PlanetMath: distance of non-parallel lines

(more info)

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distance of non-parallel lines

(Derivation)

As an application of the vector product we derive the expression of the distance between two non-parallel straight lines in $\mathbb{R}^3$ .

Suppose that the position vectors of the points of the two non-parallel lines are expressed in parametric forms

$\displaystyle \vec{r} = \vec{a}\!+\!s\vec{u}$

and

$\displaystyle \vec{r} = \vec{b}\!+\!t\vec{v},$

where

and

are parameters. A common normal vector of the lines is the cross product $\vec{u}\times\vec{v}$ of the direction vectors of the lines, and it may be normed to a unit vector

$\displaystyle \vec{n} := \frac{\vec{u}\!\times\!\vec{v}}{\vert\vec{u}\!\times\!\vec{v}\vert}$

by dividing it by its length, which is distinct from 0 because of the non-parallelity. The vectors $\vec{a}$ and $\vec{b}$ are the position vectors of certain points

and

on the lines, and thus their difference $\vec{a}\!-\!\vec{b}$ is the vector from

. If we project $\vec{a}\!-\!\vec{b}$ on the unit normal $\vec{n}$ , the obtained vector

$\displaystyle \vec{d} := [(\vec{a}\!-\!\vec{b})\!\cdot\!\vec{n}]\,\vec{n}$

has the sought length $d = \vert(\vec{a}\!-\!\vec{b})\!\cdot\!\vec{n}\vert$ , i.e.

$\displaystyle d = \frac{\vert(\vec{a}\!-\!\vec{b})\cdot(\vec{u}\!\times\!\vec{v})\vert}{\vert\vec{u}\!\times\!\vec{v}\vert}.$

For illustrating that

is the minimal distance between points of the two lines we underline, that the construction of

guarantees that it connects two points on the lines and is perpendicular to both lines -- thus any displacement of its end point makes it longer.

Notes. The numerator is the absolute value of a triple scalar product. If the lines intersect each other, then the connecting vector $\vec{a}\!-\!\vec{b}$ is at right angles to the common normal vector $\vec{n}$ of their plane and thus the dot product of these vectors vanishes, i.e. also . If the lines do not intersect, they are called agonic lines or skew lines; then .