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AP EAMCET202119 Aug 2021Evening ShiftMathematicsThree Dimensional GeometryActual

Find the equation of the plane passing through the point ( 2 , 1 , 3 ) and perpendicular to the planes x - 2 y + 2 z + 3 = 0 and 3 x - 2 y + 4 z - 4 = 0 .

Options

  1. A2 x − y − 2 z + 3 = 0
  2. Bx − 2 y + 2 z − 3 = 0
  3. C2 x − y + 2 z − 3 = 0
  4. D2 x + y − 2 z − 3 = 0

Correct answer

A. 2 x − y − 2 z + 3 = 0

Step-by-step solution

We can conclude, the normal to the plane S 1 i.e. x - 2 y + 2 z + 3 = 0 is (say) r 1 → = i ^ - 2 j ^ + 2 k ^ and the normal to the plane S 2 i.e. 3 x - 2 y + 4 z - 4 = 0 is (say) r 2 → = 3 i ^ - 2 j ^ + 4 k ^ So, the normal to the plane S 3 (which is perpendicular to the plane S 1 and S 2 ) is r 3 → = i ^ j ^ k ^ 1 - 2 2 3 - 2 4 = - 4 i ^ + 2 j ^ + 4 k ^ So, the equation of plane S 3 is - 4 x + 2 y + 4 z + c = 0     . . . 1 As, it is passing through the point 2 ,   1 ,   3 So,

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