Vector Algebra Question 202
Question: A point O is the centre of a circle circumscribed about a triangle ABC. Then $ \overrightarrow{OA} $ sin 2A+ $ \overrightarrow{OB} $ sin 2B + $ \overrightarrow{OC} $ sin 2C is equal to
Options:
A) $ (\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})sin2A $
B) $ 3,\overrightarrow{OG} $ , where G is the centroid of triangle ABC
C) $ \overrightarrow{0} $
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] The position vector of the point O with respect to itself is $ \frac{\overrightarrow{OA}\sin 2A+\overrightarrow{OB}\sin 2B+\overrightarrow{OC}\sin 2C}{\sin 2A+\sin 2B+\sin 2C} $
$ \Rightarrow \frac{\overrightarrow{OA}\sin 2A+\overrightarrow{OB}\sin 2B+\overrightarrow{OC}\sin 2C}{\sin 2A+\sin 2B+\sin 2C}=\vec{0} $ Or $ \overrightarrow{OA}\sin 2A+\overrightarrow{OB}\sin 2B+\overrightarrow{OC}\sin 2C=\vec{0} $
BETA
