3D Geometry: Application Problems (7)

Question

Find the shortest distance between the two skew lines:

$L_1: \vec{r} = (1, 2, 3) + t(2, 1, -1)$ and $L_2: \vec{r} = (2, -1, 1) + s(1, 2, 1)$.

Solution — Step by Step

Final answer: Shortest distance = $\mathbf{2\sqrt{3}/3 \approx 1.155 \text{ units}}$.

Why This Works

The shortest distance between two skew lines is along the direction perpendicular to both of them — that’s exactly $\vec{b}_1 \times \vec{b}_2$. We then project the connecting vector $(\vec{a}_2 - \vec{a}_1)$ onto this perpendicular direction.

If the lines intersect, $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0$ — the connecting vector lies in the plane of both directions, so its component along the perpendicular is zero.

Alternative Method

We can parametrize both lines and minimize $|\vec{r}_1(t) - \vec{r}_2(s)|^2$ by setting partial derivatives w.r.t. $t$ and $s$ to zero. This gives two linear equations, more work but no need to memorize the formula.

Common Mistake