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JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    Let ABC be a triangle whose circumventer is at P. If the position vectors of A,B,C and P are \[\overrightarrow{a}\,,\,\overrightarrow{b},\,\,\overrightarrow{c}\]and\[\frac{\overrightarrow{a}\,,\,+\overrightarrow{b},\,+\,\overrightarrow{c}}{4}\]respectively, then the position vector of the orthocenter of this triangle, is   JEE Main Online Paper (Held On 10 April 2016)

    A) \[\overrightarrow{0}\]                                     

    B) \[-\left( \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{2} \right)\]

    C) \[\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\]                   

    D) \[\left( \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{2} \right)\]

    Correct Answer: D

    Solution :

                 Position vector of the centroid of \[\Delta ABC\] is \[=\left( \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3} \right)\] Now we known that centroid divides the line joining orthocenter to circum centre divided by centriod divided by centroid in the ratio in 2 : 1orthocenter\[=3\left( \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3} \right)-2\left( \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{4} \right)=\left( \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{2} \right)\]


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