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MHT CET20249 May 2024Morning ShiftMathematicsThree Dimensional GeometryActual

The distance of the point (1,3,-7) from the plane passing through the point (1,-1,-1) having normal perpendicular to both the lines x-1 1 = y+2 -2 = z-4 3 and x-2 2 = y+1 -1 = z+7 -1 is

Options

  1. A10 83 units.
  2. B5 83 units.
  3. C10 74 units.
  4. D20 74 units.

Correct answer

A. 10 83 units.

Step-by-step solution

Normal vector n = | array ccc i & j & k 1 & -2 & 3 2 & -1 & -1 array | aligned & = i (2+3)- j (-1-6)+ k (-1+4) & =5 i +7 j +3 k aligned Let A (1,-1,-1) a = i - j - k Equation of the plane is aligned & 5(x-1)+7(y+1)+3(z+1)=0 & 5 x+7 y+3 z+5=0 aligned Distance of (1,3,-7) from the above plane is d= | 5(1)+7(3)+3(-7)+5 25+49+9 |= 10 83 units

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