3D Geometry
Family of planes and distance
Grade 12
Question:
<p>Let \(P\) be the plane, which contains the line of intersection of the planes, \(x + y + z - 6 = 0\) and \(2x + 3y + z + 5 = 0\) and it is perpendicular to the \(xy\)-plane. Then the distance of the point \((0, 0, 256)\) from \(P\) is ______ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: A plane containing the line of intersection of two planes can be written as a linear combination: P₁ + λP₂ = 0. The perpendicularity condition to the xy-plane means the coefficient of z must be zero (plane is vertical).
Step 1: Write the family of planes containing the line of intersection. Any plane containing the intersection of x + y + z - 6 = 0 and 2x + 3y + z + 5 = 0 is: (x + y + z - 6) + λ(2x + 3y + z + 5) = 0 (1 + 2λ)x + (1 + 3λ)y + (1 + λ)z + (-6 + 5λ) = 0 Step 2: Apply the perpendicularity condition to xy-plane. A plane perpendicular to xy-plane has no z-component in its intersection with xy-plane, meaning the normal vector is perpendicular to k̂. This requires the plane to be vertical, so the coefficient of z must be zero (for the plane to not depend on z alone). Actually, perpendicular to xy-plane means the normal is parallel to xy-plane, so: coefficient of z = 0 1 + λ = 0 ⟹ λ = -1 Step 3: Find the plane equation. Substituting λ = -1: (1 - 2)x + (1 - 3)y + (-6 - 5) = 0 -x - 2y - 11 = 0 x + 2y + 11 = 0 Step 4: Calculate distance from (0, 0, 256) to plane x + 2y + 11 = 0. Distance = |1(0) + 2(0) + 11|/√(1^2 + 2^2) = 11/√5 = 11√5/5 = 11 × 2.236.../5 = 24.596.../5 ≈ 4.919 ∴ Answer: 4.919
Correct Answer: 4