Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are $\vec a, \vec b, \vec c$ and $\frac{\vec a+\vec b+\vec c}{4}$ respectively, then the position vector of the orthocentre of this triangle, is

$\frac{\vec a+\vec b+\vec c}{2}$

Let G be the centroid of A ABC. Then, the position vector of G is $\frac{\vec a+\vec b+\vec c}{3}$. Let the  position vector of orthocentre O be $\vec α$.

We know that O, G and P are collinear and G divides OP in the ratio 2: 1.

$∴\frac{2\left(\frac{\vec a+\vec b+\vec c}{4}\right)+1.\vec α}{2+1}=\frac{\vec a+\vec b+\vec c}{3}$

$⇒\frac{\vec a+\vec b+\vec c}{2}+\vec α=\vec a+\vec b+\vec c⇒\vec α=\frac{\vec a+\vec b+\vec c}{2}$