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JEE Advanced2017MathematicsVector AlgebraActual

Let O be the origin and let P Q R be an arbitrary triangle. The point S is such that O P → . O Q → + O R → . O S → = O R → . O P → + O Q → . O S → = O Q → . O R → + O P → . O S → then triangle P Q R has S as its

Options

  1. AIncentre
  2. BOrthocentre
  3. CCircumcentre
  4. DCentroid

Correct answer

B. Orthocentre

Step-by-step solution

Let position vector of P p → , Q q → , R r → a n d S r → with respect to O o → Now, O P → . O Q → + O R → . O S → = O R → . O P → + O Q → . O S → ⇒ p → . q → + r → . s → = r → . p → + q → . s → ⇒ p → - s → . q → - r → = 0 .....(i) Also, O R → . O P → + O Q → . O S → = O Q → . O R → + O P → . O S → ⇒ r → . p → + q → . s → = q → . r → + p → . s → ⇒ r → - s → . p → - q → = 0 .....(ii) Also O P → . O Q → + O R → . O S → = O Q → . O R → + O P → . O S → ⇒ p → . q → + r → . s → = q → . r → + p → . s → ⇒ q → - s → . p → -

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