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The shortest distance between skew lines r = a₁ + λb₁ and r = a₂ + μb₂ is
A[(b₁ · b₂) × (a₂ + a₁)] / |b₁ × b₂|
B|b₁ × b₂| · |a₂ - a₁| / (b₁ + b₂)
C|(b₁ × a₂) · (b₂ × a₁)| / |b₁ - b₂|
D|(a₂ - a₁) · (b₁ × b₂)| / |b₁ × b₂|
Answer & Solution
Correct answer: D. |(a₂ - a₁) · (b₁ × b₂)| / |b₁ × b₂|
1. The common perpendicular direction to both lines is along b₁ × b₂.
2. Project the connecting vector (a₂ - a₁) onto the unit vector along b₁ × b₂.
3. Magnitude of projection: |(a₂ - a₁) · (b₁ × b₂)| / |b₁ × b₂|.
4. This gives the shortest distance between the two skew lines.
_Source: NCERT Class 12 Maths Part 2 Ch 11 "Three Dimensional Geometry", §11.5.1_
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