3D Geometry
Angle between Line and Plane
Grade 12
Question:
<p>An angle between the plane, \(x + y + z = 5\) and the line of intersection of the planes, \(3x + 4y + z - 1 = 0\) and \(5x + 8y + 2z + 14 = 0\), is</p>
<p>\(\sin^{-1}\left(\dfrac{\sqrt{3}}{17}\right)\)</p>
<p>\(\cos^{-1}\left(\dfrac{\sqrt{3}}{17}\right)\)</p>
<p>\(\cos^{-1}\left(\dfrac{\sqrt{3}}{17}\right)\)</p>
<p>\(\sin^{-1}\left(\dfrac{\sqrt{3}}{17}\right)\)</p>
Step-by-Step Solution
Key Concept: The angle between a plane and a line equals 90° minus the angle between the line's direction vector and the plane's normal vector. The line of intersection direction comes from the cross product of the two given planes' normal vectors.
Step 1: Find the direction vector of the line of intersection. Normal to plane 1: n_1 = (3, 4, 1) Normal to plane 2: n_2 = (5, 8, 2) Direction vector d = n_1 × n_2 = | i j k | |3 4 1| |5 8 2| = i (8-8) - j (6-5) + k (24-20) d = (0, -1, 4) Step 2: Find the normal to the given plane. Normal to plane x + y + z = 5: n = (1, 1, 1) Step 3: Apply the line-plane angle formula. sin θ = | d · n | / (| d | | n |) d · n = 0(1) + (-1)(1) + 4(1) = 3 | d | = √(0 + 1 + 16) = √17 | n | = √(1 + 1 + 1) = √3 sin θ = |3| / (√17 · √3) = 3/√51 = 3/(3√17) = 1/√17 = √17/17 ∴ θ = sin⁻^1(√17/17) or approximately 14.04°
Correct Answer: A