P is the circumcentre of triangle ABC. If the | vedclass

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(B) Let the position vector of the orthocentre be $\bar{h}$.We know that for any triangle,the centroid $G$ divides the line segment joining the orthoc

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

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(B) Let the position vector of the orthocentre be $\bar{h}$.We know that for any triangle,the centroid $G$ divides the line segment joining the orthocentre $H$ and the circumcentre $P$ in the ratio $2:1$.The position vector of the centroid $G$ is given by $\bar{g} = \frac{\bar{a}+\bar{b}+\bar{c}}{3}$.Using the section formula,$\bar{g} = \frac{1(\bar{h}) + 2(\bar{p})}{1+2} = \frac{\bar{h} + 2\bar{p}}{3}$.Equating the two expressions for $\bar{g}$:$\frac{\bar{a}+\bar{b}+\bar{c}}{3} = \frac{\bar{h} + 2\bar{p}}{3}$.$\bar{h} + 2\bar{p} = \bar{a}+\bar{b}+\bar{c}$.Given $\bar{p} = \frac{\bar{a}+\bar{b}+\bar{c}}{4}$,substitute this into the equation:$\bar{h} + 2\left(\frac{\bar{a}+\bar{b}+\bar{c}}{4}\right) = \bar{a}+\bar{b}+\bar{c}$.$\bar{h} + \frac{\bar{a}+\bar{b}+\bar{c}}{2} = \bar{a}+\bar{b}+\bar{c}$.$\bar{h} = (\bar{a}+\bar{b}+\bar{c}) - \frac{\bar{a}+\bar{b}+\bar{c}}{2} = \frac{\bar{a}+\bar{b}+\bar{c}}{2}$.

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

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