Q:

Find the two points where the shortest distance occurs on two lines

Find the point P on $\vec{AB}$ and point Q on $\vec{CD}$ such that $\vec{PQ}$ is the shortest distance
between the lines AB and CD, given $\vec{AB} = \begin{pmatrix}
1\\ 
0\\ 
2\\
\end{pmatrix}
+ u\begin{pmatrix}
-2\\
2\\
1\\
\end{pmatrix}     
,\vec{CD} = \begin{pmatrix}
0\\ 
1\\ 
1\\
\end{pmatrix}
+ v\begin{pmatrix}
2\\
-1\\
-2\\
\end{pmatrix}     
$ the normal vector $n=\begin{pmatrix}
-3\\
-2\\
-2\\
\end{pmatrix} 
$ 
and the shortest distance is $\frac{3}{\sqrt{17}}$. I figured all this out from 4 given points but don't know how to find points P and Q. Please help, I'm stuck...

A:

Once you know that the normal vector is $n$, the vector equation
$$
\begin{pmatrix}
1\\ 
0\\ 
2\\
\end{pmatrix}
+ u\begin{pmatrix}
-2\\
2\\
1\\
\end{pmatrix}     
+w\begin{pmatrix}
-3\\
-2\\
-2\\
\end{pmatrix} 
 = \begin{pmatrix}
0\\ 
1\\ 
1\\
\end{pmatrix}
+ v\begin{pmatrix}
2\\
-1\\
-2\\
\end{pmatrix}     
$$
is equivalent to a system of three linear equations in three unknowns, which indeed has the unique solution
$$
u=\frac{19}{17},v=-\frac{15}{17},w=\frac{3}{17}.
$$
That tells you that the points on $\vec{AB}$ and $\vec{CD}$ are
$$
\begin{pmatrix}
1\\ 
0\\ 
2\\
\end{pmatrix}
+ \frac{19}{17}\begin{pmatrix}
-2\\
2\\
1\\
\end{pmatrix}     
\quad\text{and}\quad\begin{pmatrix}
0\\ 
1\\ 
1\\
\end{pmatrix}
-\frac{15}{17}\begin{pmatrix}
2\\
-1\\
-2\\
\end{pmatrix},
$$
respectively. (It also tells you that the distance between the two lines is the norm of $\frac3{17}(-3\ {-2}\ {-2})$, or $\frac3{\sqrt{17}}$ as you indicated.)

