Ôn tập chương I

2AB+3AC=5BC

=> =5a

\(\overrightarrow{AB}=\left(4;3\right)\) ; \(\overrightarrow{AC}=\left(6;1\right)\Rightarrow2\overrightarrow{AC}=\left(12;2\right)\)

\(\Rightarrow\overrightarrow{u}=\left(12-4;2-3\right)=\left(8;-1\right)\)

Gọi \(M\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}=\left(-4-x;-y\right)\\\overrightarrow{MB}=\left(-x;3-y\right)\\\overrightarrow{MC}=\left(2-x;1-y\right)\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{MA}+2\overrightarrow{MB}+3\overrightarrow{MC}=\left(2-6x;9-6y\right)\)

\(\Rightarrow\left\{{}\begin{matrix}2-6x=0\\9-6y=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow M\left(\dfrac{1}{3};\dfrac{3}{2}\right)\)

a) Tọa độ vecto u = (-16;4)

Ta có: \(\dfrac{-3}{1}\ne\dfrac{0}{2}\Rightarrow\overrightarrow{a}\) và \(\overrightarrow{b}\) ko cùng phương

b. Đặt \(\overrightarrow{c}=x.\overrightarrow{a}+y.\overrightarrow{b}\)

\(\Rightarrow\left(-1;3\right)=x.\left(1;2\right)+y.\left(-3;0\right)=\left(x-3y;2x\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x-3y=-1\\2x=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{5}{6}\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{c}=\dfrac{3}{2}\overrightarrow{a}+\dfrac{5}{6}\overrightarrow{b}\)

- Với \(m=0\) ko thỏa mãn

- Với \(m\ne0\) hai vecto cùng phương khi:

\(\dfrac{m^2+m-2}{m}=\dfrac{4}{2}\Leftrightarrow m^2+m-2=2m\)

\(\Rightarrow m^2-3m-2=0\Rightarrow m=\dfrac{3\pm\sqrt{17}}{2}\)

\(\left[{}\begin{matrix}2x_U-3x_A+x_B=0\\2y_U-3y_A+y_B=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x_U=6\\2y_U=2\end{matrix}\right.\Rightarrow\overrightarrow{u}=\left(3;1\right)\)

\(b.\left[{}\begin{matrix}3x_U+2x_A+3x_B=3x_C\\3y_U+2y_A+3y_B=3y_C\end{matrix}\right.\left[{}\begin{matrix}3x_U=1\\3y_U=-31\end{matrix}\right.\Rightarrow\overrightarrow{u}=\left(\dfrac{1}{3};-\dfrac{31}{3}\right)\)