<p>If the angle between the line \(x - \dfrac{y-1}{-2} - \dfrac{z-3}{\lambda}\) and the plane \(x + 2y + 3z = 4\) is \(\cos^{-1}\!\left(\sqrt{5/14}\right)\), then \(\lambda\) equals</p>
Step-by-Step Solution
Key Concept: The angle θ between a line and plane satisfies sin(θ) = |a·n|/(|a||n|), where a is the direction vector and n is the normal. Here, sin²(θ) = 1 - cos²(θ), so we must first find sin from the given cosine value.
Step 1: Extract direction vector and normal vector. Line: direction vector a = (1, -2, λ) Plane: normal vector n = (1, 2, 3) Step 2: Find sin(θ) from given angle. Given: angle = cos⁻^1(√(5/14)) So cos(θ) = √(5/14), thus sin^2(θ) = 1 - 5/14 = 9/14 Therefore sin(θ) = 3/√14 Step 3: Apply line-plane angle formula. sin(θ) = | a · n |/(| a || n |) 3/√14 = |1(1) + (-2)(2) + λ(3)|/(√(1+4+λ^2)·√(1+4+9)) 3/√14 = |1 - 4 + 3λ|/(√(5+λ^2)·√14) 3/√14 = |3λ - 3|/(√(5+λ^2)·√14) Step 4: Simplify and solve for λ. 3 = |3λ - 3|/√(5+λ^2) 3√(5+λ^2) = |3λ - 3| 9(5+λ^2) = 9(λ-1)^2 45 + 9λ^2 = 9λ^2 - 18λ + 9 45 = -18λ + 9 18λ = -36 λ = -2 ∴ Answer: D
Correct Answer: D