Shortest Distance Between Two Lines

What "shortest distance" means. Given two lines $L_1$ and $L_2$ in space, the shortest distance between them is the length of the shortest possible line segment joining a point of $L_1$ to a point of $L_2$ -- equivalently, the length of the unique segment that is perpendicular to both lines simultaneously (the common perpendicular). For two intersecting lines this distance is trivially $0$; for two skew lines it is a genuine positive length; for two parallel lines it reduces to the ordinary perpendicular distance between them.

Shortest distance between two skew lines -- derivation. Let $L_1:\ \vec r=\vec a_1+\lambda\vec b_1$ and $L_2:\ \vec r=\vec a_2+\mu\vec b_2$ be two skew lines. Since $\vec b_1\times\vec b_2$ is perpendicular to both $\vec b_1$ and $\vec b_2$ (a defining property of the cross product), it points along the direction of the common perpendicular to both lines -- so the shortest distance is exactly the length of the projection of the vector $\vec a_2-\vec a_1$ (joining a point of $L_1$ to a point of $L_2$) onto this common-perpendicular direction $\vec b_1\times\vec b_2$. The scalar projection of a vector $\vec v$ onto a direction $\vec n$ is $\dfrac{\vec v\cdot\vec n}{|\vec n|}$, so

The absolute value is essential: the scalar triple product $(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)$ can come out negative depending on the order the two lines are labelled or the direction chosen for $\vec b_1,\vec b_2$, but a distance is always taken to be non-negative -- a sign flip here reflects only a labelling choice, never an error in the lines themselves. If $\vec b_1\times\vec b_2=\vec 0$, the lines are parallel and this formula cannot be used (division by a zero vector) -- the parallel-line formula below applies instead. If the numerator itself works out to $0$ while $\vec b_1\times\vec b_2\ne\vec 0$, the lines are not skew after all -- a zero shortest distance between non-parallel lines means they actually intersect (Section 6).

In Cartesian form. If $L_1$ passes through $(x_1,y_1,z_1)$ with direction ratios $a_1,b_1,c_1$ and $L_2$ through $(x_2,y_2,z_2)$ with direction ratios $a_2,b_2,c_2$, the same formula, written using the scalar triple product as a $3\times3$ determinant, is

$d=\left|\frac{\begin{vmatrix}x_2-x_1 & y_2-y_1 & z_2-z_1\\ a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\end{vmatrix}}{\left|(b_1c_2-b_2c_1,\ c_1a_2-c_2a_1,\ a_1b_2-a_2b_1)\right|}\right|.$ …