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JEE Main2016MathematicsThree Dimensional GeometryActual

The distance of the point 1 , - 2 , 4 from the plane passing through the point 1 , 2 , 2 and perpendicular to the planes x - y + 2 z = 3 and 2 x - 2 y + z + 12 = 0 , is :

Options

  1. A2
  2. B2
  3. C2 2
  4. D1 2

Correct answer

C. 2 2

Step-by-step solution

Let equation of plane be a x - 1 + b y - 2 + c z - 2 = 0 ....... i Since ( i ) is perpendicular to the given planes, then Dot product of the normal vectors of the perpendicular planes is 0 . a - b + 2 c = 0 2 a - 2 b + c = 0 Solving above equations c = 0 and a = b equation of plane i becomes x + y - 3 = 0 Distance from 1 ,   - 2 ,   4 will be D = 1 - 2 - 3 1 + 1 = 4 2 = 2 2

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