MathematicsHard202×since 2002Q5687Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are a→,b→,c→\overrightarrow a ,\overrightarrow b ,\overrightarrow ca,b,c and a→+b→+c→4{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}4a+b+c respectively, then the position vector of the orthocentre of this triangle, is :Aa→+b→+c→{\overrightarrow a + \overrightarrow b + \overrightarrow c }a+b+cB−(a→+b→+c→2)- \left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)−(2a+b+c)C0→\overrightarrow 00D(a→+b→+c→2)\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)(2a+b+c)Check answerSkip