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For r = a1 + λb1 and r = a2 + μb2 (skew), the shortest distance is:
A|(b1 x b2) x (a2 - a1)| / |b1|
B|(b1 . b2) x (a2 - a1)| / |b1 . b2|
C|(b1 x b2) . (a2 - a1)| / |b1 x b2|
D|(a1 x a2) . (b2 - b1)| / |a1 x a2|
Answer & Solution
Correct answer: C. |(b1 x b2) . (a2 - a1)| / |b1 x b2|
The distance is the projection of a2 - a1 on the unit vector along b1 x b2, the common perpendicular.
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