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MHT CET202526 Apr 2025Evening ShiftMathematicsThree Dimensional GeometryActual

The unit vectors perpendicular to the plane determined by the points A (1,-1,2) ~B (2,0,-1) ~C (0,2,1) is

Options

  1. A( 3 i + j + k 11 )
  2. B( - i +2 j + k 6 )
  3. C( 2 i + j + k 6 )
  4. D( i + j + k 3 )

Correct answer

C. ( 2 i + j + k 6 )

Step-by-step solution

Perpendicular unit vectors to the plane determined by points A(1,-1,2) , B(2,0,-1) , and C(0,2,1) can be found by computing the cross product of two vectors in the plane. The vectors AB = (2-1) i + (0-(-1)) j + (-1-2) k = i + j - 3 k and AC = (0-1) i + (2-(-1)) j + (1-2) k = - i + 3 j - k lie in the plane. The cross product N = AB AC yields a vector perpendicular to the plane: N = vmatrix i & j & k 1 & 1 & -3 -1 & 3 & -1 vmatrix = i (1 (-1) - (-3) 3) - j (1 (-1) - (-3) (-1)) + k (1 3 - 1 (-1)) N = i (-1 + 9) - j (-

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