Cho $\overrightarrow{i}(0;1), \overrightarrow{j}(-1;2). $ Xác định tọa độ của vectơ $\overrightarrow {a}$, biết:
a) $\overrightarrow {a}=-2\overrightarrow {i}+\overrightarrow {j}$
b) $\overrightarrow {a}=-4\overrightarrow {i}-3\overrightarrow {j}$
c) $\overrightarrow {a}=3\overrightarrow {i}$
d) $\overrightarrow {a}=-5\overrightarrow {j}$
$a.$
$\overrightarrow{a}= -2\overrightarrow{i}+\overrightarrow{j}\Leftrightarrow \left\{ \begin{array}{l} x_\overrightarrow{a}=-2.0-1=-1\\ y_\overrightarrow{a} =-2.1+2=0\end{array} \right.\Leftrightarrow \overrightarrow{a}(-1;0)$
$b.$
$\overrightarrow{a}= -4\overrightarrow{i}-3\overrightarrow{j}\Leftrightarrow \left\{ \begin{array}{l} x_\overrightarrow{a}=-4.0-3.1=-3\\ y_\overrightarrow{a} =-4.1-3.2=-10\end{array} \right.\Leftrightarrow \overrightarrow{a}(-3;10)$ 
$c.$
$\overrightarrow{a}= 3\overrightarrow{i}\Leftrightarrow \left\{ \begin{array}{l} x_\overrightarrow{a}=3.0=0\\ y_\overrightarrow{a} =3.1=3\end{array} \right.\Leftrightarrow \overrightarrow{a}(0;3)$ 
$d.$ 
$\overrightarrow{a}= -5\overrightarrow{j}\Leftrightarrow \left\{ \begin{array}{l} x_\overrightarrow{a}=-5.(-1)=5\\ y_\overrightarrow{a} =-5.2=-10\end{array} \right.\Leftrightarrow \overrightarrow{a}(-5;-10)$  

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