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MHT CET202312 May 2023Evening ShiftMathematicsThree Dimensional GeometryActual

The shortest distance (in units) between the lines x+1 3 = y+2 1 = z+1 2 and r =(2 i -2 j +3 k )+ ( i +2 j ) is

Options

  1. A8 3 5
  2. B1 3 5
  3. C7 3 5
  4. D2 3 5

Correct answer

A. 8 3 5

Step-by-step solution

Given lines are: x+1 3 = y+2 1 = z+1 2 and x-2 1 = y+2 2 = z-3 0 Required distance = | array lll 3 & 0 & 4 3 & 1 & 2 1 & 2 & 0 array | (6-1)^2+(0-2)^2+(0-4)^2 aligned & = | 3(0-4)+0+4(6-1) 25+4+16 | & = | 8 45 | & = 8 3 5 aligned

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