Cho tứ diện ABCD có G là trọng tâm tam giác BCD. Đặt \(\overrightarrow x = \overrightarrow {AB} ,\,\overrightarrow {\,y} = \overrightarrow {AC} ,\,\,\overrightarrow z = \overrightarrow {AD} \). Khẳng định nào sau đây là đúng?
\(\overrightarrow {AG} = \frac{1}{3}\left( {\overrightarrow x + \overrightarrow y + \overrightarrow z } \right)\).\(\overrightarrow{AG}=-\dfrac{1}{3}\left(\overrightarrow{x}+\overrightarrow{y}+\overrightarrow{z}\right)\).\(\overrightarrow{AG}=\dfrac{2}{3}\left(\overrightarrow{x}+\overrightarrow{y}+\overrightarrow{z}\right)\).\(\overrightarrow{AG}=-\dfrac{2}{3}\left(\overrightarrow{x}+\overrightarrow{y}+\overrightarrow{z}\right)\).Hướng dẫn giải:
Ta có: \(\overrightarrow{AG}=\overrightarrow{AM}+\overrightarrow{MG}=\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{2}+\dfrac{\overrightarrow{MD}}{3}\)
\(=\dfrac{\overrightarrow{x}+\overrightarrow{y}}{2}+\dfrac{\overrightarrow{MA}+\overrightarrow{AD}}{3}=\dfrac{\overrightarrow{x}+\overrightarrow{y}}{2}+\dfrac{-\dfrac{\overrightarrow{x}+\overrightarrow{y}}{2}+\overrightarrow{z}}{3}\)
\(=\dfrac{\overrightarrow{x}+\overrightarrow{y}+\overrightarrow{z}}{3}.\)
