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(D) Given the equations for the position vector $\bar{x}$ of point $P$: $1$) $\bar{x} \cdot (\bar{c} - \bar{b}) = \bar{a} \cdot \bar{c} - \bar{a} \cdo

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Orthocentre

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(D) Given the equations for the position vector $\bar{x}$ of point $P$: $1$) $\bar{x} \cdot (\bar{c} - \bar{b}) = \bar{a} \cdot \bar{c} - \bar{a} \cdot \bar{b} \implies (\bar{x} - \bar{a}) \cdot (\bar{c} - \bar{b}) = 0$. This implies that the vector $\vec{AP}$ is perpendicular to the side $BC$. $2$) $\bar{x} \cdot (\bar{a} - \bar{c}) = \bar{a} \cdot \bar{b} - \bar{b} \cdot \bar{c} \implies (\bar{x} - \bar{b}) \cdot (\bar{a} - \bar{c}) = 0$. This implies that the vector $\vec{BP}$ is perpendicular to the side $AC$. Since $P$ is a point such that $\vec{AP} \perp BC$ and $\vec{BP} \perp AC$,$P$ is the intersection of the altitudes of the triangle $ABC$. Therefore,$P$ is the orthocentre of the triangle $ABC$.

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