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NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

Equation of the plane passing through the point of intersection of the line x - 1 3 = y - 2 1 = z - 3 2 and x - 3 1 = y - 1 2 = z - 2 3 and perpendicular to the line x + 5 2 = y - 3 3 = z + 1 1 is

Options

  1. A2 x + 3 y + z + 7 = 0
  2. B2 x - 3 y - z + 22 = 0
  3. C2 x + 3 y + z - 22 = 0
  4. D2 x + 3 y + z + 13 = 0

Correct answer

C. 2 x + 3 y + z - 22 = 0

Step-by-step solution

Any general point on the line x - 1 3 = y - 2 1 = z - 3 2 is 3 λ + 1 , λ + 2,2 λ + 3 For the point of intersection, this point must satisfy x - 3 1 = y - 1 2 = z - 2 3 Hence, 3 λ - 2 1 = λ + 1 2 = 2 λ + 1 3 6 λ - 4 = λ + 1 ⇒ λ = 1 So, the point is 4,3 , 5 Also, a vector parallel to x + 5 2 = y - 3 3 = z + 1 1 is 2 i ^ + 3 j ^ + k ^ Hence, the equation of the required plane is 2 x - 4 + 3 y - 3 + 1 z - 5 = 0 2 x + 3 y + z - 22 = 0

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