Question

If ABCDEF is a regular hexagon, then $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}$ equals
(a) $2\overrightarrow{AB}$
(b) $\overrightarrow{0}$
(c) $3\overrightarrow{AB}$
(d) $4\overrightarrow{AB}$

Answer 1

(d) $4\overrightarrow{AB}$


$\begin{array}{l}\overrightarrow{\mathrm{AD}}=2\overrightarrow{\mathrm{BC}}\\ \overrightarrow{\mathrm{EB}}=2\overrightarrow{\mathrm{FA}}\\ \overrightarrow{\mathrm{FC}}=2\overrightarrow{\mathrm{AB}}\end{array}$

$\begin{array}{cll}\overrightarrow{AD}+\overrightarrow{EB}& =& 2\left(\overrightarrow{BC}+\overrightarrow{FA}\right)\hspace{0.17em}\\ & =& 2\left(\overrightarrow{AO}+\overrightarrow{FA}\right)\left(\because \hspace{0.17em}\overrightarrow{BC}=\overrightarrow{AO}\right)\end{array}$

In triangle AOF,

$\begin{array}{l}\overrightarrow{FA}+\overrightarrow{AO}+\overrightarrow{FO}=0\\ \therefore \hspace{0.17em}\hspace{0.17em}\overrightarrow{FA}+\overrightarrow{AO}=-\overrightarrow{FO}\\ \therefore \hspace{0.17em}\hspace{0.17em}\overrightarrow{AD}+\overrightarrow{EB}=-2\overrightarrow{FO}\end{array}$

And $\overrightarrow{AB}=-\overrightarrow{FO}$

$\begin{array}{l}\therefore \hspace{0.17em}\hspace{0.17em}\overrightarrow{AD}+\overrightarrow{EB}=2\overrightarrow{AB}\\ \therefore \hspace{0.17em}\hspace{0.17em}\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}=2\overrightarrow{AB}+2\overrightarrow{AB}=4\overrightarrow{AB}\end{array}$