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JEE Main2015MathematicsThree Dimensional GeometryActual

The distance of the point 1 , 0 , 2 from the point of intersection of the line x - 2 3 = y + 1 4 = z - 2 12 and the plane x - y + z =16 , is

Options

  1. A1 3
  2. B2 1 4
  3. C8
  4. D3 2 1

Correct answer

A. 1 3

Step-by-step solution

Let x - 2 3 = y + 1 4 = z - 2 12 = t General points is ( 2 + 3 t , -   1 + 4 t , 2 + 12 t ) It lies on the plane x - y + z = 16 . ⇒ 2 + 3 t + 1 - 4 t + 2 + 12 t = 16 ⇒ t = 1 Hence, the point of intersection will be ( 2 + 3 ( 1 ) , - 1 + 4 ( 1 ) , 2 + 12 ( 1 ) ) = 5 , 3 , 14 Distance from ( 1 , 0 , 2 ) = 5 - 1 2 + 3 - 0 2 + 1 4 - 2 2 = 4 2 + 3 2 + 1 2 2 = 1 3

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