想必现在有很多小伙伴对于在平面直角坐标系$xOy$中,已知四边形$ABCD$是平行四边形,$\overrightarrow{AB}=\left ( {1,-2} \right ),\overrightarrow{AD}=\left ( {2,1} \right )$,则$\overrightarrow{AC}\cdot \overrightarrow{BD}=$$\left ( {\, \, \, \, \, } \right )$$A、$$-2$$B、$$-1$$C、$$0$$D、$$1$","title_text":"在平面直角坐标系$xOy$中,已知四边形$ABCD$是平行四边形,$\overrightarrow{AB}=\left ( {1,-2} \right ),\overrightarrow{AD}=\left ( {2,1} \right )$,则$\overrightarrow{AC}\cdot \overrightarrow{BD}=$$\left ( {\, \, \, \, \, } \right )$$A、$$-2$$B、$$-1$$C、$$0$$D、$$1$方面的知识都比较想要了解,那么今天小好小编就为大家收集了一些关于在平面直角坐标系$xOy$中,已知四边形$ABCD$是平行四边形,$\overrightarrow{AB}=\left ( {1,-2} \right ),\overrightarrow{AD}=\left ( {2,1} \right )$,则$\overrightarrow{AC}\cdot \overrightarrow{BD}=$$\left ( {\, \, \, \, \, } \right )$$A、$$-2$$B、$$-1$$C、$$0$$D、$$1$","title_text":"在平面直角坐标系$xOy$中,已知四边形$ABCD$是平行四边形,$\overrightarrow{AB}=\left ( {1,-2} \right ),\overrightarrow{AD}=\left ( {2,1} \right )$,则$\overrightarrow{AC}\cdot \overrightarrow{BD}=$$\left ( {\, \, \, \, \, } \right )$$A、$$-2$$B、$$-1$$C、$$0$$D、$$1$方面的知识分享给大家,希望大家会喜欢哦。
依题意可得:$overrightarrow{AC}=overrightarrow{AB}+overrightarrow{AD}$,$overrightarrow{BD}=overrightarrow{AD}-overrightarrow{AB}$
因为$overrightarrow{AB}=left ( {1,-2} right )$,$overrightarrow{AD}=left ( {2,1} right )$
所以$left | {overrightarrow{AB}} right |=sqrt {5}$,$left | {overrightarrow{AD}} right |=sqrt {5}$
所以$overrightarrow{AC}cdot overrightarrow{BD}=left ( {overrightarrow{AB}+overrightarrow{AD}} right )left ( {overrightarrow{AD}-overrightarrow{AB}} right )=left | {overrightarrow{AD}} right |^{2}-left | {overrightarrow{AB}} right |^{2}=5-5=0$
综上所述,答案选择:$C$
本文到此结束,希望对大家有所帮助。