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\(\overrightarrow{c}=2\left(2;1\right)+3\left(3;-2\right)=\left(4+9;2-6\right)=\left(13;-4\right)\)
Gọi \(M\left(a;b\right)\)
\(\Rightarrow\overrightarrow{MB}=\left(2-a;3-b\right)\Rightarrow2\overrightarrow{MB}=\left(4-2a;6-2b\right)\)
\(\overrightarrow{MC}=\left(-1-a;-2-b\right)\Rightarrow3\overrightarrow{MC}=\left(-3-3a;-6-3b\right)\)
\(\Rightarrow2\overrightarrow{MB}+3\overrightarrow{MC}=\left(1-5a;-5b\right)=\overrightarrow{0}\)
\(\Rightarrow\left\{{}\begin{matrix}1-5a=0\\-5b=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{5}\\b=0\end{matrix}\right.\) \(\Rightarrow M\left(\frac{1}{5};0\right)\)
Để tứ giác OABC là hbh<=> \(\overrightarrow{OA}=\overrightarrow{CB}\)
\(\Leftrightarrow\left(x_A;y_A\right)=\left(x_B-x_C;y_B-y_C\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}6-x_C=1\\5-y_C=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_C=5\\y_C=4\end{matrix}\right.\Rightarrow C\left(5;4\right)\)
ta có\(\overrightarrow{OA}\)=(1;1)
\(\overrightarrow{CB}\)=(6-Cx;5-Cy)
để tứ giác OABC là hbh thì \(\overrightarrow{OA}=\overrightarrow{CB}\)
=>\(\left\{{}\begin{matrix}6-C_X=1\\5-C_Y=1\end{matrix}\right.\)=>C(5;4)
\(\overrightarrow{OA}-\overrightarrow{OB}\)
\(\overrightarrow{OA}\left(x_A-x_O;y_A-y_O\right)=\left(2;3\right)\)
\(\overrightarrow{OB}=\left(x_B-x_O;y_B-y_O\right)=\left(4;-1\right)\)
\(\Rightarrow\overrightarrow{OA}-\overrightarrow{OB}=\left(2-4;3+1\right)=\left(-2;4\right)\)
\(\overrightarrow{OA}=\left(2;3\right)\) ; \(\overrightarrow{OB}=\left(4;-1\right)\)
\(\Rightarrow\overrightarrow{OA}-\overrightarrow{OB}=\left(2-4;3+1\right)=\left(-2;4\right)\)
