Let ABC be a triangle whose circumcentre is a | vedclass

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(C) Let $\vec{G}$ be the position vector of the centroid of $	riangle ABC$. We know that $\vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$.Let $\vec{

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

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(C) Let $\vec{G}$ be the position vector of the centroid of $	riangle ABC$. We know that $\vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$.Let $\vec{H}$ be the position vector of the orthocentre and $\vec{P}$ be the position vector of the circumcentre. We are given $\vec{P} = \frac{\vec{a} + \vec{b} + \vec{c}}{4}$.In any triangle,the centroid $\vec{G}$ divides the line segment joining the orthocentre $\vec{H}$ and the circumcentre $\vec{P}$ in the ratio $2:1$. That is,$\vec{G} = \frac{2\vec{P} + 1\vec{H}}{2+1} = \frac{2\vec{P} + \vec{H}}{3}$.Therefore,$3\vec{G} = 2\vec{P} + \vec{H}$,which implies $\vec{H} = 3\vec{G} - 2\vec{P}$.Substituting the values of $\vec{G}$ and $\vec{P}$:$\vec{H} = 3\left(\frac{\vec{a} + \vec{b} + \vec{c}}{3}ight) - 2\left(\frac{\vec{a} + \vec{b} + \vec{c}}{4}ight)$$\vec{H} = (\vec{a} + \vec{b} + \vec{c}) - \frac{\vec{a} + \vec{b} + \vec{c}}{2}$$\vec{H} = \frac{\vec{a} + \vec{b} + \vec{c}}{2}$.

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:

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