Problem 90: Planar vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy: $|\vec{a}|=1, \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=1,|\vec{a}-\vec{b}+\vec{c}| \leq 2 \sqrt{2}$, then the range of values for $\vec{a} \cdot$ $\overrightarrow{\mathrm{c}}$ is $\qquad$ _.
Question 90, Answer: Let $\vec{a}=(1,0), \vec{b}=(1, x), \vec{c}=(1-x y, y)$, then $\vec{a}-\vec{b}+\vec{c}=(1-$ $\mathrm{xy}, \mathrm{y}-\mathrm{x})$, and $\vec{a} \cdot \vec{c}=1-\mathrm{xy}$. Therefore, according to the condition:
$$
8 \geq|\vec{a}-\vec{b}+\vec{c}|^{2}=(1-x y)^{2}+(y-x)^{2}=1+x^{2}+y^{2}-4 x y+x^{2} y^{2}
$$

According to (*) we know: $8 \geq 1+2 x y-4 x y+x^{2} y^{2}=(x y-1)^{2} \Rightarrow x y \leq 2 \sqrt{2}+1$, hence $\mathrm{a} \cdot \vec{c}=1-$ $\mathrm{xy} \geq-2 \sqrt{2}$.

Also, according to (*) we know: $8 \geq 1-2 x y-4 x y+x^{2} y^{2}=(x y-3)^{2}-8 \Rightarrow x y \geq-1$, hence $a \cdot c=1-$ $\mathrm{xy} \leq 2$.
Therefore, the range of $\vec{a} \cdot \vec{c}$ is $[-2 \sqrt{2}, 2]$.