Question

Cho hình bình hành ABCD, A' là trọng tâm của tâm giác BCD. Biểu diễn vectơ AA' theo hai vectơ AB và AC.

Answer 1

- Đặt \(A^,=G\) ( cho dễ đánh :))

Có : \(\overrightarrow{AG}=\overrightarrow{AB}+\overrightarrow{BG}=\overrightarrow{AB}+\frac{2}{3}\overrightarrow{BH}\)

Mà \(\overrightarrow{BH}=\frac{1}{2}\left(\overrightarrow{BD}+\overrightarrow{BC}\right)\)

=> \(\overrightarrow{AG}=\overrightarrow{AB}+\frac{1}{3}\left(\overrightarrow{BD}+\overrightarrow{BC}\right)\)

\(=\overrightarrow{AB}+\frac{1}{3}\left(\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{BA}+\overrightarrow{AC}\right)\)

Lại có : Tứ giác ABCD là hình bình hành .

=> \(\overrightarrow{AD}=\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}\)

=> \(\overrightarrow{AG}=\overrightarrow{AB}+\frac{1}{3}\left(\overrightarrow{BA}+\overrightarrow{BA}+\overrightarrow{AC}+\overrightarrow{BA}+\overrightarrow{AC}\right)\)

\(=\overrightarrow{AB}+\frac{1}{3}\left(-3\overrightarrow{AB}+2\overrightarrow{AC}\right)=\overrightarrow{AB}-\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\)

\(=0\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\)

Answer 2

Ta có \(\overrightarrow{A'B}+\overrightarrow{A'C}+\overrightarrow{A'D}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{AB}-\overrightarrow{AA'}+\overrightarrow{AC}-\overrightarrow{AA'}+\overrightarrow{AD}-\overrightarrow{AA'}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}=3\overrightarrow{AA'}\)

\(\Leftrightarrow\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{BC}=3\overrightarrow{AA'}\)

\(\Leftrightarrow\overrightarrow{BA}+\overrightarrow{AC}+\overrightarrow{AB}+\overrightarrow{AC}=3\overrightarrow{AA'}\)

\(\Leftrightarrow0\overrightarrow{AB}+2\overrightarrow{AC}=\overrightarrow{AA'}\)

Vậy \(\overrightarrow{AA'}=0\overrightarrow{AB}+2\overrightarrow{AC}\)

Answer 3

nhầm nha bạn \(0\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}=\overrightarrow{AA'}\)