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Let P 1 : x + y + 2 z - 4 = 0 and P 2 : 2 x - y + 3 z + 5 = 0 be two planes. Let A 1 , 3 , 4 and B 3 , 2 , 7 be two points in space. The equation of a third plane P 3 through the line of intersection of P 1 and P 2 and parallel to A B is

Options

  1. Ax - 4 y - 2 z + 3 = 0
  2. Bx - 4 y - 2 z + 9 = 0
  3. C2x - 3 y + 4 z + 9 = 0
  4. D3 y + z - 13 = 0

Correct answer

D. 3 y + z - 13 = 0

Step-by-step solution

Equation of P 3 is P 1 + λ P 2 = 0 x 1 + 2 λ + y 1 - λ + z 2 + 3 λ - 4 + 5 λ = 0 A B → = < 2 , - 1,3 > a normal vector to plane P 3 is < 1 + 2 λ , 1 - λ , 2 + 3 λ > If A B is parallel to P 3 = 0 ⇒ 2 1 + 2 λ - 1 1 - λ + 3 2 + 3 λ = 0 ⇒ 2 + 4 λ - 1 + λ + 6 + 9 λ = 0 ⇒ 14 λ + 7 = 0 ⇒ λ = - 1 2 Equation of required plane is 0 + y × 3 2 + z 1 2 - 13 2 = 0 i.e. 3 y + z - 13 = 0

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