Cho điểm $M\left( { - 3;1} \right)$, khi đó:

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

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

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

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

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

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

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

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

$\overrightarrow a .\overrightarrow b  = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|$.

$\overrightarrow a .\overrightarrow b  = 0$.

$\overrightarrow a .\overrightarrow b  =  - 1$.

$\overrightarrow a .\overrightarrow b =  - \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|$.

$\overrightarrow u  =  - \overrightarrow i$

$\overrightarrow u  =  - \overrightarrow j$

$\overrightarrow u  = \overrightarrow i$

$\overrightarrow u  = \overrightarrow j$

$\overrightarrow {AB}  = \left( {{y_A} - {x_A};{y_B} - {x_B}} \right)$.

$\overrightarrow {AB}  = \left( {{x_A} + {x_B};{y_A} + {y_B}} \right)$

$\overrightarrow {AB}  = \left( {{x_A} - {x_B};{y_A} - {y_B}} \right)$.

$\overrightarrow {AB}  = \left( {{x_B} - {x_A};{y_B} - {y_A}} \right)$.

$\left( {2;4} \right)$.

$\left( {5;6} \right)$.

$\left( {15;10} \right)$.

$\left( {50;6} \right)$.