JEE Main & Advanced
Mathematics
Vector Algebra
Question Bank
Modulus of vector Algebra of vectors
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 \[\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}=2\,\overrightarrow{BC}+2\,\overrightarrow{FA}+2\,\overrightarrow{AB}\] \[=2(\overrightarrow{FA}+\overrightarrow{AB}+\overrightarrow{BC})=2(\overrightarrow{FC})=2(2\overrightarrow{AB})=4\,\overrightarrow{AB}\].