The shortest distance between skew lines is one of the highest-frequency JEE questions in 3D geometry, and the STP is central to its computation.

Formula: For lines r=a1+tb1 and r=a2+sb2, the shortest distance SD = |[a2-a1, b1, b2]|/|b1 x b2|.

Derivation: The common perpendicular to both lines has direction b1 x b2. The shortest distance is the projection of the displacement vector (a2-a1) onto this direction: SD = |(a2-a1).(b1 x b2)|/|b1 x b2|.

When [a2-a1, b1, b2] = 0: The lines are coplanar. If b1 x b2 != 0, they intersect. If b1 x b2 = 0, they are parallel.

Parallel lines: When b1 is parallel to b2 (same direction b), the formula gives 0/0. Use instead: SD = |(a2-a1) x b|/|b|.

Finding the actual closest points: Solve the system (a1+tb1-a2-sb2).b1=0 and (a1+tb1-a2-sb2).b2=0 for t and s. Substitute to get the two closest points.

Direction of the common perpendicular: b1 x b2. This is perpendicular to both lines.

Computational strategy: (1) Identify a1, a2, b1, b2 from the given equations. (2) Compute a2-a1. (3) Compute b1 x b2. (4) If b1 x b2 = 0, use the parallel formula. (5) Otherwise, compute the STP and divide by |b1 x b2|.

Common error: confusing the signs or components of a2-a1. Always be consistent: a2 is the point on the second line, a1 on the first.