Distance Between a Line and a Point

Topics: Line - 3D Space

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Let $q\in {\mathbb{R}}^3$ and $l_1=\{p+tu\mid t\in \mathbb{R}\}$. The distance between $l_1$ and $q$, denoted by $\text{d}(l_1,q)$ is:

$$\text{d}(l_1,q)=\frac{\Vert u\times (p-q)\Vert }{\Vert u\Vert }$$

Reasoning behind the formula

The formula comes from the fact that the (shortest) distance $d(l_1,q)$ between a point and a line is the length of the segment that goes from the point to the line and is perpendicular to it.

If we think of a vector that goes from $p$ to $q$ (i.e. $p-q$), then we can calculate this shortest distance by dividing the area of the parallelogram (generated by the $p-q$ vector and the line vector, $\Vert u\times (p-q)\Vert$) by the length of the line vector ($\Vert u\Vert$), because this shortest distance is the height of such parallelogram.

If $u$ is a unit vector

In the case that $u$, the defining vector for $l_1$, is a unit vector, then the distance between $l_1$ and $q$ becomes:

$$\text{d}(l_1,q)=\Vert u\times (p-q)\Vert$$

Notice that, if we set $v=p-q$, then this coincides with the area $\Vert u\times v\Vert$ of the parallelogram generated by $u$ and $v$.