Question

If $ABCDEF$ is a regular hexagon, then $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}$ is equal to

• A. $0$
• B. $2\overrightarrow{AB}$
• C. $3\overrightarrow{AB}$
• D. $4\overrightarrow{AB}$

Answer 1

Answer: (D)

$ABCDEF$ is a regular hexagon. We know from the hexagon that $\overrightarrow{AD}$ is parallel to $\overrightarrow{BC}$.

$\Rightarrow \overrightarrow{AD}=2\overrightarrow{BC}$
Similarly, $\overrightarrow{EB}$ is parallel to $\overrightarrow{FA}$.
$\Rightarrow \overrightarrow{EB}=2\overrightarrow{FA}$
and $\overrightarrow{FC}$ is parallel to $\overrightarrow{AB}$.
$\Rightarrow \overrightarrow{FC}=2\overrightarrow{AB}$
Thus, $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}=2\overrightarrow{BC}+2\overrightarrow{FA}+\overrightarrow{AB}$
$=2\left(\overrightarrow{FA}+\overrightarrow{AB}+\overrightarrow{BC}\right)=2\left(\overrightarrow{FC}\right)=2\left(2\overrightarrow{AB}\right)=4\left(\overrightarrow{AB}\right)$