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 the value of \( \lambda \) is:

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

Hint:

Use the formula for the angle between a line and a plane, which involves the dot product between the direction vector of the line and the normal vector of the plane.

Answer 1

The Correct Option is C

We are given the line equation in symmetric form: \[ \frac{x + 1}{1} = \frac{y
- 1}{2} = \frac{z
- 2}{2} \] and the plane equation: \[ 2x
- y + \sqrt{\lambda}z + 4 = 0 \] We are also given that the sine of the angle \( \theta \) between the line and the plane is \( \sin \theta = \frac{1}{3} \). Our goal is to find the value of \( \lambda \). Step 1: Identify the direction ratios of the line and the normal vector to the plane The given line equation can be written in parametric form as: \[ x =
-1 + t, \quad y = 1 + 2t, \quad z = 2 + 2t \] Thus, the direction ratios of the line are \( \mathbf{d_1} = (1, 2, 2) \). The plane equation is given by: \[ 2x
- y + \sqrt{\lambda}z + 4 = 0 \] The normal vector to the plane is \( \mathbf{n} = (2,
-1, \sqrt{\lambda}) \).
Step 2: Use the formula for the angle between the line and the plane The formula for the sine of the angle \( \theta \) between the line and the plane is given by: \[ \sin \theta = \frac{|\mathbf{d_1} \cdot \mathbf{n}|}{|\mathbf{d_1}| |\mathbf{n}|} \] Where \( \mathbf{d_1} \) is the direction vector of the line and \( \mathbf{n} \) is the normal vector to the plane. We are given that \( \sin \theta = \frac{1}{3} \).
Step 3: Calculate the dot product \( \mathbf{d_1} \cdot \mathbf{n} \) The dot product of \( \mathbf{d_1} = (1, 2, 2) \) and \( \mathbf{n} = (2,
-1, \sqrt{\lambda}) \) is: \[ \mathbf{d_1} \cdot \mathbf{n} = 1 \times 2 + 2 \times (
-1) + 2 \times \sqrt{\lambda} = 2
- 2 + 2\sqrt{\lambda} = 2\sqrt{\lambda} \]
Step 4: Calculate the magnitudes of \( \mathbf{d_1} \) and \( \mathbf{n} \) The magnitude of \( \mathbf{d_1} = (1, 2, 2) \) is: \[ |\mathbf{d_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 \] The magnitude of \( \mathbf{n} = (2,
-1, \sqrt{\lambda}) \) is: \[ |\mathbf{n}| = \sqrt{2^2 + (
-1)^2 + (\sqrt{\lambda})^2} = \sqrt{4 + 1 + \lambda} = \sqrt{5 + \lambda} \]

Step 6: Set up the equation using \( \sin \theta = \frac{1}{3} \) Using the formula for \( \sin \theta \), we have: \[ \frac{|\mathbf{d_1} \cdot \mathbf{n}|}{|\mathbf{d_1}| |\mathbf{n}|} = \frac{1}{3} \] Substituting the values: \[ \frac{|2\sqrt{\lambda}|}{3 \times \sqrt{5 + \lambda}} = \frac{1}{3} \] Simplifying: \[ \frac{2\sqrt{\lambda}}{3 \sqrt{5 + \lambda}} = \frac{1}{3} \] Multiplying both sides by 3: \[ \frac{2\sqrt{\lambda}}{\sqrt{5 + \lambda}} = 1 \] Squaring both sides: \[ \frac{4\lambda}{5 + \lambda} = 1 \] \[ 4\lambda = 5 + \lambda \] \[ 4\lambda
- \lambda = 5 \] \[ 3\lambda = 5 \] \[ \lambda = \frac{5}{3} \] Thus, the value of \( \lambda \) is \( \frac{5}{3} \).