3D Geometry: Exam-Pattern Drill (8)

Question

Find the shortest distance between the lines

$\vec{r}_1 = (1, 2, 3) + t(2, 3, 4)$ $\vec{r}_2 = (2, 4, 5) + s(3, 4, 5)$

This is a JEE Main pattern question testing the skew-line distance formula.

Solution — Step by Step

Final answer: $d = 1/\sqrt{6}$.

Why This Works

The cross product $\vec{b}_1 \times \vec{b}_2$ gives a vector perpendicular to both lines. Projecting the displacement between any two points (one on each line) onto this perpendicular gives the shortest distance. The denominator normalises the cross product to unit length.

If the lines are parallel, $\vec{b}_1 \times \vec{b}_2 = \vec{0}$ and this formula breaks. Use the parallel-line distance formula instead.

Alternative Method

Set up the distance squared as a function of $t$ and $s$, take partial derivatives, set to zero, and solve. This is the calculus approach — slow for an MCQ, but bulletproof.

Common Mistake

Forgetting the absolute value in the numerator. The signed scalar triple product can be negative, but distance is always positive. Always wrap with $|\cdot|$.