Example 8 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 $B_{1} E F$.

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Solution: As shown in Figure 10, in the rectangular coordinate system, we have
$$
\begin{array}{l}
D(0,0,0), \\
B_{1}(1,1,1), \\
E\left(1, \frac{1}{2}, 0\right), \\
E\left(\frac{1}{2}, 1,0\right) .
\end{array}
$$

Thus, $B_{1} E=\left(0,-\frac{1}{2},-1\right)$,
$$
\boldsymbol{B}_{1} \boldsymbol{F}=\left(-\frac{1}{2}, 0,-1\right) \text {. }
$$

Let the normal vector of plane $B_{1} E F$ be $\boldsymbol{n}=(x, y, z)$.
Since $\boldsymbol{n} \perp B_{1} E, \boldsymbol{n} \perp B_{1} F$, 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, $D E=\left(1, \frac{1}{2}, 0\right)$, so the distance from point $D$ to plane $B_{1} E F$ is
$$
d=\frac{|n \cdot D E|}{|n|}=\frac{\left|(-2,-2,1) \cdot\left(1, \frac{1}{2}, 0\right)\right|}{3}=1 \text {. }
$$