Question

If $ABCDEF$ is regular hexagon, then $\overset{\rightarrow}{AD} + \overset{\rightarrow}{EB} + \overset{\rightarrow}{FC} =$

• A. 0
• B. $2\overset{\rightarrow}{AB}$
• C. $\mathbf{a}.\mathbf{i} = 4,$
• D. $4\overset{\rightarrow}{AB}$

Answer 1

Answer: (D)

$4\overset{\rightarrow}{AB}$

Answer 2

A regular hexagon ABCDEF.

We know from the hexagon that $\overset{\rightarrow}{AD}$ is parallel to $\overset{\rightarrow}{BC}$ or $\overset{\rightarrow}{AD} = 2\overset{\rightarrow}{BC}$; $\overset{\rightarrow}{EB}$is parallel to $\overset{\rightarrow}{FA}$ or $\overset{\rightarrow}{EB} = 2\overset{\rightarrow}{FA}$, and $\overset{\rightarrow}{FC}$ is parallel to $\overset{\rightarrow}{AB}$ or $\overset{\rightarrow}{FC} = 2\overset{\rightarrow}{AB}$.

Thus $\overset{\rightarrow}{AD} + \overset{\rightarrow}{EB} + \overset{\rightarrow}{FC} = 2\overset{\rightarrow}{BC} + 2\overset{\rightarrow}{FA} + 2\overset{\rightarrow}{AB}$

$= 2(\overset{\rightarrow}{FA} + \overset{\rightarrow}{AB} + \overset{\rightarrow}{BC}) = 2(\overset{\rightarrow}{FC}) = 2(2\overset{\rightarrow}{AB}) = 4\overset{\rightarrow}{AB}$.