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)

$4\overrightarrow{AB}$

Answer 2

$ABCDEF$ 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, $\overrightarrow{AD} + \overrightarrow{EB} + \overrightarrow{FC} = 2 \overrightarrow{BC} + 2 \overrightarrow{FA} + \overrightarrow{AB}$ $= 2\left(\overrightarrow{FA} + \overrightarrow{AB} + \overrightarrow{BC}\right) = 2\left( \overrightarrow{FC}\right) = 2\left(2 \overrightarrow{AB}\right) = 4\left( \overrightarrow{AB}\right)$