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A B C E D G
\(\text{a) Ta có : }2\overrightarrow{CD}=3\overrightarrow{DB}\\ \Rightarrow\overrightarrow{DC}=-\frac{3}{2}\overrightarrow{DB}\\ \Rightarrow D;B;C\text{ thẳng hàng },D\text{ nằm giữa }B;C\left(\frac{3}{2}< 0\right)\\ \Rightarrow\overrightarrow{BC}=\overrightarrow{BD}+\overrightarrow{DC}=\overrightarrow{BD}+\frac{3}{2}\overrightarrow{BD}=\frac{5}{2}\overrightarrow{BD}\\ 5\overrightarrow{EB}=2\overrightarrow{EC}\\ \Rightarrow\overrightarrow{EB}=\frac{2}{5}\overrightarrow{EC}\\ \Rightarrow E;B;C\text{ thẳng hàng },B\text{ nằm giữa }E;C\left(\frac{2}{5}>0;EB< EC\right)\\ \Rightarrow\overrightarrow{BC}=\overrightarrow{EC}-\overrightarrow{EB}=\overrightarrow{EC}-\frac{2}{5}\overrightarrow{EC}=\frac{3}{5}\overrightarrow{EC}\)
\(\Rightarrow\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}\\ =\overrightarrow{AB}+\frac{2}{5}\overrightarrow{BC}=\overrightarrow{AB}+\frac{2}{5}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\\ =\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}-\frac{2}{5}\overrightarrow{AB}=\frac{3}{5}\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\)
\(\overrightarrow{AE}=\overrightarrow{EC}+\overrightarrow{CA}\\ =\frac{5}{3}\overrightarrow{BC}-\overrightarrow{AC} =\frac{5}{3}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)-\overrightarrow{AC}\\ =\frac{5}{3}\overrightarrow{AC}-\frac{5}{3}\overrightarrow{AB}-\overrightarrow{AC}=\frac{2}{3}\overrightarrow{AC}-\frac{5}{3}\overrightarrow{AB}\)
\(b\text{) Theo tính chất trọng tâm }\Delta:3\overrightarrow{AG}=\overrightarrow{AA}+\overrightarrow{AB}+\overrightarrow{AC}\\ =\overrightarrow{0}+\overrightarrow{AB}+\overrightarrow{AC}\\ =\left(\frac{9}{4}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}\right)-\left(\frac{1}{2}\overrightarrow{AC}+\frac{5}{4}\overrightarrow{AC}\right)\\ =\frac{15}{4}\left(\frac{3}{5}\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\right)-\frac{3}{4}\left(\frac{2}{3}\overrightarrow{AC}+\frac{5}{3}\overrightarrow{AC}\right)\\ =\frac{15}{4}\overrightarrow{AD}-\frac{3}{4}\overrightarrow{AE}\)
\(\Rightarrow\overrightarrow{AG}=\frac{5}{4}\overrightarrow{AD}-\frac{1}{4}\overrightarrow{AE}\)
Lời giải:
a) Vì $M$ là trung điểm của $EF$ nên \(\overrightarrow {ME}+\overrightarrow{MF}=0\), tương tự \(\overrightarrow{NB}+\overrightarrow{NC}=0\)
Từ đkđb ta cũng có \(AE=\frac{1}{3}AB;AF=\frac{3}{5}AC\)
Ý 1:
\(\left\{\begin{matrix} \overrightarrow{AM}=\overrightarrow{AE}+\overrightarrow{EM}\\ \overrightarrow{AM}=\overrightarrow{AF}+\overrightarrow{FM}\end{matrix}\right. \)
\(\Rightarrow 2\overrightarrow{AM}=\overrightarrow{AE}+\overrightarrow{AF}-(\overrightarrow{ME}+\overrightarrow{MF})=\overrightarrow{AE}+\overrightarrow{AF}\)
\(=\frac{1}{3}\overrightarrow{AB}+\frac{3}{5}\overrightarrow{AC}\)\(\Leftrightarrow \overrightarrow{AM}=\frac{1}{6}\overrightarrow{AB}+\frac{3}{10}\overrightarrow{AC}\)
Ý 2:
\(\left\{\begin{matrix} \overrightarrow{MN}=\overrightarrow{ME}+\overrightarrow{EB}+\overrightarrow{BN}\\ \overrightarrow{MN}=\overrightarrow{MF}+\overrightarrow{FC}+\overrightarrow{CN}\end{matrix}\right.\Rightarrow 2\overrightarrow{MN}=(\overrightarrow{ME}+\overrightarrow{MF})+\overrightarrow{EB}+\overrightarrow{FC}-(\overrightarrow{NB}+\overrightarrow{NC})\)
\(\Leftrightarrow 2\overrightarrow{MN}=\overrightarrow{EB}+\overrightarrow{FC}=\frac{2}{3}\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\)
\(\Leftrightarrow \overrightarrow{MN}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{5}\overrightarrow{AC}\)
b)
Theo đkđb ta có: \(\overrightarrow{BG}=3\overrightarrow{CG}\)
\(\left\{\begin{matrix} \overrightarrow{AG}=\overrightarrow{AB}+\overrightarrow{BG}\\ \overrightarrow{AG}=\overrightarrow{AC}+\overrightarrow{CG}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \overrightarrow{AG}=\overrightarrow{AB}+\overrightarrow{BG}\\ 3\overrightarrow{AG}=3\overrightarrow{AC}+3\overrightarrow{CG}\end{matrix}\right.\)
\(\Rightarrow 2\overrightarrow{AG}=3\overrightarrow{AC}-\overrightarrow{AB}\Rightarrow \overrightarrow{AG}=\frac{3}{2}\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AB}\)
Lại có:
\(\overrightarrow{EG}=\overrightarrow{EA}+\overrightarrow{AG}=\frac{-1}{3}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AB}=\frac{3}{2}\overrightarrow{AC}-\frac{5}{6}\overrightarrow{AB}\)
\(\overrightarrow{FG}=\overrightarrow{FA}+\overrightarrow{AG}=\frac{-3}{5}\overrightarrow{AC}+\frac{3}{2}\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AB}=\frac{9}{10}\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AB}\)
c) Từ phần b ta thấy \(\frac{3}{5}\overrightarrow{EG}=\overrightarrow{FG}\Rightarrow E,G,F\) thẳng hàng.
\(\overrightarrow{DE}=\overrightarrow{DA}+\overrightarrow{AE}=-2\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\)
\(\overrightarrow{DG}=\overrightarrow{DA}+\overrightarrow{AG}=-2\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=-\frac{5}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=\frac{5}{6}\left(-2\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\right)\)
\(\Rightarrow\overrightarrow{DG}=\frac{5}{6}\overrightarrow{DE}\Rightarrow\overrightarrow{DE}=\frac{6}{5}\overrightarrow{DG}\Rightarrow x=\frac{6}{5}\)
\(\overrightarrow{AD}=2\overrightarrow{DB}\Rightarrow\overrightarrow{AD}=\dfrac{2}{3}\overrightarrow{AB}\) ; \(\overrightarrow{CE}=3\overrightarrow{EA}\Rightarrow\overrightarrow{AE}=\dfrac{1}{4}\overrightarrow{AC}\)
Lại có M là trung điểm DE
\(\Rightarrow\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{AE}\right)=\dfrac{1}{2}\left(\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{8}\overrightarrow{AC}\)
I là trung điểm BC \(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(\Rightarrow\overrightarrow{MI}=\overrightarrow{MA}+\overrightarrow{AI}=\overrightarrow{AI}-\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{8}\overrightarrow{AC}=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)