想必現(xiàn)在有很多小伙伴對于在平面直角坐標系$xOy$中,已知四邊形$ABCD$是平行四邊形,$\overrightarrow{AB}=\left ( {1,-2} \right ),\overrightarrow{AD}=\left ( {2,1} \right )$,則$\overrightarrow{AC}\cdot \overrightarrow{BD}=$$\left ( {\, \, \, \, \, } \right )$$A、$$-2$$B、$$-1$$C、$$0$$D、$$1$","title_text":"在平面直角坐標系$xOy$中,已知四邊形$ABCD$是平行四邊形,$\overrightarrow{AB}=\left ( {1,-2} \right ),\overrightarrow{AD}=\left ( {2,1} \right )$,則$\overrightarrow{AC}\cdot \overrightarrow{BD}=$$\left ( {\, \, \, \, \, } \right )$$A、$$-2$$B、$$-1$$C、$$0$$D、$$1$方面的知識都比較想要了解,那么今天小好小編就為大家收集了一些關于在平面直角坐標系$xOy$中,已知四邊形$ABCD$是平行四邊形,$\overrightarrow{AB}=\left ( {1,-2} \right ),\overrightarrow{AD}=\left ( {2,1} \right )$,則$\overrightarrow{AC}\cdot \overrightarrow{BD}=$$\left ( {\, \, \, \, \, } \right )$$A、$$-2$$B、$$-1$$C、$$0$$D、$$1$","title_text":"在平面直角坐標系$xOy$中,已知四邊形$ABCD$是平行四邊形,$\overrightarrow{AB}=\left ( {1,-2} \right ),\overrightarrow{AD}=\left ( {2,1} \right )$,則$\overrightarrow{AC}\cdot \overrightarrow{BD}=$$\left ( {\, \, \, \, \, } \right )$$A、$$-2$$B、$$-1$$C、$$0$$D、$$1$方面的知識分享給大家,希望大家會喜歡哦。
依題意可得:$overrightarrow{AC}=overrightarrow{AB}+overrightarrow{AD}$,$overrightarrow{BD}=overrightarrow{AD}-overrightarrow{AB}$
因為$overrightarrow{AB}=left ( {1,-2} right )$,$overrightarrow{AD}=left ( {2,1} right )$
所以$left | {overrightarrow{AB}} right |=sqrt {5}$,$left | {overrightarrow{AD}} right |=sqrt {5}$
所以$overrightarrow{AC}cdot overrightarrow{BD}=left ( {overrightarrow{AB}+overrightarrow{AD}} right )left ( {overrightarrow{AD}-overrightarrow{AB}} right )=left | {overrightarrow{AD}} right |^{2}-left | {overrightarrow{AB}} right |^{2}=5-5=0$
綜上所述,答案選擇:$C$
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