If the points P and Q are respectively the circumcenter and the orthocentre of a ∆ABC, then PA→+PB→+PC→ is equal to _______

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) PQ→. Let P be origin. Now from the above triangle we can write PA→=a→, PA→=b→, PC→=c→. Also, we know that centroid G divides orthocentre Q and Circumcentre P in the ratio 2:1. Now PG→=(a→+b→+c→)/3 ⇒PQ→=3(a→+b→+c→)/3 ⇒PQ→=a→+b→+c→=PA→+PB→+PC→ Therefore, the required answer is PQ→.

If the points $P$ and $Q$ are respectively the circumcenter and the orthocentre of a $∆ A B C$ , then $\vec{P A} + \vec{P B} + \vec{P C}$ is equal to _______

Solution

Relevant Questions