Example 15 Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $1$, and $E$ is the midpoint of $D_{1} C_{1}$, find the following:
(1) Find the distance between the skew lines $D_{1} B$ and $A_{1} E$;
(2) Find the distance from $B_{1}$ to the plane $A_{1} B E$;
(3) Find the distance from $D_{1} C$ to the plane $A_{1} B E$;
(4) Find the distance between the planes $A_{1} D B$ and $D_{1} C B_{1}$.
(1) As shown in Figure $20-14$, establish a spatial rectangular coordinate system, then $\overrightarrow{A_{1} E}=(-1$, $\left.\frac{1}{2}, 0\right), \overrightarrow{D_{1} B}=(1,1,-1)$.
Let $\boldsymbol{n}=(x, y, z)$ be a vector perpendicular to both $\overrightarrow{A_{1} E}$ and $\overrightarrow{D_{1} B}$, then
$$
\left\{\begin{array} { l } 
{ n \cdot \vec { A _ { 1 } E } = 0 , } \\
{ n \cdot \vec { D _ { 1 } B } = 0 , }
\end{array} \quad \text { i.e., } \left\{\begin{array} { l } 
{ - x + \frac { 1 } { 2 } y = 0 , } \\
{ x + y - z = 0 , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
y=2 x, \\
z=3 x .
\end{array}\right.\right.\right.
$$

Assume $\boldsymbol{n}=(1,2,3)$, and select the vector between points $A_{1} E$ and $B D_{1}$ as $\overrightarrow{D_{1} A_{1}}=(1,0,0)$, then the distance $d$ between $A_{1} E$ and $B D_{1}$ is $d=\frac{\left|\overrightarrow{D_{1} A_{1}} \cdot n\right|}{|n|}=\frac{1}{\sqrt{14}}=\frac{\sqrt{14}}{14}$.
(2) Let $\boldsymbol{n}=(x, y, z)$ be the normal vector of plane $A_{1} B E$, then $\left\{\begin{array}{l}\boldsymbol{n} \cdot \overrightarrow{A_{1} E}=0, \\ \boldsymbol{n} \cdot \overrightarrow{A_{1} B}=0,\end{array}\right.$ i.e., $\left\{\begin{array}{l}-x+\frac{1}{2} y=0, \\ y-z=0,\end{array}\right.$ i.e., $\left\{\begin{array}{l}y=2 x \\ z=2 x\end{array}\right.$

Assume $n=(1,2,2)$, and select the vector between points $B_{1}$ and plane $A_{1} B E$ as $\overrightarrow{A_{1} B_{1}}=(0,1,0)$, then by the formula, the distance $d$ from point $B_{1}$ to plane $A_{1} B E$ is $d=\frac{\left|\overrightarrow{A_{1} B_{1}} \cdot \boldsymbol{n}\right|}{n}=\frac{2}{3}$.
(3) From (2), the normal vector of plane $A_{1} B E$ can be set as $\boldsymbol{n}=(1,2,2)$, and the vector between points $D_{1} C$ and plane $A_{1} B E$ as $\overrightarrow{D_{1} A_{1}}=(1,0,0)$, hence the distance $d$ from line $D_{1} C$ to plane $A_{1} B E$ is $d=\frac{\left|\overrightarrow{D_{1} A_{1}} \cdot n\right|}{|n|}=\frac{1}{3}$.
(4) Select the normal vectors of planes $A_{1} D B$ and $D_{1} C B_{1}$: $\boldsymbol{n}=\overrightarrow{A C_{1}}=(-1,1,1)$, and any vector between two points $\boldsymbol{p}=\overrightarrow{D_{1} A_{1}}=$ $(1,0,0)$, hence the distance $d$ between planes $A_{1} D B$ and $D_{1} C B_{1}$ is $d=\frac{|\boldsymbol{p} \cdot \boldsymbol{n}|}{|\boldsymbol{n}|}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$.