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2021年高考数学新高考Ⅰ-10

  2021-06-14 21:41:55  

(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.

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