Cho \(\overrightarrow a = (0,1)\),\(\overrightarrow b  = ( - 1;2)\),\(\overrightarrow c  = ( - 3; - 2)\). Tọa độ của \(\overrightarrow u  = 3\overrightarrow a  + 2\overrightarrow b  - 4\overrightarrow c \)

\(\left( {i;j} \right)\)

\(\left( {\overrightarrow i ;\overrightarrow j } \right)\)

\(\left( {\overrightarrow m ;\overrightarrow n } \right)\)

\(\left( {m;n} \right)\)

$\left\{ \begin{array}{l}{u_1} = {u_2}\\{v_1} = {v_2}\end{array} \right.$.

$\left\{ \begin{array}{l}{u_1} =  - {v_1}\\{u_2} =  - {v_2}\end{array} \right.$

$\left\{ \begin{array}{l}{u_1} = {v_1}\\{u_2} = {v_2}\end{array} \right.$.

$\left\{ \begin{array}{l}{u_1} = {v_2}\\{u_2} = {v_1}\end{array} \right.$.

\(\overrightarrow u  =  - \overrightarrow i \)

\(\overrightarrow u  =  - \overrightarrow j \)

\(\overrightarrow u  = \overrightarrow i \)

\(\overrightarrow u  = \overrightarrow j \)

\(\overrightarrow {OM}  = \left( { - 3; - 1} \right)\)

\(\overrightarrow {OM}  = \left( {3;1} \right)\)         

\(\overrightarrow {MO}  = \left( { - 3;1} \right)\)

\(\overrightarrow {OM}  = \left( { - 3;1} \right)\)

\(\overrightarrow {OM}  =  - 2\overrightarrow i  + 4\overrightarrow j \)

\(\overrightarrow {OM}  = 2\overrightarrow i  - 4\overrightarrow j \)

\(\overrightarrow {OM}  = 2\overrightarrow j  - 4\overrightarrow i \)

\(\overrightarrow {OM}  =  - 2\overrightarrow i  - 4\overrightarrow j \)

Hai vectơ $\overrightarrow u  = \left( {2; - 1} \right){\rm{ và  }}\overrightarrow v  = \left( { - 1;2} \right)$đối nhau

Hai vectơ $\overrightarrow u  = \left( {2; - 1} \right){\rm{ và  }}\overrightarrow v  = \left( { - 2; - 1} \right)$đối nhau.

Hai vectơ $\overrightarrow u  = \left( {2; - 1} \right){\rm{ và  }}\overrightarrow v  = \left( { - 2;1} \right)$đối nhau.

Hai vectơ $\overrightarrow u  = \left( {2; - 1} \right){\rm{ và  }}\overrightarrow v  = \left( {2;1} \right)$đối nhau.

$I\left( {\dfrac{{{x_A} - {x_B}}}{2};\dfrac{{{y_A} - {y_B}}}{2}} \right)$.

$I\left( {\dfrac{{{x_A} + {x_B}}}{2};\dfrac{{{y_A} + {y_B}}}{2}} \right)$.

$I\left( {\dfrac{{{x_A} + {x_B}}}{3};\dfrac{{{y_A} + {y_B}}}{3}} \right)$.

$I\left( {\dfrac{{{x_A} + {y_A}}}{2};\dfrac{{{x_B} + {y_B}}}{2}} \right)$.