Gọi `BE` là đường trung tuyến `\DeltaABC`
`\Rightarrow BE = \sqrt[4a^2 -a^2]= \sqrt(3)a`
Ta có: `|\vec{AB} + \vec{CG}|= |\vec{AB} + \vec{CG}|`
`= |\vec{AB}+ \vec{GA} +\vec{GB}|`
`=| 2\vec{GB}|= |2. 2/3\vec{EB}|`
`= |4/3. \sqrt(3)a| = (4\sqrt(3))/3a`
Vậy: `|\vec{AB}+ \vec{CG}|= (4\sqrt(3))/3a`