Cho hình bình hành ABCD tâm O. Đặt \(\overrightarrow {AO} = \overrightarrow a ,\,\,\overrightarrow {BO} = \overrightarrow b \). Hãy biểu diễn các vectơ \[\overrightarrow {AB} \,,\,\,\overrightarrow {BC} \,,\,\,\overrightarrow {CD} \,,\,\,\overrightarrow {DA} \] theo hai vectơ \(\overrightarrow a ,\overrightarrow b \).

Vì A ∈ Ox; B ∈ Oy nên ta gọi A(x_A; 0); B(0; y_B).

Ta có M là trung điểm AB nên \[\left\{ {\begin{array}{*{20}{c}}{{x_A} + {x_B} = 2{x_M}}\\{{y_A} + {y_B} = 2{y_M}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{x_A} + 0 = 2.3}\\{0 + {y_B} = 2.2}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{x_A} = 6}\\{{y_B} = 4}\end{array}} \right.\]

Suy ra (AB): \(\frac{x}{6} + \frac{y}{4} = 1\) = 1 hay 4x + 6y – 24 = 0.

Ta có: \(\overrightarrow {AH} + \overrightarrow {AB} = 2\overrightarrow {AG} \).

Suy ra: \(\overrightarrow {AH} = - \overrightarrow {AB} + \frac{4}{3}\overrightarrow {AM} = - \overrightarrow {AB} + \frac{2}{3}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right) = - \frac{1}{3}\overrightarrow {AB} + \frac{2}{3}\overrightarrow {AC} \).

Vậy: \(\overrightarrow {AH} = - \frac{1}{3}\overrightarrow b + \frac{2}{3}\overrightarrow c \).

Tương tự:

• \[\overrightarrow {CH} = \frac{2}{3}\overrightarrow {CA} - \frac{1}{3}\overrightarrow {CB} = - \frac{2}{3}\overrightarrow {AC} - \frac{1}{3}\left( {\overrightarrow {AB} - \overrightarrow {AC} } \right)\]

\[ = - \frac{1}{3}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right) = - \frac{1}{3}\left( {\overrightarrow b + \overrightarrow c } \right)\]

• \[\overrightarrow {MH} = \overrightarrow {MC} + \overrightarrow {CH} = \frac{1}{2}\overrightarrow {BC} - \frac{1}{3}\left( {\overrightarrow b + \overrightarrow c } \right)\]

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

\[ = \frac{1}{2}\left( {\overrightarrow c - \overrightarrow b } \right) - \frac{1}{3}\left( {\overrightarrow b + \overrightarrow c } \right) = - \frac{5}{6}\overrightarrow b + \frac{1}{6}\overrightarrow c \]