Example 3. As shown in Figure 6. Let the right prism $A B C D-A_{1} B_{1} C_{1} D_{1}$ have a height of 1, and the base be an isosceles trapezoid, with the longer base $|A B|=7$, the leg length being 2, and the base angle being $60^{\circ}$. Let $E$ be on $A B$ such that $|E B|=1$. Find the distance between $A_{1} D_{1}$ and $B_{1} E$, the angle they form, and the position of their common perpendicular.
It is evident that $\left|B_{1} E\right|=\sqrt{2}$, hence
$$
\left(\begin{array}{l}
x^{2}+y^{2}+d^{2}-2 x y \cos \theta=\left|A_{1} E\right|^{2}=37, \\
(\sqrt{2}-x)^{2}+(2-y)^{2}+d^{2}-2(\sqrt{2} \\
-x)(2-y) \cos \theta=\left|B_{1} D_{1}\right|^{2}=39, \\
(\sqrt{2}-x)^{2}+y^{2}+d^{2}+2(\sqrt{2}-x) y \\
\cdot \cos \theta=\left|A_{1} B_{1}\right|^{2}=49, \\
x^{2}+(2-y)^{2}+d^{2}+2 x(2-y) \cos \theta \\
=\left|D_{1} E\right|^{2}=29
\end{array}\right.
$$

Solving, we get
$$
\left\{\begin{array}{l}
x=-2 \sqrt{2}, \\
y=2, \\
\cos \theta=\sqrt{2} \\
d=\sqrt{21} .
\end{array}\right.
$$

Finally, there are four points to note:
(1) The solution must be correctly interpreted. For example, in Example 3, point $F$ actually coincides with $D_{1}$. From $x=-2 \sqrt{2}$, we know that point $G$ is actually on the extension of $B_{1} E$,
$$
|E G|=2 \sqrt{2} \text {. }
$$
(2) The general order for solving the system of equations is: eliminate $d^{2}$, $\cos \theta$, and then solve for $x, y$.
(3) Sometimes, the initially obtained $x, y$ may have more than one solution. In such cases, some solutions should be eliminated based on the condition $|\cos \theta|>0$ (not $d^{2}>0$).
(4) The system of equations must have a unique solution, because the problem essentially involves finding the angle and distance between two opposite edges in a tetrahedron with six known edge lengths, and the position of the common perpendicular, which of course is uniquely determined.