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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.
ta có : 2vecto AM =vecto AB +vectoAC
2 vecto BN = vectoBA +vectoBC
2 vecto CP =vecto CA + vectoCB
=> 2vecto AM +2 vecto BN + 2 vecto CP =vecto AB +vecto AC +vecto BA +vecto BC +vecto CA + vecto CB =vecto 0
=> vecto AM + vecto BN + vecto CP =vecto 0
MÌNH LÀM VẬY CŨNG KO BIẾT ĐÚNG KO NỮA
Câu 2:
Vì G là trọng tâm nên \(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)
hay \(\overrightarrow{GC}=-\overrightarrow{a}-\overrightarrow{b}\)
\(\overrightarrow{BC}=\overrightarrow{BG}+\overrightarrow{GC}=-\overrightarrow{b}-\overrightarrow{a}-\overrightarrow{b}=-\overrightarrow{a}-2\overrightarrow{b}\)
=>m=-1; n=-2
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 B C D I M
a)
\(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AD}\right)=\dfrac{1}{2}\left(\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\right)=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\).
b)
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+x\overrightarrow{BC}\)\(=\overrightarrow{AB}+x\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\left(1-x\right)\overrightarrow{AB}+x\overrightarrow{AC}\).
c) A, M, I thẳng hàng khi và chỉ khi hai véc tơ \(\overrightarrow{AM};\overrightarrow{AI}\) cùng phương
hay \(\dfrac{1-x}{\dfrac{1}{2}}=\dfrac{x}{\dfrac{3}{8}}\Leftrightarrow\dfrac{3}{8}\left(1-x\right)=\dfrac{1}{2}x\)
\(\Leftrightarrow\dfrac{7}{8}x=\dfrac{3}{8}\)\(\Leftrightarrow x=\dfrac{3}{7}\).
a: \(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}=\dfrac{2}{3}\cdot\dfrac{1}{2}\cdot\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\dfrac{1}{3}\cdot\overrightarrow{AB}+\dfrac{1}{3}\cdot\overrightarrow{AC}\)