Question 89: If planar vectors $\vec{a} 、 \vec{b} 、 \vec{c}$ satisfy $\vec{a} \cdot(\vec{a}+\vec{c})=0,|\vec{a}+\vec{b}-2 \vec{c}|=2$, try to find the maximum value of $\vec{a} \cdot \vec{b}$.
Question 89, Answer: According to the conditions: $\vec{a} \cdot(\vec{a}+\vec{c})=\vec{a}^{2}+\vec{a} \cdot \vec{c}=0$. Let $\vec{a}+\vec{b}-2 \vec{c}=\vec{d}$, then $|\vec{d}|=2$,
$$
\begin{aligned}
\text { and } \vec{b} & =\vec{d}+2 \vec{c}-\vec{a} \text { . Therefore, we know that } \\
& \vec{a} \cdot \vec{b}=\vec{a} \cdot(\vec{d}+2 \vec{c}-\vec{a})=\vec{a} \cdot \vec{d}+2 \vec{a} \cdot \vec{c}-\vec{a}^{2}=\vec{a} \cdot \vec{d}-3 \vec{a}^{2} \\
& \leq|\vec{a}| \cdot|\vec{d}|-3|\vec{a}|^{2}=2|\vec{a}|-3|\vec{a}|^{2} \leq \frac{1}{3}
\end{aligned}
$$

Equality holds if and only if $|\vec{a}|=\frac{1}{3},|\vec{d}|=2$, and $\vec{a}$ and $\vec{d}$ are in the same direction.