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Lời giải:
Với $I$ là trung điểm của $BC$ thì \(\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{0}\)
Ta có:
\(\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AI}+\overrightarrow{IB}+\overrightarrow{AI}+\overrightarrow{IC}\)
\(=2\overrightarrow{AI}+(\overrightarrow{IB}+\overrightarrow{IC})\)
\(=2\overrightarrow{AI}\)
\(\Rightarrow \overrightarrow{AI}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\) (đpcm)
b) Gọi giao điểm của $AG$ với $BC$ là $T$
\(\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AG}+\overrightarrow{GB}+\overrightarrow{AG}+\overrightarrow{GC}\)
\(=2\overrightarrow{AG}+\overrightarrow{GB}+\overrightarrow{GC}=2\overrightarrow{AG}+\overrightarrow{GI}+\overrightarrow{IB}+\overrightarrow{GI}+\overrightarrow{IC}\)
\(=2\overrightarrow{AG}+2\overrightarrow{GI}\)
Theo tính chất đường trung tuyến thì \(\overrightarrow{AG}=2\overrightarrow{GI}\) nên:
\(\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AG}+\overrightarrow{AG}=3\overrightarrow{AG}\)
\(\Rightarrow \overrightarrow{AG}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
Câu 1:
vecto AM+vecto BN+vecto CP
=1/2(vecto AB+vecto AC+vecto BA+vecto BC+vecto CA+vecto CB)
=1/2*vecto 0
=vecto 0
\(\overrightarrow{JA}=-\frac{2}{3}\overrightarrow{JC}\Rightarrow\overrightarrow{JA}=\frac{2}{5}\overrightarrow{CA}\)
\(\overrightarrow{IB}=\overrightarrow{BA}\Rightarrow\overrightarrow{IA}=2\overrightarrow{BA}\)
a/ \(\overrightarrow{IJ}=\overrightarrow{IA}+\overrightarrow{AJ}=2\overrightarrow{BA}-\frac{2}{5}\overrightarrow{CA}=\frac{2}{5}\overrightarrow{AC}-2\overrightarrow{AB}\)
b/Theo tính chất trọng tâm \(3\overrightarrow{AG}=\overrightarrow{AB}+\overrightarrow{AC}\Rightarrow\overrightarrow{AG}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{IG}=\overrightarrow{IA}+\overrightarrow{AG}=2\overrightarrow{BA}+\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AC}-\frac{5}{3}\overrightarrow{AB}\)
\(a,\) \(\overrightarrow{IA}=2\overrightarrow{IB}-4\overrightarrow{IC}\)
\(\overrightarrow{IA}=2\overrightarrow{IB}-2\overrightarrow{IC}-2\overrightarrow{IC}=2\overrightarrow{CB}-2\overrightarrow{IC}\)
\(=2\left(\overrightarrow{AB}-\overrightarrow{AC}\right)-2\left(\overrightarrow{AC}-\overrightarrow{AI}\right)\)
\(\overrightarrow{IA}=2\overrightarrow{AB}-2\overrightarrow{AC}-2\overrightarrow{AC}+2\overrightarrow{AI}\)
\(\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}\)
\(b,\overrightarrow{IJ}=\overrightarrow{AJ}-\overrightarrow{AI}=\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}=\dfrac{4}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(1\right)\)
\(\overrightarrow{JG}=\overrightarrow{AG}-\overrightarrow{AJ}=\dfrac{2}{3}\overrightarrow{AM}-\dfrac{2}{3}\overrightarrow{AB}\)\((\) \(\) \(M\) \(trung\) \(điểm\) \(BC)\)
\(\overrightarrow{JG}=\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{3}-\dfrac{2}{3}\overrightarrow{AB}=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=-\dfrac{1}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\overrightarrow{IJ}=-4\overrightarrow{JG}\Rightarrow I,J,G\) \(thẳng\) \(hàng\)
a: CI+BI=CB
=>\(\dfrac{3}{2}BI+BI=CB\)
=>\(\dfrac{5}{2}BI=CB\)
=>\(BI=\dfrac{2}{5}BC\)
=>\(CI=\dfrac{3}{2}\cdot BI=\dfrac{3}{2}\cdot\dfrac{2}{5}CB=\dfrac{3}{5}CB\)
\(\overrightarrow{AI}=\overrightarrow{AB}+\overrightarrow{BI}\)
\(=\overrightarrow{AB}+\dfrac{2}{5}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{5}\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\)
\(=\dfrac{3}{5}\overrightarrow{AB}+\dfrac{2}{5}\overrightarrow{AC}\)
JB=2/5JC mà J không nằm trong đoạn thẳng BC
nên B nằm giữa J và C
=>JB+BC=JC
=>\(BC+\dfrac{2}{5}JC=JC\)
=>\(BC=\dfrac{3}{5}JC\)
\(\dfrac{JB}{BC}=\dfrac{\dfrac{2}{5}JC}{\dfrac{3}{5}JC}=\dfrac{2}{5}:\dfrac{3}{5}=\dfrac{2}{3}\)
=>\(JB=\dfrac{2}{3}BC\)
\(\overrightarrow{AJ}=\overrightarrow{AB}+\overrightarrow{BJ}\)
\(=\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}-\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{BA}-\dfrac{2}{3}\overrightarrow{AC}=\dfrac{5}{3}\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{AC}\)
b:
Gọi giao điểm của AG với BC là M
G là trọng tâm của ΔABC
nên AG cắt BC tại trung điểm M của BC
=>\(AG=\dfrac{2}{3}AM\)
Xét ΔABC có AM là trung tuyến
nên \(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
=>\(\overrightarrow{AG}=\dfrac{2}{3}\cdot\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
Đặt \(\overrightarrow{AG}=x\cdot\overrightarrow{AI}+y\cdot\overrightarrow{AJ}\)
\(\overrightarrow{AG}=\dfrac{1}{3}\cdot\overrightarrow{AB}+\dfrac{1}{3}\cdot\overrightarrow{AC};\overrightarrow{AI}=\dfrac{3}{5}\cdot\overrightarrow{AB}+\dfrac{2}{5}\cdot\overrightarrow{AC};\overrightarrow{AJ}=\dfrac{5}{3}\overrightarrow{AB}-\dfrac{2}{3}\cdot\overrightarrow{AC}\)
Ta có hệ phương trình sau:
\(\left\{{}\begin{matrix}\dfrac{1}{3}=x\cdot\dfrac{3}{5}+y\cdot\dfrac{5}{3}\\\dfrac{1}{3}=x\cdot\dfrac{2}{5}+y\cdot\dfrac{-2}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\cdot\dfrac{3}{5}+y\cdot\dfrac{5}{3}=\dfrac{1}{3}\\x\cdot\dfrac{2}{5}+y\cdot\dfrac{-2}{3}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+25y=5\\6x-10y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}18x+50y=10\\18x-30y=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}80y=-5\\6x-10y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{1}{16}\\6x=10y+5=-\dfrac{5}{8}+5=\dfrac{35}{8}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{1}{16}\\x=\dfrac{35}{48}\end{matrix}\right.\)
Vậy: \(\overrightarrow{AG}=\dfrac{35}{48}\overrightarrow{AI}-\dfrac{1}{16}\overrightarrow{AJ}\)
