3D Geometry: Diagram-Based Questions (11)

Question

Find the shortest distance between the lines $L_1: \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $L_2: \vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})$.

Solution — Step by Step

Why This Works

For parallel lines, the perpendicular distance is the same at every point. We pick any vector connecting the two lines (here $\vec{a}_2 - \vec{a}_1$) and project out the part along $\vec{b}$ — what’s left is the perpendicular component.

The cross-product trick gives us the perpendicular component’s magnitude in one shot: $|\vec{u} \times \vec{b}|/|\vec{b}|$ is the area of the parallelogram divided by the base, which is the height.

Alternative Method

We can drop a perpendicular from any point on $L_2$ (say $\vec{a}_2$) onto $L_1$. Set up the foot of perpendicular by minimising distance — gets to the same answer through more algebra.

Common Mistake