3D Geometry
Line and Plane Intersection
Grade 12
Question:
<p>If the coordinates of the point where the line \(x - 2y + z - 1 = 0 = x + 2y - 2z - 5\) intersects the plane \(x + y - 2z = 7\) is \((\alpha, \beta, \gamma)\), then:</p>
<p>(a) \(\alpha + \beta + \gamma = 7\)</p>
<p>(b) \(\alpha - \beta + \gamma + 1 = 0\)</p>
<p>(c) \(\alpha^2 + \beta^2 + \gamma^2 = 21\)</p>
<p>(d) \(\alpha\beta + \beta\gamma + \gamma\alpha = 2\)</p>
Step-by-Step Solution
Key Concept: Find the line of intersection of two planes by solving them simultaneously, then substitute the parametric form into the third plane to find the specific intersection point.
Step 1: Find the line formed by intersection of the two planes: Given: x - 2y + z - 1 = 0 ... (1) and x + 2y - 2z - 5 = 0 ... (2) Subtract (1) from (2): 4y - 3z - 4 = 0 → z = (4y - 4)/3 Add (1) and (2): 2x - z - 6 = 0 → z = 2x - 6 Equating: (4y - 4)/3 = 2x - 6 → 4y = 6x - 14 → y = (3x - 7)/2 Step 2: Write parametric form. Let x = t: x = t, y = (3t - 7)/2, z = 2t - 6 Step 3: Substitute into the plane equation x + y - 2z = 7: t + (3t - 7)/2 - 2(2t - 6) = 7 t + (3t - 7)/2 - 4t + 12 = 7 2t + 3t - 7 - 8t + 24 = 14 -3t + 17 = 14 t = 1 Step 4: Find the coordinates: α = 1, β = (3(1) - 7)/2 = -2, γ = 2(1) - 6 = -4 ∴ The point is (1, -2, -4) Verification: 1 + (-2) - 2(-4) = 1 - 2 + 8 = 7 ✓
Correct Answer: A,B,C