3D Geometry
Equation of a Plane
Grade 12
Question:
<p>The equation of the plane through the line of intersection of \(4x + 7y + 4z + 81 = 0\) and \(5x + 3y + 10z = 25\) and perpendicular to \(4x + 7y + 4z + 81 = 0\) is:</p>
<p>\(x - 4y + 6z = 106\)</p>
<p>\(x + 4y - 6z = 106\)</p>
<p>\(x - 4y - 6z = 106\)</p>
<p>\(x + 4y + 6z = 106\)</p>
Step-by-Step Solution
Key Concept: Any plane through the line of intersection of two planes can be written as P₁ + λP₂ = 0. The perpendicularity condition requires the normal vectors to satisfy n₁·n₂ = 0, which determines λ.
Step 1: Write the family of planes through the line of intersection: (4x + 7y + 4z + 81) + λ(5x + 3y + 10z - 25) = 0 (4 + 5λ)x + (7 + 3λ)y + (4 + 10λ)z + (81 - 25λ) = 0 Step 2: For perpendicularity to plane 4x + 7y + 4z + 81 = 0, the normal vectors must be orthogonal: n_1 = (4 + 5λ, 7 + 3λ, 4 + 10λ) and n_0 = (4, 7, 4) n_1·n_0 = 0: 4(4 + 5λ) + 7(7 + 3λ) + 4(4 + 10λ) = 0 16 + 20λ + 49 + 21λ + 16 + 40λ = 0 81 + 81λ = 0 ⟹ λ = -1 Step 3: Substitute λ = -1: (4 - 5)x + (7 - 3)y + (4 - 10)z + (81 + 25) = 0 -x + 4y - 6z + 106 = 0 or x - 4y + 6z - 106 = 0 ∴ Answer: A
Correct Answer: A