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JEE Main2015MathematicsThree Dimensional GeometryActual

A plane containing the point ( 3 , 2 , 0 ) and the line x - 1 1 = y - 2 5 = z - 3 4 also contains the point

Options

  1. A( 0 , 7 , - 10 )
  2. B( 0 , 7 , 10 )
  3. C( 0 , 3 , 1 )
  4. D( 0 , - 3 , 1 )

Correct answer

B. ( 0 , 7 , 10 )

Step-by-step solution

The Direction ratio perpendicular to the plane would be cross product of i ^ + 5 j ^ + 4 k ^ and 3 - 1 i ^ + 2 - 2 j ^ + 0- 3 k ^ = i ^ j ^ k ^ 1 5 4 2 0 - 3 = - 15 i ^ + 11 j ^ - 10 k ^ ∴ Equation of plane would be - 15 x + 11 y - 10 z = λ As plane contain 3,2 , 0 - 45 + 22 - 0 = λ - 23 = λ - 15 x + 11 y - 10 z + 23 = 0 This passes through ( 0,7 , 10 )

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