Example 2 As shown in Figure 2, $AC$ and $CE$ are two diagonals of the regular hexagon $ABCDEF$. Points $M$ and $N$ internally divide $AC$ and $CE$ such that $AM$ : $AC = CN : CE = r$. If points $B$, $M$, and $N$ are collinear, find $r$.
(23rd IMO)
Solution: Let $\overrightarrow{A C}=2 \vec{a}, \overrightarrow{A F}=2 \vec{b}$, then
$$
\begin{array}{l}
\overrightarrow{A M}=r \overrightarrow{A C}=2 r \vec{a}, \overrightarrow{A B}=\vec{a}-\vec{b} \text {. } \\
\text { Hence } \overrightarrow{B M}=\overrightarrow{B A}+\overrightarrow{A M}=\vec{b}+(2 r-1) \cdot \vec{a} \text {. } \\
\text { Also } \because \overrightarrow{C E}=\frac{1}{2} \overrightarrow{C A}+\frac{3}{2} \overrightarrow{A F}=3 \vec{b}-\vec{a} \text {, } \\
\therefore \overrightarrow{C N}=r \overrightarrow{C E}=3 r \vec{b}-r \vec{a} \text {. } \\
\therefore \overrightarrow{M N}=\overrightarrow{M C}+\overrightarrow{C N} \\
=2(\mathbf{i}-r) \vec{a}+3 r \vec{b}-r \vec{a} \\
=3 \vec{b}+(2-3 r) \vec{a} \text {. } \\
\end{array}
$$
Since $B, M, N$ are collinear,
$$
\therefore \frac{1}{3 r}=\frac{2 r-1}{2-3 r} \text {. Hence } r=\frac{\sqrt{3}}{3} \text {. }
$$