Vector Algebra Question 273
Question: If $ ABCDEF $ is regular hexagon, then $ \overrightarrow{AD},+\overrightarrow{EB}+\overrightarrow{FC}= $
[Karnataka CET 2002]
Options:
A) 0
B) $ 2\overrightarrow{AB} $
C) $ 3\overrightarrow{AB} $
D) $ 4\overrightarrow{AB} $
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Answer:
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} $ .
BETA
