Example 4 As shown in Figure $20-4$, in a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with an edge length of 1, $E$ and $F$ are the midpoints of $A_{1} B_{1}$ and $C D$ respectively. Find the distance from point $B$ to the plane $A E C_{1} F$.
Solve: Taking $D$ as the origin, establish a spatial rectangular coordinate system as shown in the figure, then $A(1, 0,0), F\left(0, \frac{1}{2}, 0\right), E\left(1, \frac{1}{2}, 1\right)$,
so $\overrightarrow{A E}=\left(0, \frac{1}{2}, 1\right), \overrightarrow{A F}=\left(-1, \frac{1}{2}, 0\right)$.
Let the normal vector of plane $A E C_{1} F$ be $\boldsymbol{n}=(1, \lambda, \mu)$,
then we have: $\boldsymbol{n} \cdot \overrightarrow{A E}=0, \boldsymbol{n} \cdot \overrightarrow{A F}=0$,
so $\left\{\begin{array}{l}\frac{1}{2} \lambda+\mu=0 \\ -1+\frac{1}{2} \lambda=0\end{array}\right.$, thus $\left\{\begin{array}{l}\lambda=2 \\ \mu=-1\end{array}\right.$, so $\boldsymbol{n}=(1,2,-1)$. Also, $\overrightarrow{A B}=(0,1,0)$, so the distance from point $B$ to the section $A E C_{1} F$ is $\frac{\overrightarrow{A B} \cdot \boldsymbol{n}}{|\overrightarrow{A B}||\boldsymbol{n}|}=\frac{2}{1 \times \sqrt{6}}=\frac{\sqrt{6}}{3}$.