Example 13 In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $B C$ respectively. Find the distance from point $D$ to the plane $B_{1} E F$.
(2002 Hunan Province High School Competition Question)
Solve: Establish a spatial rectangular coordinate system as shown in Figure 20-12, then $D(0,0,0) 、 B_{1}(1,1,1) 、 E\left(1, \frac{1}{2}, 0\right) 、 F\left(\frac{1}{2}, 1,0\right)$.
Thus, $\overrightarrow{B_{1} E}=\left(0,-\frac{1}{2},-1\right), \overrightarrow{B_{1} F}=\left(-\frac{1}{2}, 0,-1\right)$.
Let the normal vector of plane $B_{1} E F$ be $\boldsymbol{n}=(x, y, z)$.
Since $\boldsymbol{n} \perp \overrightarrow{B_{1} E}, \boldsymbol{n} \perp \overrightarrow{B_{1} F}$, we get $-\frac{1}{2} y-z=0,-\frac{1}{2} x-z=0$.
Let $z=1$, then $x=-2, y=-2$. Therefore, $\boldsymbol{n}=(-2,-2,1)$. Also, $\overrightarrow{D E}=\left(1, \frac{1}{2}, 0\right)$, so the distance from point $D$ to plane $B_{1} E F$ is
$$
d=\frac{|n \cdot \overrightarrow{D E}|}{|n|}=\frac{\left|(-2,-2,1) \cdot\left(1, \frac{1}{2}, 0\right)\right|}{3}=1 .
$$