Let ABC be a triangle whose circumcentre is at P.

Question

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 $\dfrac{\vec a+\vec b+\vec c}4$ respectively, then the position vector of the orthocentre of this triangle is:

Accepted Answer

Centroid $\vec G=\dfrac{\vec a+\vec b+\vec c}3$. Since $G$ divides the segment from orthocentre $H$ to circumcentre $C_p$ in ratio $2:1$, $\vec G=\dfrac{2\vec C_p+\vec H}3$, so:

$$\vec H = 3\vec G-2\vec C_p = (\vec a+\vec b+\vec c)-\frac{\vec a+\vec b+\vec c}2 = \frac{\vec a+\vec b+\vec c}2$$

Solutions · 1