Question

The angle between the line \( x = \frac{y-1}{2} = \frac{z-3}{\lambda} \) and the plane \( x + 2y + 3z = 6 \) is \( \cos^{-1} \sqrt{\frac{5}{14}} \), then the value of \( \lambda \) is

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

Hint:

Angle between line and plane uses $\sin \theta = \frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}$, unlike line-line or plane-plane which use $\cos \theta$.

Answer 1

The Correct Option is A

Step 1: Identify Vectors
Line DRs: $\vec{b} = (1, 2, \lambda)$. Plane Normal DRs: $\vec{n} = (1, 2, 3)$.
Step 2: Formula for Angle
The angle $\alpha$ between a line and a plane is given by $\sin \alpha = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}$.
Given $\alpha = \cos^{-1} \sqrt{5/14} \implies \cos^2 \alpha = 5/14$.
$\sin^2 \alpha = 1 - 5/14 = 9/14 \implies \sin \alpha = 3/\sqrt{14}$.
Step 3: Calculation
$\frac{1(1) + 2(2) + 3\lambda}{\sqrt{1+4+\lambda^2}\sqrt{1+4+9}} = \frac{3}{\sqrt{14}}$.
$\frac{5+3\lambda}{\sqrt{5+\lambda^2}\sqrt{14}} = \frac{3}{\sqrt{14}} \implies 5+3\lambda = 3\sqrt{5+\lambda^2}$.
Step 4: Solve for \(\lambda\)
Squaring: $25 + 9\lambda^2 + 30\lambda = 9(5 + \lambda^2) \implies 25 + 30\lambda = 45$.
$30\lambda = 20 \implies \lambda = 2/3$.
Final Answer:(A)