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NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

Let P 1 : x + y + 2 z = 3 and P 2 : x - 2 y + z = 4 be two planes. Let A 2 , 4 , 5 and B 4 , 3 , 8 be two points in space. The equation of plane P 3 through the line of intersection of P 1 and P 2 such that the length of the projection upon it of the line segment A B is the least, is

Options

  1. A2 x - y + 3 z = 7
  2. B3 y + z + 1 = 0
  3. Cx + 3 y + z + 2 = 0
  4. D3 x - 3 y + 4 z - 11 = 0

Correct answer

A. 2 x - y + 3 z = 7

Step-by-step solution

Equation of plane P 3 is P 1 + λ P 2 = 0 x 1 + λ + y 1 - 2 λ + z 2 + λ = 3 + 4 λ A B → = < 2 , - 1 , 3 > Length of projection of A B → on plane is least ⇒ A B → must be perpendicular to the plane ⇒ 1 + λ 2 = 1 - 2 λ - 1 = 2 + λ 3 ⇒ λ = 1 So, the equation of the plane is 2 x - y + 3 z = 7

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