Equation of a plane at a distance √(2/21) from the origin, which contains the line of intersection of the planes x - y - z - 1 = 0 and 2x + y - 3z + 4 = 0, is :

Solution

Maths sir · 4 steps

Write every plane through the line as P₁ + λ P₂ = 0, then let the distance condition fix λ.

1. Family of planes. Any plane through the line of intersection is (x - y - z - 1) + λ(2x + y - 3z + 4) = 0. Collecting terms gives (1+2λ)x + (λ-1)y - (1+3λ)z + (4λ - 1) = 0.
2. Distance from origin. The distance is (|4λ-1|)/√((1+2λ)²+(λ-1)²+(1+3λ)²). The three squares add up to 14λ²+8λ+3. Set this distance equal to √(2/21) and square both sides.

   21(4λ-1)² = 2(14λ²+8λ+3) ⇒ 308λ² - 184λ + 15 = 0

3. Solve for lambda. The discriminant is 184² - 4 × 308 × 15 = 15376 = 124², so λ = (184 ± 124)/616. That gives λ = 1/2 or λ = 15/154.
4. So. With λ = 1/2 the plane is 2x - 1/2y - 5/2z + 1 = 0, that is 4x - y - 5z + 2 = 0. Check: its distance from the origin is 2/√42 = √(2/21). The answer is (d).

The idea to keep. Plane through the intersection of two planes? Write P₁ + λ P₂ = 0 straight away. The one extra condition (a distance, a point, parallel to a line) pins down λ.

Answer from the NTA final answer key. Solution by PrepWiser.