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NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

The projection of the line x - 1 2 = y + 1 1 = z - 2 3 on a plane P is x - 1 1 = y + 1 2 = z - 2 1 , then the equation of the plane P is

Options

  1. A5 x - 8 y + 11 z = 35
  2. B5 x + 8 y – 21 z + 45 = 0
  3. C5 x + 8 y + 11 z = 35
  4. D5 x – 8 y + 21 z = 45

Correct answer

A. 5 x - 8 y + 11 z = 35

Step-by-step solution

The required plane passes through 1 , – 1,2 The required plane is parallel to i ^ j ^ k ^ 2 1 3 1 2 1 = i ^ - 5 - j ^ - 1 + k ^ 3 = - 5 i ^ + j ^ + 3 k ^ So, normal vector to the plane is i ^ j ^ k ^ - 5 1 3 1 2 1 = i ^ - 5 - j ^ - 8 + k ^ - 11 = - 5 i ^ + 8 j ^ - 11 k ^ Equation of the required plane is - 5 x - 1 + 8 y + 1 - 11 z - 2 = 0 - 5 x + 5 + 8 y + 8 - 11 z + 22 = 0 - 5 x + 8 y - 11 z + 35 = 0 5 x - 8 y + 11 z - 35 = 0

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