Shortest distance between $\vec{V_1}$ and $\vec{V_2}$ with $d=|\frac{(\vec{V_1}\times\vec{V_2})\cdot\vec{P_1P_2}}{|\vec{V_1}\times\vec{V_2}|}|$. I can find the shortest distance $d$ between two skew lines $\vec{V_1}$ and $\vec{V_2}$  in 3D space with $d=\left|\frac{(\vec{V_1}\times\vec{V_2})\cdot\vec{P_1P_2}}{|\vec{V_1}\times\vec{V_2}|}\right|$. But how do I calculate the actual points $A(x,y,z)$ and $B(x,y,z)$ on those two lines where said shortest distance $d$ is located? Thanks in advance.
 A: The minimum can be found with analytical method described by @Lubin in a comment. If you want to avoid explicit differentiation, you might take a shortcut: the shortest line segment between two lines is perpendicular to both lines.
Assuming $\vec A_0$ and $\vec B_0$ are points on each line, and their respective vectors are $\vec a$ and $\vec b$, so the point of one line is $$\vec A(t)=\vec A_0+t\vec a $$ and the point of the other one is $$\vec B(s)=\vec B_0+s\vec b$$
you have the segment $\overline{AB}$ must be perpendicular to $\vec a$ and to $\vec b$, which is equivalent to zero value of respective scalar products:
$$\begin{cases}
(\vec A(t)-\vec B(s))\cdot \vec a = 0 \\
(\vec A(t)-\vec B(s))\cdot \vec b = 0
\end{cases}$$
which results in a system of two linear equations with unknown $t,s$.
Solve it, plug $t$ and $s$ values into $\vec A(t)$ and $\vec B(s)$ definitions and you're done.
PS.
Don't forget to consider a special case of the two lines parallel.
EDIT
$$\begin{cases}
(\vec A_0+t\vec a-\vec B_0-s\vec b)\cdot \vec a = 0 \\
(\vec A_0+t\vec a-\vec B_0-s\vec b)\cdot \vec b = 0
\end{cases}$$
$$\begin{cases}
t(\vec a\cdot \vec a)-s(\vec a\cdot \vec b) = (\vec B_0-\vec A_0)\cdot \vec a \\
t(\vec a\cdot \vec b)-s(\vec b\cdot \vec b) = (\vec B_0-\vec A_0)\cdot \vec b
\end{cases}$$
A: Here's a solution, which I have also added to Wikipedia (https://en.wikipedia.org/wiki/Skew_lines#Nearest_Points), done completely using vectors without a need to solve equations.
Expressing the two lines as vectors:
Line 1: $\mathbf{v_1}=\mathbf{p_1}+t_1\mathbf{d_1}$
Line 2: $\mathbf{v_2}=\mathbf{p_2}+t_2\mathbf{d_2}$
The cross product of $\mathbf{d_1}$ and $\mathbf{d_2}$ is perpendicular to the lines.
$\mathbf{n}= \mathbf{d_1} \times \mathbf{d_2}$
The plane formed by the translations of Line 2 along $\mathbf{n}$ contains the point $\mathbf{p_2}$ and is perpendicular to $\mathbf{n_2}= \mathbf{d_2} \times \mathbf{n}$.
Therefore, the intersecting point of Line 1 with the above mentioned plane, which is also the point on Line 1 that is nearest to Line 2 is given by
$\mathbf{c_1}=\mathbf{p_1}+ \frac{(\mathbf{p_2}-\mathbf{p_1})\cdot\mathbf{n_2}}{\mathbf{d_1}\cdot\mathbf{n_2}} \mathbf{d_1}$
Similarly, the point on Line 2 nearest to Line 1 is given by (where $\mathbf{n_1}= \mathbf{d_1} \times \mathbf{n}$)
$\mathbf{c_2}=\mathbf{p_2}+ \frac{(\mathbf{p_1}-\mathbf{p_2})\cdot\mathbf{n_1}}{\mathbf{d_2}\cdot\mathbf{n_1}} \mathbf{d_2}$
Now, $\mathbf{c_1}$ and $\mathbf{c_2}$ form the shortest line segment joining Line 1 and Line 2.
