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MHT CET202210 Aug 2022Morning ShiftMathematicsVector AlgebraActual

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

Options

  1. AIncentre.
  2. BCentroid.
  3. COrthocentre.
  4. DCircumcentre.

Correct answer

C. Orthocentre.

Step-by-step solution

aligned & O P O Q + O R O S = O R O P + O Q O S & O P ( O Q - O R )= O S ( O Q - O R ) & O P R Q = O S R Q & R Q ( O P - O S )=0 & R Q P S =0 & P S Q R aligned Similarly Q S P R and R S P Q i.e., S is the orthocenter

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