3D Geometry: Numerical Problems Set (1)

Question

Find the shortest distance between the lines $\vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k} + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$ and $\vec{r}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k} + \mu(3\hat{i} + 4\hat{j} + 5\hat{k})$.

Solution — Step by Step

The shortest distance is $\dfrac{1}{\sqrt{6}}$ units.

Why This Works

Two skew lines in 3D have a unique shortest distance, measured along the common perpendicular. The vector $\vec{b}_1\times\vec{b}_2$ is perpendicular to both direction vectors, so projecting the joining vector $\vec{a}_2-\vec{a}_1$ onto this normal gives the perpendicular separation directly.

If $\vec{b}_1 \times \vec{b}_2 = \vec{0}$, the lines are parallel and a different formula applies. If the scalar triple product in the numerator is zero, the lines are coplanar and intersect — distance is zero.

Alternative Method

Parameterise the points on each line, write the squared distance as a function of $\lambda$ and $\mu$, and minimise. Tedious for 3D — the cross-product method is much faster.

Common Mistake

Forgetting the modulus on the numerator. The shortest distance is non-negative; if your scalar triple product comes out negative, just take its magnitude. CBSE 2024 had a four-mark question on exactly this setup.