Cho tam giác ABC, gọi A1, B1, C1 lần lượt là trung điểm của các cạnh BC, CA, AB. Đặt \(\overrightarrow {B{B_1}} = \overrightarrow u ;\,\,\overrightarrow {C{C_1}} = \overrightarrow v \). Tính \(\overrightarrow {BC} ;\overrightarrow {CA} ;\,\;\overrightarrow {AB} \) theo \(\overrightarrow u ;\overrightarrow v \).
\[\overrightarrow {BC} = \overrightarrow {B{B_1}} + \overrightarrow {{B_1}{C_1}} + \overrightarrow {{C_1}C} = \overrightarrow u - \frac{1}{2}\overrightarrow {BC} - \overrightarrow v \]
Suy ra: \[\overrightarrow {BC} = \frac{2}{3}\overrightarrow u - \frac{2}{3}\overrightarrow v \]
Ta có: \[\overrightarrow {A{A_1}} + \overrightarrow {B{B_1}} + \overrightarrow {C{C_1}} = \frac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {BA} + \overrightarrow {BC} + \overrightarrow {CA} + \overrightarrow {CB} } \right) = \frac{1}{2}\left( {\overrightarrow 0 + \overrightarrow 0 + \overrightarrow 0 } \right) = 0\]
Suy ra: \[\overrightarrow {A{A_1}} = - \overrightarrow u - \overrightarrow v \]
\[\overrightarrow {CA} = \overrightarrow {C{A_1}} + \overrightarrow {{A_1}A} = \frac{{ - 1}}{2}BC - \overrightarrow {A{A_1}} = \frac{{ - 1}}{2}\left( {\frac{2}{3}\overrightarrow u - \frac{2}{3}\overrightarrow v } \right) + \overrightarrow u + \overrightarrow v = \frac{2}{3}\overrightarrow u + \frac{4}{3}\overrightarrow v \]
\[\overrightarrow {AB} = \overrightarrow {CB} - \overrightarrow {CA} = - \left( {\frac{2}{3}\overrightarrow u - \frac{2}{3}\overrightarrow v } \right) - \left( {\frac{2}{3}\overrightarrow u + \frac{4}{3}\overrightarrow v } \right) = \frac{{ - 4}}{3}\overrightarrow u - \frac{2}{3}\overrightarrow v \].