3D Geometry: Edge Cases and Subtle Traps (9)

Question

Find the shortest distance between the lines $\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$ and $\vec{r} = 2\hat{i} + 4\hat{j} + 5\hat{k} + \mu(3\hat{i} + 4\hat{j} + 5\hat{k})$.

Solution — Step by Step

Why This Works

The shortest distance between two skew lines is along their common perpendicular. The vector $\vec{b}_1 \times \vec{b}_2$ points along this common perpendicular (perpendicular to both directions). Project $\vec{a}_2 - \vec{a}_1$ (the connector between any two specific points on the lines) onto this perpendicular direction, and take the magnitude. That’s the shortest distance.

If $\vec{b}_1 \times \vec{b}_2 = 0$, the lines are parallel — different formula. If the dot product is zero but the cross product is not, the lines intersect (distance is 0).

Alternative Method

Parametrise: minimise $|P_1 P_2|^2$ where $P_1$ is on line 1 and $P_2$ is on line 2. Take partial derivatives with respect to $\lambda$ and $\mu$, set them to zero, and find the critical $\lambda, \mu$. Then compute the distance. This is heavy algebra but equivalent.

Common Mistake

Forgetting the absolute value in the formula. The dot product can come out negative; distance is always positive. Also, students sometimes forget to divide by $|\vec{b}_1 \times \vec{b}_2|$, treating it as a vector instead of a scalar. The output should be a number with units of length.