Question

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

• A. $\frac{1}{3}$
• B. $\frac{4}{5}$
• C. $\frac{2}{3}$
• D. $\frac{2}{5}$

Hint:

Angle between line and plane uses $\sin \theta$, not $\cos \theta$.

Answer 1

The Correct Option is C

Step 1: Formula
$\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Given $\theta = \sin^{-1} \sqrt{1 - (\sqrt{5/14})^2} = \sin^{-1} \sqrt{9/14} = \sin^{-1} \frac{3}{\sqrt{14}}$.
Step 2: Vectors
$\vec{b} = (1, 2, \lambda)$, $\vec{n} = (1, 2, 3)$.
$\vec{b} \cdot \vec{n} = 1 + 4 + 3\lambda = 5 + 3\lambda$.
$|\vec{b}| = \sqrt{5+\lambda^2}$, $|\vec{n}| = \sqrt{14}$.
Step 3: Calculation
$\frac{3}{\sqrt{14}} = \frac{5+3\lambda}{\sqrt{5+\lambda^2}\sqrt{14}} \implies 3\sqrt{5+\lambda^2} = 5+3\lambda$.
Square both sides: $9(5+\lambda^2) = 25 + 9\lambda^2 + 30\lambda \implies 45 + 9\lambda^2 = 25 + 9\lambda^2 + 30\lambda$.
$20 = 30\lambda \implies \lambda = \frac{2}{3}$.
Final Answer: (C)