Question

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that
\[\overrightarrow{OP}.\overrightarrow{OQ}+\overrightarrow{OR}.\overrightarrow{OS}=\overrightarrow{OR}.\overrightarrow{OP}+\overrightarrow{OQ}.\overrightarrow{OS}=\overrightarrow{OQ}.\overrightarrow{OR}+\overrightarrow{OP}.\overrightarrow{OS}\]
Then the triangle PQR has S as its
A. Circumcentre
B. Orthocentre
C. Incentre
D. Centroid

Answer 1

Hint: First of all, draw a triangle PQR. Join the vertices of the triangle from origin. Take a point S and join it with origin. Try to find out the relation between sides of the triangle in vector form. In this case use the vector law of addition. In any triangle ABC vector law of addition says that $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{0}$.
And properties of vector \[\overrightarrow{AB}=-\overrightarrow{BA}\]

Complete step-by-step answer:
Let PQR be any arbitrary triangle and O be the origin so we have the following figure, S be any point











From question it is given that
\[\begin{align}
  & \overrightarrow{OP}.\overrightarrow{OQ}+\overrightarrow{OR}.\overrightarrow{OS}=\overrightarrow{OR}.\overrightarrow{OP}+\overrightarrow{OQ}.\overrightarrow{OS} \\
 & \Rightarrow \overrightarrow{OP}.\overrightarrow{OQ}+\overrightarrow{OR}.\overrightarrow{OS}-\overrightarrow{OR}.\overrightarrow{OP}+\overrightarrow{OQ}.\overrightarrow{OS}=0 \\
\end{align}\]
We take $\overrightarrow{OP}$as common from first and third term and $\overrightarrow{OS}$from second and fourth term so we can write
$\begin{align}
  & \left( \overrightarrow{OQ}-\overrightarrow{OR} \right).\overrightarrow{OP}-\overrightarrow{OS}.\left( \overrightarrow{OQ}-\overrightarrow{OR} \right)=0 \\
 & \Rightarrow \left( \overrightarrow{OQ}-\overrightarrow{OR} \right).\left( \overrightarrow{OP}-\overrightarrow{OS} \right)=0----(a) \\
\end{align}$
As we know that from triangle law addition in any triangle OQR
\[\begin{align}
  & \overrightarrow{OQ}+\overrightarrow{QR}+\overrightarrow{RO}=\overrightarrow{0} \\
 & \Rightarrow \overrightarrow{QR}=-(\overrightarrow{OQ}+\overrightarrow{RO}) \\
 & \Rightarrow -\overrightarrow{RQ}=-(\overrightarrow{OQ}-\overrightarrow{OR}) \\
 & \Rightarrow \overrightarrow{RQ}=\overrightarrow{OQ}-\overrightarrow{OR} \\
\end{align}\]
Similarly, we can write
$\overrightarrow{OP}-\overrightarrow{OS}=\overrightarrow{PS}$
Hence equation $(a)$ can be written as
$\overrightarrow{RQ.}\overrightarrow{PS}=\overrightarrow{0}$
As we know that if DOT product of two vectors is zero the both vectors are perpendicular
So, we can write
$\overrightarrow{RQ}\bot \overrightarrow{PS}-----(A)$
Similarly, we can solve the second equality as
\[\begin{align}
  & \overrightarrow{OR.}\overrightarrow{OP}+\overrightarrow{OQ}.\overrightarrow{OS}=\overrightarrow{OQ}.\overrightarrow{OR}+\overrightarrow{OP}.\overrightarrow{OS} \\
 & \Rightarrow \overrightarrow{OR}(\overrightarrow{OP}-\overrightarrow{OQ})-\overrightarrow{OS}(\overrightarrow{OP}-\overrightarrow{OQ})=\overrightarrow{0} \\
 & \Rightarrow (\overrightarrow{OP}-\overrightarrow{OQ}).(\overrightarrow{OR}-\overrightarrow{OS})=\overrightarrow{0} \\
\end{align}\]
In above we use triangle law of addition
Further we can write
$\overrightarrow{PQ}.\overrightarrow{RS}=\overrightarrow{0}$
Hence, we can write
$\overrightarrow{PQ}\bot \overrightarrow{RS}-----(B)$
Hence, we can say from (A) and (B)
S is the orthocentre of triangle PQR. Hence option B is correct.

NOTE: The orthocentre of a triangle is the point of intersection of altitude from vertices to the side containing other two angles. In case of right angle triangle, it is the vertex at which angle is $90{}^\circ $.In case of acute angle triangle Orthocentre lies inside the triangle and in case of obtuse triangle it lies outside the triangle.