Multiple choice

A point $O$ is the centre of a circle circumscribed about a triangle $ABC$, then $\vec{OA}\sin 2A + \vec{OB}\sin 2B + \vec{OC} \sin 2C $ is equal to

  1. $(\vec{OA} + \vec{OB} + \vec{OC})\sin 2A$
  2. $3\vec{OG}$, where $G$ is the centroid of triangle $ABC$
  3. $\vec 0$

  4. None of these
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C Correct answer
Explanation

Using the properties of a circumcenter O, the vector OA sin 2A + OB sin 2B + OC sin 2C is a known identity in triangle geometry. Since sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C, the vector sum simplifies to the zero vector.

AI explanation

The given expression represents the vector equation of the circumcentre O for triangle ABC. Since O is the origin for the position vectors OA, OB, and OC, substituting the circumcentre into the equation yields the zero vector. Therefore, the vector sum simplifies to 0.

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