Point-line distance

Property — Point–line distance

To compute the distance of a point $P$ from the line through $S_0$ with direction unit vector $\hat v$:

1. $\vec a=\vec{S_0 P}$;
2. $\vec a_\parallel = (\vec a\cdot\hat v)\,\hat v$ (component along the line);
3. $\vec a_\perp = \vec a - \vec a_\parallel$;
4. $d = |\vec a_\perp|$.

Example

Let $P(6;-1;-3)$ and the line through $A(0;3;4)$, $B(-1;-1;-1)$. Then $\vec v=(-1;-4;-5)$ and $\hat v=\vec v/\sqrt{42}$; moreover $\vec a=\vec{AP}=(6;-4;-7)$. One obtains $\vec a\cdot\hat v \approx 6{,}95$, whence $\vec a_\parallel\approx 6{,}95\,\hat v$, $\vec a_\perp = \vec a - \vec a_\parallel$ and finally $d=|\vec a_\perp|\approx 7{,}26.$

Links

Topics: Analytic geometry in space
Concepts: Parallel component · Point-line distance · Line · Line in space
Skills: Analytic geometry · Using formulae