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
- A \(2 \overrightarrow{ QP }\)
- B \(\overrightarrow{Q P}\)
- C \(2 \overrightarrow{ PQ }\)
- D \(\overrightarrow{ PQ }\)
Answer & Solution
Correct Answer
(D) \(\overrightarrow{ PQ }\)
Step-by-step Solution
Detailed explanation
\(\overline{ PA }+\overline{ PB }+\overline{ PC }=\overline{ a }+\overline{ b }+\overline{ c }\) \(\overline{ PG }=\frac{\overline{ a }+\overline{ b }+\overline{ c }}{3}\) \(\Rightarrow \overline{ a }+\overline{ b }+\overline{ c }=3 \overline{ PG }=\overline{ PQ }\)
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