（5分）已知$O$为坐标原点，点$P_{1}(\cos \alpha ,\sin \alpha )$，$P_{2}(\cos \beta ,-\sin \beta )$，$P_{3}(\cos (\alpha +\beta )$，$\sin (\alpha +\beta ))$，$A(1,0)$，则(　　)
A．$\vert \overrightarrow{O{P}_{1}}\vert =\vert \overrightarrow{O{P}_{2}}\vert$              
B．$\vert \overrightarrow{A{P}_{1}}\vert =\vert \overrightarrow{A{P}_{2}}\vert$              
C．$\overrightarrow{OA}\cdot \overrightarrow{O{P}_{3}}=\overrightarrow{O{P}_{1}}\cdot \overrightarrow{O{P}_{2}}$              
D．$\overrightarrow{OA}\cdot \overrightarrow{O{P}_{1}}=\overrightarrow{O{P}_{2}}\cdot \overrightarrow{O{P}_{3}}$
分析：由已知点的坐标分别求得对应向量的坐标，然后逐一验证四个选项得答案．
解：$\because P_{1}(\cos \alpha ,\sin \alpha )$，$P_{2}(\cos \beta ,-\sin \beta )$，$P_{3}(\cos (\alpha +\beta )$，$\sin (\alpha +\beta ))$，$A(1,0)$，
$\therefore$$\overrightarrow{O{P}_{1}}=(\cos \alpha ,\sin \alpha )$，$\overrightarrow{O{P}_{2}}=(\cos \beta ,-\sin \beta )$，
$\overrightarrow{O{P}_{3}}=(\cos (\alpha +\beta )$，$\sin (\alpha +\beta ))$，$\overrightarrow{OA}=(1,0)$，
$\overrightarrow{A{P}_{1}}=(\cos \alpha -1,\sin \alpha )$，$\overrightarrow{A{P}_{2}}=(\cos \beta -1,-\sin \beta )$，
则$\vert \overrightarrow{O{P}_{1}}\vert =\sqrt{co{s}^{2}\alpha +si{n}^{2}\alpha }=1$，$\vert \overrightarrow{O{P}_{2}}\vert =\sqrt{co{s}^{2}\beta +(-\sin \beta )^{2}}=1$，则$\vert \overrightarrow{O{P}_{1}}\vert =\vert \overrightarrow{O{P}_{2}}\vert$，故$A$正确；
$\vert \overrightarrow{A{P}_{1}}\vert =\sqrt{(\cos \alpha -1)^{2}+si{n}^{2}\alpha }=\sqrt{co{s}^{2}\alpha +si{n}^{2}\alpha -2\cos \alpha +1}=\sqrt{2-2\cos \alpha }$，
$\vert \overrightarrow{A{P}_{2}}\vert =\sqrt{(\cos \beta -1)^{2}+(-\sin \beta )^{2}}=\sqrt{co{s}^{2}\beta +si{n}^{2}\beta -2\cos \beta +1}=\sqrt{2-2\cos \beta }$，
$\vert \overrightarrow{A{P}_{1}}\vert \ne \vert \overrightarrow{A{P}_{2}}\vert$，故$B$错误；
$\overrightarrow{OA}\cdot \overrightarrow{O{P}_{3}}=1\times \cos (\alpha +\beta )+0\times \sin (\alpha +\beta )=\cos (\alpha +\beta )$，
$\overline{O{P}_{1}}\cdot \overrightarrow{O{P}_{2}}=\cos \alpha \cos \beta -\sin \alpha \sin \beta =\cos (\alpha +\beta )$，
$\therefore$$\overrightarrow{OA}\cdot \overrightarrow{O{P}_{3}}=\overrightarrow{O{P}_{1}}\cdot \overrightarrow{O{P}_{2}}$，故$C$正确；
$\overrightarrow{OA}\cdot \overrightarrow{O{P}_{1}}=1\times \cos \alpha +0\times \sin \alpha =\cos \alpha$，
$\overrightarrow{O{P}_{2}}\cdot \overrightarrow{O{P}_{3}}=\cos \beta \cos (\alpha +\beta )-\sin \beta \sin (\alpha +\beta )=\cos [\beta +(\alpha +\beta )]=\cos (\alpha +2\beta )$，
$\therefore$$\overrightarrow{OA}\cdot \overrightarrow{O{P}_{1}}\ne \overrightarrow{O{P}_{2}}\cdot \overrightarrow{O{P}_{3}}$，故$D$错误．
故选：$AC$．
点评：本题考查平面向量数量积的性质及运算，考查同角三角函数基本关系式及两角和的三角函数，考查运算求解能力，是中档题．