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NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

The equation of the plane which passes through the point of intersection of x - 1 3 = y - 2 1 = z - 3 2 and x - 3 1 = y - 1 2 = z - 2 3 and perpendicular to 4 i ^ + 3 j ^ + 5 k ^ , is

Options

  1. A4 x + 3 y + 5 z = 50
  2. B4 x + 3 y + 5 z = 25
  3. C4 x + 3 y + 5 z = 1 0
  4. D4 x - 3 y + 5 z = 50

Correct answer

A. 4 x + 3 y + 5 z = 50

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 This must satisfy the given 2 n d line ⇒ 3 λ - 2 1 = λ + 1 2 = 2 λ + 1 3 ⇒ λ = 1 So, the point of intersection of the two lines are 4 , 3 , 5 Equation of the required plane is 4 x - 4 + 3 y - 3 + 5 z - 5 = 0 ⇒ 4 x - 16 + 3 y - 9 + 5 z - 25 = 0 ⇒ 4 x + 3 y + 5 z - 50 = 0

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