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MHT CET202619 April 2026Evening ShiftMathematicsThree Dimensional GeometryActual

The shortest distance between the lines r = (4 i - j ) + ( i + 2 j - 3 k ) and r = ( i - j + 2 k ) + ( i + 4 j - 5 k ) is...

Options

  1. A1 2
  2. B1 2
  3. C1 3
  4. D1 3

Correct answer

C. 1 3

Step-by-step solution

The given lines are r = a ₁ + b ₁ and r = a ₂ + b ₂ , where: a ₁ = 4 i - j b ₁ = i + 2 j - 3 k a ₂ = i - j + 2 k b ₂ = i + 4 j - 5 k The shortest distance d between two skew lines is given by: d = |( a ₂ - a ₁) ( b ₁ b ₂)| | b ₁ b ₂| First, find a ₂ - a ₁ : a ₂ - a ₁ = ( i - j + 2 k ) - (4 i - j ) = -3 i + 2 k Next, find the cross product b ₁ b ₂ : b ₁ b ₂ = vmatrix i & j & k 1 & 2 & -3 1 & 4 & -5 vmatrix b ₁ b ₂ = i (-10 + 12) - j (-5 + 3) + k (4 - 2) = 2 i + 2 j + 2 k Find the magnitude of b ₁ b ₂ : | b ₁ b ₂| =

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