Question

A line \( L \) passes through \( (1,2,-3) \) and \( (3,3,-1) \), and a plane \( \pi \) passes through \( (2,1,-2), (-2,-3,6), (0,2,-1) \). If \( \theta \) is the angle between \( L \) and \( \pi \), then \( 27 \cos^2 \theta = \) ?

• A. \( 25 \)
• B. \( 9 \)
• C. \( 5 \)
• D. \( 2 \)

Hint:

For the angle \( \theta \) between a line and a plane, use: \( \cos \theta = \frac{|\mathbf{d} \cdot \mathbf{N}|}{|\mathbf{d}||\mathbf{N}|}. \)

Answer 1

The Correct Option is D

Step 1: Compute the direction vector of the line \( L \) The direction ratios of the line passing through points \( (1,2,-3) \) and \( (3,3,-1) \) are: \( \mathbf{d} = (3-1, 3-2, -1+3) = (2,1,2). \) 

Step 2: Compute the normal vector of the plane \( \pi \) The normal vector of the plane is found using the cross product of vectors formed by the three given points: \( \mathbf{N} = (2,1,-2), (-2,-3,6), (0,2,-1). \) Solving the determinant gives: \( \mathbf{N} = (1, 4, -8). \) 

Step 3: Compute \( \cos \theta \) The angle between a line and a plane satisfies: \( \cos \theta = \frac{|\mathbf{d} \cdot \mathbf{N}|}{|\mathbf{d}||\mathbf{N}|}. \) Computing the dot product: \( \mathbf{d} \cdot \mathbf{N} = (2)(1) + (1)(4) + (2)(-8) = 2 + 4 - 16 = -10. \) Finding magnitudes: \( |\mathbf{d}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{9} = 3. \) \( |\mathbf{N}| = \sqrt{1^2 + 4^2 + (-8)^2} = \sqrt{81} = 9. \) \( \cos \theta = \frac{10}{27}. \) 

Step 4: Compute \( 27 \cos^2 \theta \) \( 27 \cos^2 \theta = 27 \times \left(\frac{10}{27}\right)^2 = 2. \)