（5分）设 a、 b、 c 是单位向量，且 a · b =…——2009 高考数学第 6 题答案解析

【考点】90：平面向量数量积的性质及其运算．
【专题】16：压轴题．
【分析】由题意可得 $|\vec{a}+\vec{b}|=\sqrt{2}$ ，故要求的式子即 $\vec{a} \cdot \vec{b}-(\vec{a}+\vec{b}) \cdot \vec{c}+\vec{c}^{2}=1- |\vec{a}+\vec{b}| \cdot|\vec{c}|$
$\cos <\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}, ~ \overrightarrow{\mathrm{c}}>=1-\sqrt{2} \cos <\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}, ~ \overrightarrow{\mathrm{c}}>$ ，再由余弦函数的值域求出它的最小值．

【解答】解：$\because \vec{a} , \vec{b} , \vec{c}$ 是单位向量，$\vec{a} \cdot \vec{b}=0, \therefore \vec{a} \perp \vec{b},|\vec{a}+\vec{b}|=\sqrt{2}$ ．
$\therefore(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}^{2}=0-(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}+1=1-|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}| \cdot|\overrightarrow{\mathrm{c}}|$

$$ \begin{aligned} & \cos <\vec{a}+\vec{b}, \quad \vec{c}> \\ = & 1-\sqrt{2} \cos <\vec{a}+\vec{b} \quad, \quad \vec{c}>\geq 1-\sqrt{2} . \end{aligned} $$

故选：D．
【点评】考查向量的运算法则；交换律、分配律但注意不满足结合律．