The cross product b1×b2 gives a vector perpendicular to both lines. The shortest distance between two skew lines is the projection of the vector connecting any two points (one on each line) onto this perpendicular direction.
Geometrically, think of it as: a2−a1 connects a point on line 1 to a point on line 2. The component of this vector along b1×b2 is exactly the perpendicular gap between the lines — everything else is “along” one of the lines and doesn’t contribute to the distance.
Alternative Method — Parametric approach
Set points on each line as P=(1+2s,−1+3s,1−s) and Q=(−1+5t,2+t,3t).
The vector PQ must be perpendicular to both direction vectors. This gives two equations:
PQ⋅b1=0 and PQ⋅b2=0
Solving for s and t, then computing ∣PQ∣. This works but involves more algebra. The cross product formula is faster.
In CBSE boards, the examiner expects the cross product formula. Show every step: write b1×b2 using the determinant, compute a2−a1, take the dot product, then divide. Don’t skip the intermediate cross product — it carries marks in the marking scheme.
Common Mistake
The most common error: computing a1−a2 instead of a2−a1 (or vice versa) and worrying about the sign. Since we take the absolute value of the dot product in the numerator, the direction of a2−a1 doesn’t matter. Don’t waste time checking which way to subtract — the modulus takes care of it.
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