3D Geometry
Angle between line and plane
Grade 12
Question:
<p>Given planes \(3x + 4y + z = 1\) and \(5x + 8y + 2z + 14 = 0\). The sine of the angle between the plane \(x + y + z = 5\) and the line of intersection of the two given planes is:</p>
<p>\(\sqrt{\dfrac{3}{17}}\)</p>
<p>\(\sqrt{\dfrac{2}{17}}\)</p>
<p>\(\sqrt{\dfrac{1}{17}}\)</p>
<p>\(\sqrt{\dfrac{5}{17}}\)</p>
Step-by-Step Solution
Key Concept: The line of intersection of two planes is perpendicular to both normal vectors; find its direction vector via cross product, then use the formula sin(θ) = |cos(α)| where α is the angle between the line's direction and the third plane's normal.
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| d = i (8-8) - j (6-5) + k (24-20) = (0, -1, 4) Step 2: Find the angle between the line and plane x + y + z = 5. Normal to plane 3: n_3 = (1, 1, 1) sin(θ) = | d · n_3 | / (| d | · | n_3 |) Step 3: Calculate. d · n_3 = 0(1) + (-1)(1) + 4(1) = -1 + 4 = 3 | d | = √(0^2 + 1^2 + 4^2) = √17 | n_3 | = √(1^2 + 1^2 + 1^2) = √3 sin(θ) = |3| / (√17 · √3) = 3 / √51 = 3√51 / 51 = √51 / 17 ∴ Answer: A
Correct Answer: A