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…
Best Answer
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.)