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MHT CET202613 April 2026Morning ShiftMathematicsThree Dimensional GeometryActual

The co-ordinates of the foot of the perpendicular from the origin to the plane 2x - 3y - 6z = 49 are...

Options

  1. A(2, -3, -6)
  2. B( 2 7 , -3 7 , -6 7 )
  3. C(-2, 3, 6)
  4. D( -2 7 , 3 7 , 6 7 )

Correct answer

A. (2, -3, -6)

Step-by-step solution

The equation of the plane is 2x - 3y - 6z = 49 . The direction ratios of the normal to the plane are (2, -3, -6) . The equation of the line passing through the origin and perpendicular to the plane is x 2 = y -3 = z -6 = k . Any point on this line is of the form (2k, -3k, -6k) . Since this point is the foot of the perpendicular, it must lie on the plane. Substituting these coordinates into the equation of the plane: 2(2k) - 3(-3k) - 6(-6k) = 49 4k + 9k + 36k = 49 49k = 49 k = 1 Therefore, the coordinates of the foo

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