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: \(\overrightarrow{AG}=\dfrac{35}{48}\overrightarrow{AI}-\dfrac{1}{16}\overrightarrow{AJ}\)

\(a,\) \(\overrightarrow{IA}=2\overrightarrow{IB}-4\overrightarrow{IC}\)

\(\overrightarrow{IA}=2\overrightarrow{IB}-2\overrightarrow{IC}-2\overrightarrow{IC}=2\overrightarrow{CB}-2\overrightarrow{IC}\)

\(=2\left(\overrightarrow{AB}-\overrightarrow{AC}\right)-2\left(\overrightarrow{AC}-\overrightarrow{AI}\right)\)

\(\overrightarrow{IA}=2\overrightarrow{AB}-2\overrightarrow{AC}-2\overrightarrow{AC}+2\overrightarrow{AI}\)

\(\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}\)

\(b,\overrightarrow{IJ}=\overrightarrow{AJ}-\overrightarrow{AI}=\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}=\dfrac{4}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(1\right)\)

\(\overrightarrow{JG}=\overrightarrow{AG}-\overrightarrow{AJ}=\dfrac{2}{3}\overrightarrow{AM}-\dfrac{2}{3}\overrightarrow{AB}\)\((\) \(\) \(M\)  \(trung\) \(điểm\) \(BC)\)

\(\overrightarrow{JG}=\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{3}-\dfrac{2}{3}\overrightarrow{AB}=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=-\dfrac{1}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\overrightarrow{IJ}=-4\overrightarrow{JG}\Rightarrow I,J,G\) \(thẳng\) \(hàng\)

theo câu a, ta có\(\overrightarrow{AK}\) =\(\dfrac{3}{7}\overrightarrow{AB}+\dfrac{4}{7}\overrightarrow{AC}\)

\(=\dfrac{3}{7}.\dfrac{5}{3}\overrightarrow{AI}+\dfrac{4}{7}.\dfrac{1}{2}\overrightarrow{AJ}\)

=> K,I, J thẳng hàng

a) \(\overrightarrow{BI}=\overrightarrow{BC}+\overrightarrow{CI}=\overrightarrow{BC}+\dfrac{1}{4}\overrightarrow{CA}=\overrightarrow{BA}+\overrightarrow{AC}+\dfrac{1}{4}\overrightarrow{CA}\)
\(=\overrightarrow{BA}+\overrightarrow{AC}-\dfrac{1}{4}\overrightarrow{AC}=\dfrac{3}{4}\overrightarrow{AC}+\overrightarrow{BA}=\dfrac{3}{4}\overrightarrow{AC}-\overrightarrow{AB}\).
b) Có \(\overrightarrow{BJ}=\dfrac{1}{2}\overrightarrow{AC}-\dfrac{2}{3}\overrightarrow{AB}=\dfrac{3}{2}\left(\dfrac{1}{2}\overrightarrow{AC}-\overrightarrow{AB}\right)=\dfrac{3}{2}\overrightarrow{BI}\).
Vì vậy 3 điểm B, I, J thẳng hàng.
c)
Trên cạnh AC lấy điểm K sao cho \(\overrightarrow{AK}=\dfrac{1}{2}\overrightarrow{AC}\).
Tại điểm K dựng điểm T sao cho \(\overrightarrow{KT}=-\dfrac{3}{2}\overrightarrow{AB}=\dfrac{3}{2}\overrightarrow{BA}\).
\(\overrightarrow{BJ}=\dfrac{1}{2}\overrightarrow{AC}-\dfrac{3}{2}\overrightarrow{AB}=\overrightarrow{AK}+\overrightarrow{KT}=\overrightarrow{AT}\).
Dựng điểm T sao cho \(\overrightarrow{BJ}=\overrightarrow{AT}\).


 

→IB+→IA−→IC−→CM=→0

=>\(\overrightarrow{IB}+\overrightarrow{IA}-\overrightarrow{IM}=\overrightarrow{0}\)

=>\(\overrightarrow{IB}+\overrightarrow{IA}=\overrightarrow{IM}\)

Đặt K là trung điểm AB

=>\(\overrightarrow{IB}+\overrightarrow{IA}=\overrightarrow{2IK}\)(T/c trung tuyến)

=>\(\overrightarrow{2IK}=\overrightarrow{IM}\)

=>K,M,I thẳng hàng

Vậy điểm M thuộc đoạn KI sao cho \(\dfrac{\overrightarrow{IK}}{\overrightarrow{IM}}=\dfrac{1}{2}\)