From the concept of the dot product of vectors, it is necessary to find the length of $\overrightarrow{C N}$ from the condition $\overrightarrow{M N} \cdot \overrightarrow{A B_{1}}=0$, but neither $\overrightarrow{M N}$ nor $\overrightarrow{A B_{1}}$ are known vectors, and they have no direct connection with $\overrightarrow{C N}$. Therefore, a set of basic vectors must be chosen to represent $\overrightarrow{M N}$ and $\overrightarrow{A B_{1}}$.

Take unit vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$ along the directions of $A B$, $A C$, and $A A_{1}$, respectively. Let $\overrightarrow{C N}=x \vec{k}$, then $\overrightarrow{A B_{1}}=\vec{i}+2 \vec{k}$, $\overrightarrow{M N}=\frac{1}{2}(\vec{j}-\vec{i})+x \vec{k}$.
From $\overrightarrow{A B_{1}} \cdot \overrightarrow{M N}=(\vec{i}+2 \vec{k})\left(-\frac{1}{2} \vec{i}+\frac{1}{2} \vec{j}+x \vec{k}\right)=0$,
we solve to get $x=\frac{1}{8}$.
Thus, $|\overrightarrow{C N}|=\frac{1}{8}$ when $M N \perp A B_{1}$.
Example 7 As shown in Figure 8-4, the side edge length of the regular triangular prism $A B C-A_{1} B_{1} C_{1}$ is 2, and the base edge length is $1, M$ is the midpoint of $B C$, find a point $N$ on the line $O C_{1}$ such that $M N \perp A B_{1}$.