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