# Shortest distance

Given a point A and a line $\displaystyle \vec{r} = \vec{r}_{0} + t \vec{m}$, we can find the shortest distance from the point to the line easily with

$\displaystyle d = \frac{\left \lvert \overrightharpoon{A B} \times \vec{m} \right \rvert}{\left \lvert \vec{m} \right \rvert}$.

Given two parallel planes with a normal vector $\displaystyle \vec{n}$ (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:

$\displaystyle d = \left \lvert \overrightharpoon{A B}_{n} \right \rvert = \left \lvert \overrightharpoon{A B} \cdot \hat{n} \right \rvert = \frac{\left \lvert \overrightharpoon{A B} \cdot \vec{n} \right \rvert}{\left \lvert \vec{n} \right \rvert}$.

(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 $\displaystyle \vec{n}$ by crossing their direction vectors:

$\displaystyle \vec{n} = \vec{m}_{1} \times \vec{m}_{2}$.