Find the shortest distance between the two skew lines:
L1:2x−1=3y−2=4z−3
L2:3x−2=4y−4=5z−5
Solution — Step by Step
From L1: point A=(1,2,3), direction b1=(2,3,4).
From L2: point B=(2,4,5), direction b2=(3,4,5).
The connecting vector is AB=(1,2,2).
b1×b2=i^23j^34k^45
=i^(3⋅5−4⋅4)−j^(2⋅5−4⋅3)+k^(2⋅4−3⋅3)
=i^(−1)−j^(−2)+k^(−1)=(−1,2,−1)
AB⋅(b1×b2)=(1)(−1)+(2)(2)+(2)(−1)=−1+4−2=1
d=∣b1×b2∣∣AB⋅(b1×b2)∣
The magnitude of the cross product:
∣b1×b2∣=1+4+1=6
So:
d=6∣1∣=61
Final answer: d=61 units.
Why This Works
The cross product b1×b2 is perpendicular to both lines, so it is the direction of the shortest connecting segment. Projecting AB onto this perpendicular direction gives the shortest distance.
If the cross product had been zero, the lines would be parallel (not skew), and we would use a different formula. If the dot product had come out to zero, the lines would intersect.
Alternative Method
Set up a parametric distance function:
d2(s,t)=∣A+sb1−B−tb2∣2
Differentiate with respect to s and t, set both partials to zero, solve the 2×2 system, and substitute. Same answer but ten times more algebra.
Common Mistake
Students forget to check whether the lines are skew before applying the formula. If they intersect, the shortest distance is zero — and the formula will correctly give zero, but only if you compute carefully. Always interpret the dot-product numerator: zero means intersecting, nonzero means skew.
Want to master this topic?
Read the complete guide with more examples and exam tips.