Shortest distance

Given a point A and a line $\overrightarrow{r}={\overrightarrow{r}}_0+t\overrightarrow{m}\text{,}$ we can find the shortest distance from the point to the line easily with

$d=\frac{|\overrightarrow{AB}\times \overrightarrow{m}|}{\left|\overrightarrow{m}\right|}\text{.}$

Given two parallel planes with a normal vector $\overrightarrow{n}$ (it doesn’t matter which plane you take it from) and containing points A and B, we can find the shortest distance between them using the scalar projection:

$d=\left|{\overrightarrow{AB}}_n\right|=|\overrightarrow{AB}\cdot \hat{n}|=\frac{|\overrightarrow{AB}\cdot \overrightarrow{n}|}{\left|\overrightarrow{n}\right|}\text{.}$

(Note that nonparallel planes intersect, so the shortest distance is zero.) We can also use this equation for skew lines, where one line contains A and the other has B. We calculate $\overrightarrow{n}$ by crossing their direction vectors:

$\overrightarrow{n}={\overrightarrow{m}}_1\times {\overrightarrow{m}}_2\text{.}$