If the points $$P$$ and $$Q$$ are respectively the circumcenter and the orthocentre of a $$\triangle ABC$$, then $$\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$$ is equal to

Let the position vectors of the vertices be $$\vec{A}, \vec{B}, \text{ and } \vec{C}$$.

• $$P$$ (Circumcenter): Let $$P$$ be the origin for our vector calculations ($$\vec{P} = \vec{0}$$).
• $$G$$ (Centroid): The position vector of the centroid relative to $$P$$ is given by:

$$\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}$$

• $$Q$$ (Orthocenter): We need to find the vector $$\vec{PQ}$$.

A fundamental property in triangle geometry is that the circumcenter ($$P$$), centroid ($$G$$), and orthocenter ($$Q$$) are collinear (forming the Euler Line), and the centroid divides the segment $$PQ$$ in a ratio of $$1:2$$.

Specifically:

$$\vec{G} = \frac{2\vec{P} + 1\vec{Q}}{3}$$

Since we set $$P$$ as the origin ($$\vec{P} = \vec{0}$$), this simplifies to:

$$\vec{G} = \frac{\vec{Q}}{3} \implies \vec{Q} = 3\vec{G}$$

Substitute the formula for $$\vec{G}$$ into the expression for $$\vec{Q}$$:

$$\vec{Q} = 3 \left( \frac{\vec{A} + \vec{B} + \vec{C}}{3} \right)$$

$$\vec{Q} = \vec{A} + \vec{B} + \vec{C}$$

The question asks for the value of $$\vec{PA} + \vec{PB} + \vec{PC}$$.

Because we designated $$P$$ as the origin, $$\vec{PA} = \vec{A}$$, $$\vec{PB} = \vec{B}$$, and $$\vec{PC} = \vec{C}$$.

Therefore:

$$\vec{PA} + \vec{PB} + \vec{PC} = \vec{A} + \vec{B} + \vec{C}$$

From our previous step, we know that $$\vec{A} + \vec{B} + \vec{C} = \vec{Q}$$. In vector notation relative to point $$P$$, $$\vec{Q}$$ is represented as the vector $$\vec{PQ}$$.

$$\vec{PA} + \vec{PB} + \vec{PC} = \vec{PQ}$$

Final Answer: C ($$\vec{PQ}$$)