Question

the points P and Q are respectively the circumcentre and the orthocentre of a ΔABC, then \( \overrightarrow{PA}+ \overrightarrow{PB}+ \overrightarrow{PC}\)  is 

• A. \(\overrightarrow{2QP}\)
• B. \(\overrightarrow{QP}\)
• C. \(\overrightarrow{PQ}\)
• D. \(\overrightarrow{2PQ}\)

Answer 1

The Correct Option is C

To solve the problem, we need to determine the vector sum \( \overrightarrow{PA}+ \overrightarrow{PB}+ \overrightarrow{PC} \) given that P is the circumcentre and Q is the orthocentre of triangle ΔABC.

We'll use some properties of triangle geometry to find this vector expression:

1. Circumcentre Property: The circumcentre (P) is the point equidistant from all vertices of the triangle, so vectors from the circumcentre to the vertices have a balanced, symmetrical property around the center.
2. Orthocentre Property: The orthocentre (Q) of a triangle is the common intersection point of the altitudes. It can be considered as a vector outcome of combining properties related to altitudes of the triangle.

For any triangle, there is a relation between the orthocentre (Q), circumcentre (P), and centroid (G). Specifically, the relation between these points is given by:

\[ \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = 3 \overrightarrow{PG} \] ,

where G is the centroid given by the average of the position vectors of the vertices of the triangle.

We also know the following relation due to the Euler line property:

\[ \overrightarrow{QG} = 2 \overrightarrow{GP} \]

Hence, substituting the relation based on the centroid properties and rearranging the Euler line relation, we have:

\[ \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = 3 \cdot \left(\frac{2}{3} \overrightarrow{PQ} + \frac{1}{3} \overrightarrow{PQ}\right) = \overrightarrow{PQ} \]

Thus, the vector expression \( \overrightarrow{PA}+ \overrightarrow{PB}+ \overrightarrow{PC} \) turns out to be \(\overrightarrow{PQ}\).

Therefore, the correct answer is: \(\overrightarrow{PQ}\).