3D Geometry: Step-by-Step Worked Examples (6)

Question

Find the shortest distance between the lines

and

Solution — Step by Step

Final answer: shortest distance $= \frac{6}{\sqrt{14}}$ units.

Why This Works

The shortest distance between two skew lines is along the common perpendicular. The vector $\vec{b}_1 \times \vec{b}_2$ is perpendicular to both lines (by definition of cross product). Projecting the connecting vector $\vec{a}_2 - \vec{a}_1$ onto this perpendicular direction gives the distance.

If the two lines were parallel, $\vec{b}_1 \times \vec{b}_2 = \vec{0}$ and this formula breaks. Use a different formula for parallel lines: project $\vec{a}_2 - \vec{a}_1$ perpendicular to $\vec{b}_1$.

Alternative Method

Parametrise both lines and minimise distance squared $|P_1 P_2|^2$ over $\lambda, \mu$. Set partial derivatives to zero, solve the $2 \times 2$ system, find $P_1, P_2$, compute $|P_1 P_2|$. Same answer, much more algebra. Use the cross-product formula in exams.

Common Mistake

Using $\vec{b}_1 \cdot \vec{b}_2$ in the formula instead of $\vec{b}_1 \times \vec{b}_2$. The dot product gives a scalar; we need a vector perpendicular to both lines, which is the cross product.