Question

In a regular hexagon $ABCDEF$ with center at $O$ (origin). Prove
If five forces $\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD},\overrightarrow{AE},\overrightarrow{AF}$ act at the vertex $A$ of regular hexagon $ABCDEF$. Then their resultant is equal to $6\overrightarrow{AO}$

Answer 1

$AB+AC+AD+AE+AF$

$=AB+(AC+CD)+AD+AE$ ( since $AF=CD$ )

$=AB+AD+AD+AE$

$=2AD+AB+AE$

$=2AD+AD$

$=3AD$

$=3\times 2AO$ ( since O is center $AB=OD$ )

$=6AO$

Thus $AB+AC+AD+AE+AF=6AO$

Hence, the answer is $6AO.$