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G là trung điểm BD \(\Rightarrow\overrightarrow{BG}=\overrightarrow{GD}\)
Gọi M là trung điểm BC \(\Rightarrow\) GM là đường trung bình tam giác BCD
\(\Rightarrow\overrightarrow{GM}=\frac{1}{2}\overrightarrow{DC}\Rightarrow\overrightarrow{DC}=\overrightarrow{AG}\)
\(\overrightarrow{AB}=\overrightarrow{AG}+\overrightarrow{GB}=\overrightarrow{AG}+\overrightarrow{DG}=\overrightarrow{AG}+\overrightarrow{DA}+\overrightarrow{AG}=2\overrightarrow{AG}-\overrightarrow{AD}=2\overrightarrow{a}-\overrightarrow{b}\)
\(\overrightarrow{AC}=\overrightarrow{AD}+\overrightarrow{DC}=\overrightarrow{AD}+\overrightarrow{AG}=\overrightarrow{a}+\overrightarrow{b}\)
Có: \(3\overrightarrow{MA}+4\overrightarrow{MB}=0\Leftrightarrow3\overrightarrow{MA}+4\overrightarrow{MB}+3\overrightarrow{MC}=3\overrightarrow{MC}\)
\(\Leftrightarrow3\overrightarrow{MG}+\overrightarrow{MB}=3\overrightarrow{MC}\)
\(\Leftrightarrow3\overrightarrow{MG}+\overrightarrow{MC}+\overrightarrow{CB}=3\overrightarrow{MC}\)
\(\Leftrightarrow3\overrightarrow{MG}+2\overrightarrow{CM}-2\overrightarrow{CN}=0\)
\(\Leftrightarrow3\overrightarrow{MG}+2\overrightarrow{NM}=0\)
Vậy 3 điểm M, N, G thẳng hàng.
b, theo như mình biết thì không có thương hai vec tơ.
a.\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{CB}\)
VT:\(\overrightarrow{AB}+\overrightarrow{CD}\)
=\(\overrightarrow{AC}+\overrightarrow{CB}+\overrightarrow{CA}+\overrightarrow{AD}\)
=\(\overrightarrow{AB}+\overrightarrow{CB}=0\left(đpcm\right)\)
b.\(\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{EA}=\overrightarrow{ED}+\overrightarrow{CB}\)
\(\Leftrightarrow\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{EA}+\overrightarrow{DE}+\overrightarrow{BC}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{EA}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{AE}+\overrightarrow{EA}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{0}=\overrightarrow{0}\left(LĐ\right)\)
A B C E D G
\(\text{a) Ta có : }2\overrightarrow{CD}=3\overrightarrow{DB}\\ \Rightarrow\overrightarrow{DC}=-\frac{3}{2}\overrightarrow{DB}\\ \Rightarrow D;B;C\text{ thẳng hàng },D\text{ nằm giữa }B;C\left(\frac{3}{2}< 0\right)\\ \Rightarrow\overrightarrow{BC}=\overrightarrow{BD}+\overrightarrow{DC}=\overrightarrow{BD}+\frac{3}{2}\overrightarrow{BD}=\frac{5}{2}\overrightarrow{BD}\\ 5\overrightarrow{EB}=2\overrightarrow{EC}\\ \Rightarrow\overrightarrow{EB}=\frac{2}{5}\overrightarrow{EC}\\ \Rightarrow E;B;C\text{ thẳng hàng },B\text{ nằm giữa }E;C\left(\frac{2}{5}>0;EB< EC\right)\\ \Rightarrow\overrightarrow{BC}=\overrightarrow{EC}-\overrightarrow{EB}=\overrightarrow{EC}-\frac{2}{5}\overrightarrow{EC}=\frac{3}{5}\overrightarrow{EC}\)
\(\Rightarrow\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}\\ =\overrightarrow{AB}+\frac{2}{5}\overrightarrow{BC}=\overrightarrow{AB}+\frac{2}{5}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\\ =\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}-\frac{2}{5}\overrightarrow{AB}=\frac{3}{5}\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\)
\(\overrightarrow{AE}=\overrightarrow{EC}+\overrightarrow{CA}\\ =\frac{5}{3}\overrightarrow{BC}-\overrightarrow{AC} =\frac{5}{3}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)-\overrightarrow{AC}\\ =\frac{5}{3}\overrightarrow{AC}-\frac{5}{3}\overrightarrow{AB}-\overrightarrow{AC}=\frac{2}{3}\overrightarrow{AC}-\frac{5}{3}\overrightarrow{AB}\)
\(b\text{) Theo tính chất trọng tâm }\Delta:3\overrightarrow{AG}=\overrightarrow{AA}+\overrightarrow{AB}+\overrightarrow{AC}\\ =\overrightarrow{0}+\overrightarrow{AB}+\overrightarrow{AC}\\ =\left(\frac{9}{4}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}\right)-\left(\frac{1}{2}\overrightarrow{AC}+\frac{5}{4}\overrightarrow{AC}\right)\\ =\frac{15}{4}\left(\frac{3}{5}\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\right)-\frac{3}{4}\left(\frac{2}{3}\overrightarrow{AC}+\frac{5}{3}\overrightarrow{AC}\right)\\ =\frac{15}{4}\overrightarrow{AD}-\frac{3}{4}\overrightarrow{AE}\)
\(\Rightarrow\overrightarrow{AG}=\frac{5}{4}\overrightarrow{AD}-\frac{1}{4}\overrightarrow{AE}\)