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MHT CET202313 May 2023Evening ShiftMathematicsThree Dimensional GeometryActual

The distance of the point (1,6,2) from the point of intersection of the line x-2 3 = y+1 4 = z-2 12 and the plane x-y+z=16 is

Options

  1. A11 units
  2. B12 units
  3. C13 units
  4. D14 units

Correct answer

C. 13 units

Step-by-step solution

Let x-2 3 = y+1 4 = z-2 12 = The co-ordinates of any point on the line are P (3 +2,4 -1,12 +2) This point lies on the plane x-y+z=16 3 +2-(4 -1)+12 +2=16 aligned & 11 =11 & =1 aligned P (5,3,14) Let Q (1,6,2) PQ = (1-5)^2+(6-3)^2+(2-14)^2 = 16+9+144 =13 units

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