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AP EAMCET201921 Apr 2019Evening ShiftMathematicsThree Dimensional GeometryActual

The equation of the plane in normal form which passes through the points ( - 2 , 1 , 3 ) , ( 1 , 1 , 1 ) and ( 2 , 3 , 4 ) is

Options

  1. A2 3 x + - 2 3 y + 1 3 z = 1 3
  2. B- 2 3 x + 2 3 y + - 1 3 z = 1 3
  3. C- 4 173 x + 11 173 y + - 6 173 z = 1 173
  4. D4 173 x + - 11 173 y + 6 173 z = 1 173

Correct answer

C. - 4 173 x + 11 173 y + - 6 173 z = 1 173

Step-by-step solution

It is given that the points, A ( - 2 , 1 , 3 ) ,   B ( 1 , 1 , 1 ) ,   C ( 2 , 3 , 4 ) The position vectors are, AB → = i ^ + j ^ + k ^ - - 2 i ^ + j ^ + 3 k ^ = 3 i ^ - 2 k ^ BC ¯ = 2 i ^ + 3 j ^ + 4 k ^ - i ^ + j ^ + k ^ = i ^ + 2 j ^ + 3 k ^ The normal of the plane, n → = A B → × B C → n → = i ^ j ^ k ˙ 3 0 - 2 1 2 3 By solving the above matrix, n = 4 i - 11 j ^ + 6 k ^ The equation of the plane is, 4 ( x - 2 ) - 11 ( y - 3 ) + 6 ( z - 4 ) = 0 4   x - 1

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