Vector Algebra Question 32
Question: If ABCDEF is a regular hexagon and $ \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=\lambda ,\overrightarrow{AD}, $ then $ \lambda = $
[RPET 1985]
Options:
A) 2
B) 3
C) 4
D) 6
Show Answer
Answer:
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}] $ .
BETA
