambda)$ and plane normal $\bar{\text{n}} = (1, 2, 3)$.
\[ \sin \theta = \frac{|1(1) + 2(2) + \lambda(3)|}{\sqrt{1^2+2^2+\lambda^2}\sqrt{1^2+2^2+3^2}} = \frac{|5 + 3\lambda|}{\sqrt{5+\lambda^2}\sqrt{14}} \] Squaring both sides:
\[ \frac{9}{14} = \frac{(5 + 3\lambda)^2}{(5 + \lambda^2)(14)} \implies 9(5 + \lambda^2) = 25 + 30\lambda + 9\lambda^2 \] \[ 45 + 9\lambda^2 = 25 + 30\lambda + 9\lambda^2 \implies 20 = 30\lambda \implies \lambda = \frac{2}{3} \] Step 4: Final Answer:
The value of $\lambda$ is $\frac{2}{3}$.
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