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Ta có: \(\overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC} \Leftrightarrow \overrightarrow {BC} = \overrightarrow b - \overrightarrow a \)
Lại có: vecto \(\overrightarrow {BD} ,\overrightarrow {BC} \) cùng hướng và \(\left| {\overrightarrow {BD} } \right| = \frac{1}{3}\left| {\overrightarrow {BC} } \right|\)
\( \Rightarrow \overrightarrow {BD} = \frac{1}{3}\overrightarrow {BC} = \frac{1}{3}(\overrightarrow b - \overrightarrow a )\)
Tương tự: vecto \(\overrightarrow {BE} ,\overrightarrow {BC} \) cùng hướng và \(\left| {\overrightarrow {BE} } \right| = \frac{2}{3}\left| {\overrightarrow {BC} } \right|\)
\( \Rightarrow \overrightarrow {BE} = \frac{2}{3}\overrightarrow {BC} = \frac{2}{3}(\overrightarrow b - \overrightarrow a )\)
Ta có:
\(\overrightarrow {AB} + \overrightarrow {BD} = \overrightarrow {AD} \Leftrightarrow \overrightarrow {AD} = \overrightarrow a + \frac{1}{3}(\overrightarrow b - \overrightarrow a ) = \frac{2}{3}\overrightarrow a + \frac{1}{3}\overrightarrow b \)
\(\overrightarrow {AB} + \overrightarrow {BE} = \overrightarrow {AE} \Leftrightarrow \overrightarrow {AE} = \overrightarrow a + \frac{2}{3}(\overrightarrow b - \overrightarrow a ) = \frac{1}{3}\overrightarrow a + \frac{2}{3}\overrightarrow b \)
\(\text{a) Ta có : }2\overrightarrow{CD}=3\overrightarrow{DB}\\ \Rightarrow\overrightarrow{DC}=-\frac{3}{2}\overrightarrow{DB}\\ \Rightarrow D;B;C\text{ thẳng hàng },D\text{ nằm giữa }B;C\left(\frac{3}{2}< 0\right)\\ \Rightarrow\overrightarrow{BC}=\overrightarrow{BD}+\overrightarrow{DC}=\overrightarrow{BD}+\frac{3}{2}\overrightarrow{BD}=\frac{5}{2}\overrightarrow{BD}\\ 5\overrightarrow{EB}=2\overrightarrow{EC}\\ \Rightarrow\overrightarrow{EB}=\frac{2}{5}\overrightarrow{EC}\\ \Rightarrow E;B;C\text{ thẳng hàng },B\text{ nằm giữa }E;C\left(\frac{2}{5}>0;EB< EC\right)\\ \Rightarrow\overrightarrow{BC}=\overrightarrow{EC}-\overrightarrow{EB}=\overrightarrow{EC}-\frac{2}{5}\overrightarrow{EC}=\frac{3}{5}\overrightarrow{EC}\)
\(\Rightarrow\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}\\ =\overrightarrow{AB}+\frac{2}{5}\overrightarrow{BC}=\overrightarrow{AB}+\frac{2}{5}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\\ =\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}-\frac{2}{5}\overrightarrow{AB}=\frac{3}{5}\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\)
\(\overrightarrow{AE}=\overrightarrow{EC}+\overrightarrow{CA}\\ =\frac{5}{3}\overrightarrow{BC}-\overrightarrow{AC} =\frac{5}{3}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)-\overrightarrow{AC}\\ =\frac{5}{3}\overrightarrow{AC}-\frac{5}{3}\overrightarrow{AB}-\overrightarrow{AC}=\frac{2}{3}\overrightarrow{AC}-\frac{5}{3}\overrightarrow{AB}\)
\(b\text{) Theo tính chất trọng tâm }\Delta:3\overrightarrow{AG}=\overrightarrow{AA}+\overrightarrow{AB}+\overrightarrow{AC}\\ =\overrightarrow{0}+\overrightarrow{AB}+\overrightarrow{AC}\\ =\left(\frac{9}{4}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}\right)-\left(\frac{1}{2}\overrightarrow{AC}+\frac{5}{4}\overrightarrow{AC}\right)\\ =\frac{15}{4}\left(\frac{3}{5}\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\right)-\frac{3}{4}\left(\frac{2}{3}\overrightarrow{AC}+\frac{5}{3}\overrightarrow{AC}\right)\\ =\frac{15}{4}\overrightarrow{AD}-\frac{3}{4}\overrightarrow{AE}\)
\(\Rightarrow\overrightarrow{AG}=\frac{5}{4}\overrightarrow{AD}-\frac{1}{4}\overrightarrow{AE}\)