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\(\overrightarrow{AD}=2\overrightarrow{DB}\Rightarrow\overrightarrow{AD}=\dfrac{2}{3}\overrightarrow{AB}\) ; \(\overrightarrow{CE}=3\overrightarrow{EA}\Rightarrow\overrightarrow{AE}=\dfrac{1}{4}\overrightarrow{AC}\)
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 \(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(\Rightarrow\overrightarrow{MI}=\overrightarrow{MA}+\overrightarrow{AI}=\overrightarrow{AI}-\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{8}\overrightarrow{AC}=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)
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.
Câu 2:
a: \(\overrightarrow{AB}+\overrightarrow{CD}\)
\(=\overrightarrow{AI}+\overrightarrow{IB}+\overrightarrow{CI}+\overrightarrow{ID}\)
\(=\overrightarrow{IB}+\overrightarrow{ID}=2\overrightarrow{IJ}\)
b: \(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\dfrac{1}{2}\overrightarrow{a}-\dfrac{1}{2}\overrightarrow{b}\)
\(\overrightarrow{KA}=-\overrightarrow{AK}=-\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=-\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)\)
\(=-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)
\(\overrightarrow{KD}=\overrightarrow{AD}-\overrightarrow{AK}=\overrightarrow{AD}+\overrightarrow{KA}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)
\(=\frac{1}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
a, CM=MB
b, |CM|=|MB|=1/2|CB|
c, AM=MB+BA
=1/`2.CB-AB
=-1/2.BC-a
vì ABC đều
=> -1/2.a-a
= -5/2.a
GA=2/3.MA
= -2/3.AM
=-2/3.-5/2.a
=5/3.a
GM=1/3.AM
=1/3.-5/2a
=-5/6.a
c, |AB+AC|=|CB|=a
|AB-AC|=|AB+CA|=|CB|=a
Câu 1:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)