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JEE Main202622 January 2026Morning ShiftMathematicsThree Dimensional GeometryActual

Let P ( , , ) be the point on the line x-1 2 = y+1 -3 =z at a distance 4 14 from the point (1,-1,0) and nearer to the origin. Then the shortest distance, between the lines x- 1 = y- 2 = z- 3 and x+5 2 = y-10 1 = z-3 1 , is equal to

Options

  1. A4 7 5
  2. B2 7 4
  3. C7 5 4
  4. D4 5 7

Correct answer

A. 4 7 5

Step-by-step solution

Let P(2 + 1, -3 - 1, ) Then 4 ^2 + 9 ^2 + ^2 = 16 14 = 4 -4 (nearer to origin) P(-7, 11, -4) Shortest distance = vmatrix 2 & -1 & 7 1 & 2 & 3 2 & 1 & 1 vmatrix | i j k | vmatrix 1 & 2 & 3 2 & 1 & 1 vmatrix = 28 1 + 25 + 9 = 4 7 5

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