the points P and Q are respectively the circumcentre and the orthocentre of a ΔABC, then \( \overrightarrow{PA}+ \overrightarrow{PB}+ \overrightarrow{PC}\) is
- \(\overrightarrow{2QP}\)
- \(\overrightarrow{QP}\)
- \(\overrightarrow{PQ}\)
- \(\overrightarrow{2PQ}\)
The Correct Option is C
Solution and Explanation
Step 1: Using the centroid formula Let the points \( A, B, C \) have position vectors \( \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \), respectively. Since \( P \) and \( Q \) are the circumcenter and orthocenter of the triangle, respectively, we can use the following result: \[ \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \]
Step 2: Using the centroid formula The centroid \( G \) of the triangle is given by: \[ \overrightarrow{PG} = \frac{\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}}{3} \] Thus: \[ \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 3 \overrightarrow{PG} \] Therefore: \[ \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = 3 \overrightarrow{PG} = \overrightarrow{PQ} \]
Step 3: Conclusion Thus, \( \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = \overrightarrow{PQ} \), which is the correct option (3).
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