Cho tam giác $ABC$ và đặt $\vec a = \overrightarrow {BC} ,\,\,\vec b = \overrightarrow {AC} .$ Cặp vectơ nào sau đây cùng phương?

$\overrightarrow {AI}  = \dfrac{1}{4}\left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right).$

$\overrightarrow {AI}  = \dfrac{1}{4}\left( {\overrightarrow {AB}  - \overrightarrow {AC} } \right).$

$\overrightarrow {AI}  = \dfrac{1}{4}\overrightarrow {AB}  + \dfrac{1}{2}\overrightarrow {AC} .$

$\overrightarrow {AI}  = \dfrac{1}{4}\overrightarrow {AB}  - \dfrac{1}{2}\overrightarrow {AC} .$

$\overrightarrow {AG}  = \dfrac{2}{3}\left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right).$

$\overrightarrow {AG}  = \dfrac{1}{3}\left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right).$

$\overrightarrow {AG}  = \dfrac{1}{3}\overrightarrow {AB}  + \dfrac{2}{2}\overrightarrow {AC} .$

$\overrightarrow {AI}  = \dfrac{2}{3}\overrightarrow {AB}  + 3\overrightarrow {AC} .$

$\overrightarrow {IB}  + 2\overrightarrow {IC}  + \overrightarrow {IA}  = \overrightarrow 0 .$

$\overrightarrow {IB}  + \overrightarrow {IC}  + 2\overrightarrow {IA}  = \overrightarrow 0 .$

$2\overrightarrow {IB}  + \overrightarrow {IC}  + \overrightarrow {IA}  = \overrightarrow 0 .$

$\overrightarrow {IB}  + \overrightarrow {IC}  + \overrightarrow {IA}  = \overrightarrow 0 .$

$a.$    

$\left( {1 + \sqrt 2 } \right)a.$

$a\sqrt 5 .$

$2a\sqrt 2 .$

$\overrightarrow {AC}  + \overrightarrow {BD}  = 2\overrightarrow {BC} .$

$\overrightarrow {AC}  + \overrightarrow {BC}  = \overrightarrow {AB} .$

$\overrightarrow {AC}  - \overrightarrow {BD}  = 2\,\overrightarrow {CD} .$

$\overrightarrow {AC}  - \overrightarrow {AD}  = \overrightarrow {CD} .$

$2\overrightarrow {MA}  + \overrightarrow {MB}  - 3\overrightarrow {MC}  = \overrightarrow {AC}  + 2\overrightarrow {BC} .$

$2\overrightarrow {MA}  + \overrightarrow {MB}  - 3\overrightarrow {MC}  = 2\overrightarrow {AC}  + \overrightarrow {BC} .$

$2\overrightarrow {MA}  + \overrightarrow {MB}  - 3\overrightarrow {MC}  = 2\overrightarrow {CA}  + \overrightarrow {CB} .$

$2\overrightarrow {MA}  + \overrightarrow {MB}  - 3\overrightarrow {MC}  = 2\overrightarrow {CB}  - \overrightarrow {CA} .$