Question

trong hệ trục Oxy. Cho \(\overrightarrow{a}\)= (3,-4), \(\overrightarrow{b}\)=(-1,2)

a) tìm tọa độ của vecto \(2\overrightarrow{a}\)- \(3\overrightarrow{b}\)

b) Biết \(\overrightarrow{u}=2i-3j\). Biểu diễn vecto \(\overrightarrow{u}\) theo hai vecto \(\overrightarrow{a}\) và \(\overrightarrow{b}\)

Answer 1

a: Tọa độ vecto \(2\overrightarrow{a}-3\overrightarrow{b}\) là:

\(\left\{{}\begin{matrix}x=2\cdot3-3\cdot\left(-1\right)=6+3=9\\y=2\cdot\left(-4\right)-3\cdot2=-8-6=-14\end{matrix}\right.\)

b: \(\overrightarrow{u}=2i-3j=\left(2;-3\right)\)

Đặt \(\overrightarrow{u}=x\cdot\overrightarrow{a}+y\cdot\overrightarrow{b}\)

=>\(\left\{{}\begin{matrix}3x-y=2\\-4x+2y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4\\-4x+2y=-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x=1\\3x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3x-2=\dfrac{3}{2}-2=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(\overrightarrow{u}=\dfrac{1}{2}\overrightarrow{a}-\dfrac{1}{2}\overrightarrow{b}\)