Question

If $ABCDEF$ is a regular hexagon with its center at $O$ Then $\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}$ is equal to

• A. $5\overrightarrow{AO}$
• B. $6\overrightarrow{AO}$
• C. $4\overrightarrow{AO}$
• D. None of these

Answer 1

Answer: (B)

If $\overrightarrow{AB}=\vec{a},\overrightarrow{BC}=\vec{b}$, then $\overrightarrow{AC}=\vec{a}+\vec{b}$
$\overrightarrow{AD}=2\vec{b},\overrightarrow{AE}=\overrightarrow{AD}+\overrightarrow{DE}=2\vec{b}-\vec{a}$
and $\overrightarrow{AF}=\overrightarrow{AE}+\overrightarrow{EF}=\overrightarrow{AE}+\overrightarrow{CB}=2\vec{b}-\vec{a}-\vec{b}=\vec{b}-\vec{a}$
$\therefore \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=6\overrightarrow{BC}=6\overrightarrow{AO}$