question_answer

If ABCDEF is a regular hexagon then \[\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}\] equals:

A)  zero                                     

B) \[2\overrightarrow{AB}\]

C) \[4\overrightarrow{AB}\]                            

D) \[3\overrightarrow{AB}\]

Correct Answer: C

Solution :

Let \[\overrightarrow{AB}=\vec{a},\overrightarrow{BC}=\overrightarrow{b},\overrightarrow{CD}=\vec{c},\] \[\overrightarrow{AD}=\vec{a}+\vec{b}+\vec{c}\] \[\overrightarrow{EB}=\overrightarrow{ED}+\overrightarrow{DC}+\overrightarrow{CB}\] \[=\vec{a}-\vec{c}-\vec{b}\] \[\overrightarrow{FC}=\overrightarrow{FA}+\overrightarrow{AB}+\overrightarrow{BC}=-\vec{c}+\vec{a}+\vec{b}\] \[\therefore \]  \[\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}=\vec{a}+\vec{b}+\vec{c}\] \[+\,\vec{a}-\vec{c}-\vec{b}-\vec{c}+\vec{a}+\vec{b}\] \[=\overrightarrow{3a}+\vec{b}-\vec{c},[\overrightarrow{BO}+\overrightarrow{OC}=\overrightarrow{BC},\vec{c}+\vec{a}=\vec{b}\,\therefore \,\vec{a}=\vec{b}-\vec{c}]\] \[=3\vec{a}+\vec{a}=4\vec{a}=4\overrightarrow{AB}\]