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JEE Mains · Maths · STD 12 - 10. vector algebra

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

  1. A \(2 \overrightarrow{ QP }\)
  2. B \(\overrightarrow{Q P}\)
  3. C \(2 \overrightarrow{ PQ }\)
  4. D \(\overrightarrow{ PQ }\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\overrightarrow{ PQ }\)

Step-by-step Solution

Detailed explanation

\(\overline{ PA }+\overline{ PB }+\overline{ PC }=\overline{ a }+\overline{ b }+\overline{ c }\) \(\overline{ PG }=\frac{\overline{ a }+\overline{ b }+\overline{ c }}{3}\) \(\Rightarrow \overline{ a }+\overline{ b }+\overline{ c }=3 \overline{ PG }=\overline{ PQ }\)