Question

cho vecto a = (1;2), vecto b = (-3;4), vecto c = (-2;-5) . Tìm tọa độ của vecto u biết : 

a) 2u - 3a + b = 0 

b) 3u + 2a + 3b = 3c

 

Answer 1

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

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