a: \(\overrightarrow{u}=3\overrightarrow{a}-2\overrightarrow{b}\)
=>\(\left\{{}\begin{matrix}x_{\overrightarrow{u}}=3\cdot3-2\cdot4=9-8=1\\y_{\overrightarrow{u}}=3\cdot\left(-2\right)-2\cdot1=-6-2=-8\end{matrix}\right.\)
Vậy: \(\overrightarrow{u}=\left(1;-8\right)\)
b: \(\overrightarrow{v}=\overrightarrow{a}+3\overrightarrow{b}-2\overrightarrow{c}\)
=>\(\left\{{}\begin{matrix}x_{\overrightarrow{v}}=3+3\cdot4-2\cdot0=3+12=14\\y_{\overrightarrow{v}}=-2+3\cdot1-2\cdot\left(-5\right)=-2+3+10=11\end{matrix}\right.\)
Vậy: \(\overrightarrow{v}=\left(14;11\right)\)
c: Để \(\overrightarrow{x};\overrightarrow{u}\) là hai vecto cùng phương thì \(\dfrac{m}{1}=\dfrac{m+1}{-8}\)
=>-8m=m+1
=>-9m=1
=>\(m=-\dfrac{1}{9}\)
