question_answer

If \[ABCDEF\] is regular hexagon, then \[\overrightarrow{AD}\,+\overrightarrow{EB}+\overrightarrow{FC}=\] [Karnataka CET 2002]

A) 0             

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

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

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

Correct Answer: D

Solution :

A regular hexagon ABCDEF. We know from the hexagon that \[\overrightarrow{AD}\] is parallel to \[\overrightarrow{BC}\] or \[\overrightarrow{AD}=2\,\overrightarrow{BC}\]; \[\overrightarrow{EB}\]is parallel to \[\overrightarrow{FA}\] or \[\overrightarrow{EB}=2\overrightarrow{FA}\], and \[\overrightarrow{FC}\] is parallel to \[\overrightarrow{AB}\] or \[\overrightarrow{FC}=2\,\overrightarrow{AB}\]. Thus \[\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}=2\,\overrightarrow{BC}+2\,\overrightarrow{FA}+2\,\overrightarrow{AB}\]                 \[=2(\overrightarrow{FA}+\overrightarrow{AB}+\overrightarrow{BC})=2(\overrightarrow{FC})=2(2\overrightarrow{AB})=4\,\overrightarrow{AB}\].