question_answer

If ABCDEF is a regular hexagon and \[\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=\lambda \,\overrightarrow{AD},\] then \[\lambda =\]                                      [RPET 1985]

A) 2             

B) 3

C) 4             

D) 6

Correct Answer: B

Solution :

By triangle law, \[\overrightarrow{AB}=\overrightarrow{AD}-\overrightarrow{BD},\] \[\overrightarrow{AC}=\overrightarrow{AD}-\overrightarrow{CD}\] Therefore, \[\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}\] = \[3\overrightarrow{AD}+(\overrightarrow{AE}-\overrightarrow{BD})+(\overrightarrow{AF}-\overrightarrow{CD})=3\overrightarrow{AD}\] Hence \[\lambda =3\],     [Since \[\overrightarrow{AE}=\overrightarrow{BD,}\,\overrightarrow{AF}\,=\overrightarrow{CD}]\].