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NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

Let L be the line through the intersection of the planes 3 x - y + 2 z + 1 = 0 and 3 x - 2 y + z = 3 . Then, the equation of the plane passing through 2 , 1 , 4 and perpendicular to the line L is

Options

  1. Ax + y - z = 2
  2. Bx + y - z + 1 = 0
  3. Cx + y + z - 7 = 0
  4. D2 x - 3 y + 4 z = 17

Correct answer

B. x + y - z + 1 = 0

Step-by-step solution

A vector parallel to the line of intersection is i ^ j ^ k ^ 3 - 1 2 3 - 2 1 = i ^ 3 - j ^ - 3 + k ^ - 3 = 3 i ^ + 3 j ^ - 3 k ^ Also, the line of intersection is perpendicular to the required plane. Hence, equation of the plane is 1 x - 2 + 1 y - 1 - 1 z - 4 = 0 ⇒ x + y - z + 1 = 0

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