<p>Let <em>L</em> be the line of intersection of the planes \(2x + 3y + z = 1\) and \(x + 3y + 2z = 2\). If <em>L</em> makes an angle \(\alpha\) with the positive <em>x</em>-axis, then \(\cos\alpha\) equals</p>
Step-by-Step Solution
Key Concept: The direction vector of line L is perpendicular to both plane normals, found via their cross product. The angle α with the x-axis is determined by the direction cosine along the x-component of this direction vector.
Step 1: Find the direction vector of line L using cross product of normal vectors. Normal to plane 1: n_1 = (2, 3, 1) Normal to plane 2: n_2 = (1, 3, 2) Step 2: Calculate n_1 × n_2 : n_1 × n_2 = | i j k | |2 3 1| |1 3 2| = i (6-3) - j (4-1) + k (6-3) = 3 i - 3 j + 3 k = 3(1, -1, 1) Direction vector: d = (1, -1, 1) Step 3: The angle α with positive x-axis has direction cosine: cos α = (x-component)/(magnitude of direction vector) cos α = 1/√(1^2 + (-1)^2 + 1^2) = 1/√3 Step 4: Rationalize: cos α = 1/√3 = √3/3 ∴ Answer: A
Correct Answer: A