JEE Main & Advanced Mathematics Vector Algebra Question Bank Mock Test - Vector Algebra

  • question_answer
    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

    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

    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 \[\overrightarrow{OA}\sin 2A+\overrightarrow{OB}\sin 2B+\overrightarrow{OC}\sin 2C=\vec{0}\]


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