Cho \(\Delta ABC,\)  gọi \({A_1},\,\,{B_1},\,\,{C_1}\)  lần lượt là trung điểm của \(BC,\,\,CA,\,\,AB.\)  1.Chứng minh rằng \(\overrightarrow {A{A_1}}  + \overrigh...

A \(\begin{array}{*{20}{l}}
{\overrightarrow {BC} = \frac{2}{3}\vec u - \frac{2}{3}\vec v}\\
{\overrightarrow {AB} = - \frac{4}{3}\vec u - \frac{2}{3}\vec v}\\
{\overrightarrow {CA} = \frac{2}{3}\vec u + \frac{4}{3}\vec v}
\end{array}\)

B \(\begin{array}{*{20}{l}}
{\overrightarrow {BC} = \frac{2}{3}\vec u - \frac{2}{3}\vec v}\\
{\overrightarrow {AB} =  \frac{4}{3}\vec u + \frac{2}{3}\vec v}\\
{\overrightarrow {CA} = \frac{2}{3}\vec u + \frac{4}{3}\vec v}
\end{array}\)

C \(\begin{array}{l}
\overrightarrow {BC} = \frac{2}{3}\overrightarrow u + \frac{2}{3}\overrightarrow v \\
\overrightarrow {AB} = \frac{4}{3}\overrightarrow u - \frac{2}{3}\overrightarrow v \\
\overrightarrow {CA} = \frac{2}{3}\overrightarrow u - \frac{4}{3}\overrightarrow v
\end{array}\)

D \(\begin{array}{l}
\overrightarrow {BC} = - \frac{2}{3}\overrightarrow u - \frac{2}{3}\overrightarrow v \\
\overrightarrow {AB} = \frac{4}{3}\overrightarrow u + \frac{2}{3}\overrightarrow v \\
\overrightarrow {CA} = 2\overrightarrow u + \frac{4}{3}\overrightarrow v
\end{array}\)