3D Geometry
Family of planes
Grade 12
Question:
<p>Given two planes \(P_1: 2x - y - 4 = 0\) and \(P_2: y + 2z - 4 = 0\) and point \(K(1, 1, 0)\). Let a third plane \(P_3\) pass through \(K\) and satisfy \(P_3: P_1 + \lambda P_2 = 0\). The equation of plane \(P_3\) is:</p>
<p>(A) \(x - y - z = 0\)</p>
<p>(B) \(2x - 2y - 2z = 0\)</p>
<p>(C) \(x - y - z = 1\)</p>
<p>(D) \(2x - y - z = 0\)</p>
Step-by-Step Solution
Key Concept: A plane passing through the intersection of two planes can be expressed as P₁ + λP₂ = 0. Substitute point K(1,1,0) into this family equation to find the specific value of λ, then determine P₃.
Step 1: Write the family of planes through the line of intersection of P_1 and P_2: P_3: (2x - y - 4) + λ(y + 2z - 4) = 0 Step 2: Expand and rearrange: 2x - y - 4 + λy + 2λz - 4λ = 0 2x + (-1 + λ)y + 2λz - (4 + 4λ) = 0 Step 3: Since P_3 passes through K(1, 1, 0), substitute x = 1, y = 1, z = 0: 2(1) + (-1 + λ)(1) + 2λ(0) - (4 + 4λ) = 0 2 - 1 + λ - 4 - 4λ = 0 -3 - 3λ = 0 λ = -1 Step 4: Substitute λ = -1 back into the family equation: 2x + (-1 - 1)y + 2(-1)z - (4 - 4) = 0 2x - 2y - 2z = 0 ∴ Answer: P_3: x - y - z = 0 (or equivalent form: 2x - 2y - 2z = 0)
Correct Answer: A