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 the triangle is

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$- (\frac{\vec{a} + \vec{b} + \vec{c}}{2})$

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

$\vec{0}$

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

answer is D.

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Detailed Solution

Circumcentre $\vec{O P} = \frac{\vec{a} + \vec{b} + \vec{c}}{4}$

And we know that

Centroid, $\vec{O G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$

We have orthocentre, centroid and circumcentre are collinear and G divides the line in 2:1 ratio

$\vec{O G} = \frac{2 \vec{O P} + \vec{OH}}{3}$

$\vec{O H} = 3 \vec{O G} - 2 \vec{O P}$

$O \vec{H} = (\vec{a} + \vec{b} + \vec{c}) - (\frac{\vec{a} + \vec{b} + \vec{c}}{2}) = \frac{\vec{a} + \vec{b} + \vec{c}}{2}$

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