A) 2
B) 3
C) 4
D) 5
Correct Answer: B
Solution :
[b] \[\because \,\,\overrightarrow{AB}=\overrightarrow{ED}\] and \[\overrightarrow{AF}=\overrightarrow{CD}\], so \[\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}\] \[=\overrightarrow{ED}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{CD}\] \[=(\overrightarrow{AC}+\overrightarrow{CD})+(\overrightarrow{AE}+\overrightarrow{ED})+\overrightarrow{AD}\] \[=\overrightarrow{AD}+\overrightarrow{AD}+\overrightarrow{AD}+=3\overrightarrow{AD}\therefore k=3\]You need to login to perform this action.
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