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: Understanding the Question:
We are given the angle between a line and a plane. We need to use the formula for this angle to find the unknown parameter \( \lambda \).
Step 2: Key Formula or Approach:
The angle \( \theta \) between a line with direction vector \( \vec{d} \) and a plane with normal vector \( \vec{n} \) is given by:
\[ \sin \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|} \] Step 3: Detailed Explanation:
Line: \( \frac{x}{1} = \frac{y-1}{2} = \frac{z-3}{\lambda} \implies \vec{d} = (1, 2, \lambda) \).
Plane: \( x + 2y + 3z = 6 \implies \vec{n} = (1, 2, 3) \).
Given \( \theta = \cos^{-1} \sqrt{\frac{5}{14}} \implies \cos \theta = \sqrt{\frac{5}{14}} \).
Then \( \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{5}{14}} = \sqrt{\frac{9}{14}} = \frac{3}{\sqrt{14}} \).
Using the formula:
\[ \frac{|1(1) + 2(2) + \lambda(3)|}{\sqrt{1^2 + 2^2 + \lambda^2} \sqrt{1^2 + 2^2 + 3^2}} = \frac{3}{\sqrt{14}} \] \[ \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}} = \frac{3}{\sqrt{14}} \] \[ |5 + 3\lambda| = 3\sqrt{5 + \lambda^2} \] Squaring both sides:
\[ 25 + 9\lambda^2 + 30\lambda = 9(5 + \lambda^2) \] \[ 25 + 9\lambda^2 + 30\lambda = 45 + 9\lambda^2 \] \[ 30\lambda = 20 \implies \lambda = \frac{2}{3} \] Step 4: Final Answer:
The value of \( \lambda \) is \( 2/3 \).