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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
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\begin{document}
{\bf Question}
Four points in space have coordinates $$A(1,1,0)\hspace{.2in}
B(3,0,1)\hspace{.2in} C(1,0,2)\hspace{.2in} D(1,1,3)$$ Find the
equations of two parallel planes, of which one contains $A$ and
$B$ and the other contains $C$ and $D$. Deduce the shortest
distance between the lines $AB$, $CD$.
\vspace{.25in}
{\bf Answer}
Parallel planes have the same normal vectors.
Let ${\bf a}$ be such a normal vector.
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Then $\ds {\bf a} \cdot \vec{AB} = 0$ and ${\bf a} \cdot \vec{CD}
= 0$
So ${\bf a} \cdot (2,-1,1) = 0$ and ${\bf a} \cdot (0,1,1) = 0$
$\Rightarrow 2a - b +c = 0$ and $ b+c = 0$
Choose $c = -1 \, b = 1 \, a = 1$
So ${\bf a} = (1,1,-1)$ is a normal vector.
${}$
The equation of the plane though
$AB: x+y-z = 2$
$CD: x+y - z = -1$
So the distance between the planes is $\ds \frac{3}{\sqrt3} =
\sqrt3$
\end{document}