Vector Algebra Question 39
Question: If ABCDEF is a regular hexagon and $ \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=k\overrightarrow{AD} $ , then find the value of k.
Options:
A) 2
B) 3
C) 4
D) 5
Show Answer
Answer:
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 $
BETA
