As an application of the vector product we derive the expression of the distance $d$ 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 $$\vec{r} = \vec{a}\!+\!s\vec{u}$$ and $$\vec{r} = \vec{b}\!+\!t\vec{v},$$ where $s$ and $t$ 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 $$\vec{n} := \frac{\vec{u}\!\times\!\vec{v}}{|\vec{u}\!\times\!\vec{v}|}$$ 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 $A$ and $B$ on the lines, and thus their difference $\vec{a}\!-\!\vec{b}$ is the vector from $B$ to $A$ If we project $\vec{a}\!-\!\vec{b}$ on the unit normal $\vec{n}$ the obtained vector $$\vec{d} := [(\vec{a}\!-\!\vec{b})\!\cdot\!\vec{n}]\,\vec{n}$$ has the sought length $d = |(\vec{a}\!-\!\vec{b})\!\cdot\!\vec{n}|$ i.e. $$d = \frac{|(\vec{a}\!-\!\vec{b})\cdot(\vec{u}\!\times\!\vec{v})|}{|\vec{u}\!\times\!\vec{v}|}. $$ For illustrating that $d$ is the minimal distance between points of the two lines we underline, that the construction of $d$ 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 $d = 0$ If the lines do not intersect, they are called agonic lines or skew lines; then $d > 0$