Example 4 Given a unit cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, $E$ is a moving point on $B C$, and $F$ is the midpoint of $A B$. Determine the position of point $E$ such that $C_{1} F \perp A_{1} E$.
Prove: Connect $A_{1} D, D E$, and establish a spatial rectangular coordinate system as shown in Figure 4. Then,
$$
\begin{array}{l}
D(0,0,0), \\
A_{1}(1,0,1), \\
E(a, 1,0), \\
F\left(1, \frac{1}{2}, 0\right), \\
C_{1}(0,1,1).
\end{array}
$$

Thus, $D A_{1}=(1,0,1), D E=(a, 1,0)$.
Let the normal vector of plane $A_{1} D E$ be $\boldsymbol{n}=(x, y, z)$.
If $C_{1} F \perp$ plane $D A_{1} E$, then $\boldsymbol{n}$ is collinear with $\boldsymbol{C}_{1} \boldsymbol{F}$.
Hence, $\boldsymbol{n}=\lambda\left(\boldsymbol{C}_{1} \boldsymbol{F}\right)=\lambda\left(1,-\frac{1}{2},-1\right)$
$$
=\left(\lambda,-\frac{1}{2} \lambda,-\lambda\right).
$$

By $\boldsymbol{n} \perp \boldsymbol{D E}$, we get
$$
\left(\lambda,-\frac{1}{2} \lambda,-\lambda\right) \cdot(a, 1,0)=0.
$$

This yields $a \lambda=\frac{1}{2} \lambda$, i.e., $a=\frac{1}{2}$. This shows that $E$ is the midpoint of $B C$. At this point, $C_{1} F \perp$ plane $A_{1} D C$, then $C_{1} F \perp A_{1} E$.