JEE Advanced 2017 Paper 2Question 1
Let $O$ be the origin and let $PQR$ be an arbitrary triangle. The point $S$ is such that
$$\overrightarrow{OP}\cdot\overrightarrow{OQ} + \overrightarrow{OR}\cdot\overrightarrow{OS} = \overrightarrow{OR}\cdot\overrightarrow{OP} + \overrightarrow{OQ}\cdot\overrightarrow{OS} = \overrightarrow{OQ}\cdot\overrightarrow{OR} + \overrightarrow{OP}\cdot\overrightarrow{OS}$$
Then the triangle $PQR$ has $S$ as its
Vector Algebra
Solution
Equating the first two expressions: $\overrightarrow{OS}\cdot(\overrightarrow{OR}-\overrightarrow{OQ})=\overrightarrow{OP}\cdot(\overrightarrow{OR}-\overrightarrow{OQ})$, i.e. $\overrightarrow{PS}\cdot\overrightarrow{QR}=0$. Likewise the other pair gives $\overrightarrow{QS}\cdot\overrightarrow{RP}=0$. So $S$ lies on two altitudes: $S$ is the orthocentre.