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NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

If the point of intersection of the plane 4 x - 5 y + 2 z - 6 = 0 with the line through the origin and perpendicular to the plane x - 2 y - 4 z = 4 is P , then the distance of the point P from 1 , 2 , 3 is

Options

  1. A63 units
  2. B8 units
  3. C65 units
  4. D72 units

Correct answer

C. 65 units

Step-by-step solution

Equation of line through origin and perpendicular to the plane x - 2 y - 4 z = 4 is x - 0 1 = y - 0 - 2 = z - 0 - 4 = λ So, any point on this line is λ , - 2 λ , - 4 λ ∵ the above point lies on the plane 4 x - 5 y + 2 z - 6 = 0 ⇒ 4 λ + 10 λ - 8 λ - 6 = 0 ⇒ 6 λ = 6 ⇒ λ = 1 Coordinates of the point P are 1 , – 2 , – 4 The distance of point P from 1 , 2 , 3 is 0 + 16 + 49 = 65

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