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For two skew lines (x-x₁)/a₁ = (y-y₁)/b₁ = (z-z₁)/c₁ and (x-x₂)/a₂ = (y-y₂)/b₂ = (z-z₂)/c₂, the shortest distance equals
A|sum of products of intercepts × magnitudes|
B|scalar triple product of (a₂-a₁), b₁, b₂| / |b₁ × b₂|
C(a₁ + a₂)(b₁ + b₂)(c₁ + c₂) / √sum
D(a₁ a₂ + b₁ b₂ + c₁ c₂) / √sum
Answer & Solution
Correct answer: B. |scalar triple product of (a₂-a₁), b₁, b₂| / |b₁ × b₂|
1. The connecting vector between fixed points is (x₂-x₁, y₂-y₁, z₂-z₁).
2. The cross product b₁ × b₂ has components (b₁c₂-b₂c₁, c₁a₂-c₂a₁, a₁b₂-a₂b₁).
3. The shortest distance is the magnitude of the scalar triple product divided by |b₁ × b₂|.
4. Symbolically: d = |(a₂ - a₁) · (b₁ × b₂)| / |b₁ × b₂|.
_Source: NCERT Class 12 Maths Part 2 Ch 11 "Three Dimensional Geometry", §11.5.1_
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