3D Geometry: Tricky Questions Solved (5)

Question

Find the shortest distance between the lines:

$L_1: \vec{r} = (1, 2, 3) + t(2, 3, 4), \quad L_2: \vec{r} = (2, 4, 5) + s(3, 4, 5)$

Solution — Step by Step

Why This Works

The cross product $\vec{d}_1 \times \vec{d}_2$ is perpendicular to both lines. Projecting the connecting vector onto this direction gives the perpendicular distance between the two skew lines. If the lines are parallel, $\vec{d}_1 \times \vec{d}_2 = \vec{0}$ — use a different formula instead.

Alternative Method

For coplanar (intersecting) lines, the shortest distance is zero — and the scalar triple product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{d}_1 \times \vec{d}_2)$ equals zero. So this product also tests coplanarity. Here it’s $1 \neq 0$, confirming the lines are skew.

Common Mistake

Students sometimes forget the absolute value in the numerator and report a negative distance. Distances are always non-negative — wrap the scalar triple product in $|\cdot|$ before dividing.