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JEE Main20246 Apr 2024Morning ShiftMathematicsThree Dimensional GeometryActual

The shortest distance between the lines x-3 2 = y+15 -7 = z-9 5 and x+1 2 = y-1 1 = z-9 -3 is

Options

  1. A8 3
  2. B4 3
  3. C5 3
  4. D6 3

Correct answer

B. 4 3

Step-by-step solution

aligned & x-3 2 = y+15 -7 = z-9 5 & x+1 2 = y-1 1 = z-9 -3 & S.D = | ( a ₂ a ₁ ) ( b ₁ b ₂ ) | | b ₁ b ₂ | & a₁=3,-15,9 & ~b ₁=2,-7,5 & a ₂=-1,1,9 & ~b ₂=2,1,-3 & a₂-a₁=-4,16,0 & b ₁ b ₂= | array ccc i & j & k 2 & -7 & 5 2 & 1 & -3 array |= i (16)- j (-16)+ k (16) & 16( i + j + k ) & | b ₁ b ₂ |=16 3 & ( a ₂- a ₁ ) ( b ₁- b ₂ )=16[-4+16]=(16)(12) & S.D. = (16)(12) 16 3 =4 3 & aligned

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