3D Geometry
Angle between a line and a plane
Grade 12
Question:
<p>If the angle between the line \(2(x+1) = y = z + 4\) and the plane \(2x - y + \sqrt{\lambda}z + 4 = 0\) is \(\pi/6\), then the value of \(\lambda\) is</p>
<p>\(\dfrac{135}{7}\)</p>
<p>\(\dfrac{45}{11}\)</p>
<p>\(\dfrac{45}{7}\)</p>
<p>\(\dfrac{135}{11}\)</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 of the line and n is the normal to the plane. Here sin(π/6) = 1/2 gives us the equation to solve for λ.
Step 1: Convert the line equation to standard form. From 2(x+1) = y = z+4, we get direction vector a = (1/2, 1, 1) or equivalently a = (1, 2, 2). Step 2: The normal vector to the plane 2x - y + √λz + 4 = 0 is n = (2, -1, √λ). Step 3: Use the angle formula: sin(π/6) = | a · n |/(| a |·| n |) Step 4: Calculate a · n = (1)(2) + (2)(-1) + (2)(√λ) = 2 - 2 + 2√λ = 2√λ Step 5: Calculate | a | = √(1 + 4 + 4) = √9 = 3 Step 6: Calculate | n | = √(4 + 1 + λ) = √(5 + λ) Step 7: Substitute into formula: 1/2 = |2√λ|/(3√(5 + λ)) Step 8: Cross multiply: 3√(5 + λ) = 4√λ Step 9: Square both sides: 9(5 + λ) = 16λ → 45 + 9λ = 16λ → 45 = 7λ → λ = 45/7 ∴ Answer: λ = 45/7
Correct Answer: A