9. (16 points) In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ is the midpoint of $A B$. Find the distance between the skew lines $D E$ and $B_{1} C$.

In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ is the midpoint of $A B$. Find the distance between the skew lines $D E$ and $B_{1} C$.
9. Solution 1 As shown in Figure 3, connect $A_{1} C$, $A_{1} D$, and $A_{1} E$.
Since $B_{1} C / / A_{1} D$, we know
$B_{1} C / /$ plane $A_{1} D E$.
Let the distance between the skew lines $D E$ and $B_{1} C$ be $d$. Then the distance from point $C$ to plane $A_{1} D E$ is $d$.
Notice that
$$
\begin{array}{l}
S_{\triangle C D E}=\frac{1}{2} S_{\text {quadrilateral } A B C D}=\frac{1}{2}, \\
S_{\triangle A_{1} D E}=\frac{1}{2} A_{1} D \sqrt{D E^{2}-\left(\frac{A_{1} D}{2}\right)^{2}} \\
=\frac{1}{2} \times \sqrt{2} \times \sqrt{\frac{5}{4}-\frac{2}{4}}=\frac{\sqrt{6}}{4} .
\end{array}
$$

By $V_{C-A_{1} D E}=V_{A_{1}-C D E}$, we have
$$
\frac{1}{3} \times \frac{\sqrt{6}}{4} \times d=\frac{1}{3} \times \frac{1}{2} \times 1 \Rightarrow d=\frac{\sqrt{6}}{3} .
$$

Therefore, the distance between the skew lines $D E$ and $B_{1} C$ is $\frac{\sqrt{6}}{3}$.
Solution 2 Establish a spatial rectangular coordinate system with $D_{1}$ as the origin and $D_{1} A_{1}$, $D_{1} C$, and $D_{1} D$ as the coordinate axes. We have
$$
D(0,0,1), E\left(1,-\frac{1}{2}, 1\right), B_{1}(1,1,0), C(0,1,1) \text {, }
$$

Thus, $\overrightarrow{E D}=\left(-1, \frac{1}{2}, 0\right), \overrightarrow{B_{1} C}=(-1,0,1)$,
$$
\overrightarrow{C D}=(0,-1,0) \text {. }
$$

Let $n=(x, y, z)$ be perpendicular to both $\overrightarrow{E D}$ and $\overrightarrow{B_{1} C}$, then
$$
\left\{\begin{array}{l}
n \cdot \overrightarrow{E D}=-x+\frac{1}{2} y=0, \\
n \cdot \overrightarrow{B C_{1}}=-x+z=0 .
\end{array}\right.
$$

Taking $z=1$, we get $n=(1,2,1)$.
Then the distance from point $C$ to plane $A_{1} D E$ is
$$
\begin{array}{l}
d=\frac{|\overrightarrow{C D} \cdot n|}{|n|}=\frac{|1 \times 0+2 \times(-1)+1 \times 0|}{\sqrt{1^{2}+2^{2}+1^{2}}} \\
=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3} .
\end{array}
$$