Shortest Distance Between Skew Lines

For skew lines r = a₁+λb₁ and r = a₂+μb₂, the shortest distance is the projection of (a₂−a₁) onto the common perpendicular direction b₁×b₂.

Let the skew lines be $L_1: \mathbf{r} = \mathbf{a_1}+\lambda\mathbf{b_1}$ and $L_2: \mathbf{r} = \mathbf{a_2}+\mu\mathbf{b_2}$.

The common perpendicular: A line perpendicular to both $L_1$ and $L_2$ must be parallel to $\mathbf{b_1}\times\mathbf{b_2}$ (perpendicular to both direction vectors). This common perpendicular is unique for skew lines and has the shortest length.

Shortest distance: Let $P$ on $L_1$ and $Q$ on $L_2$ be the endpoints of the common perpendicular. $\overrightarrow{PQ}$ is parallel to $\mathbf{b_1}\times\mathbf{b_2}$.

The vector $\mathbf{a_2}-\mathbf{a_1}$ connects one base point to the other. Its projection onto the common perpendicular direction gives the shortest distance:

$$d = \frac{|(\mathbf{a_2}-\mathbf{a_1})\cdot(\mathbf{b_1}\times\mathbf{b_2})|}{|\mathbf{b_1}\times\mathbf{b_2}|}$$

Cartesian form: For lines $(x-x_1)/a_1 = (y-y_1)/b_1 = (z-z_1)/c_1$ and $(x-x_2)/a_2 = (y-y_2)/b_2 = (z-z_2)/c_2$:

$$d = \frac{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}}{|\mathbf{b_1}\times\mathbf{b_2}|}$$

When lines are parallel ($\mathbf{b_1}\times\mathbf{b_2} = \mathbf{0}$): The formula breaks down. Use the point-to-line distance formula instead.

When $d = 0$: The lines are coplanar (either intersecting or parallel).