Let $L_1$ and $L_2$ be two parallel lines with the equations $\vec{r}={\vec{a}}_1+\lambda \vec{b}$ and $\vec{r}={\vec{a}}_2+\mu \vec{b}$ respectively. The shortest distance between them is:

For two parallel lines, the shortest distance $d$ is the perpendicular distance from any point on $L_2$ (with position vector ${\vec{a}}_2$) to the line $L_1$.The vector joining points on the two lines is $({\vec{a}}_2-{\vec{a}}_1)$.The distance is the magnitude of the component of $({\vec{a}}_2-{\vec{a}}_1)$ perpendicular to the direction vector $\vec{b}$.Using the cross product:
$$d=\frac{|\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)|}{|\vec{b}|}$$