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 the triangle P Q R has S as its

Question

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 the triangle P Q R has S as its (A) centroid (B) circumcentre (C) incentre (D) orthocenter

Tardigrade Admin · Accepted Answer

OP ⋅ OQ + OR ⋅ OS = OR ⋅ OP + OQ ⋅ OS ⇒ OP ⋅( OQ - OR )= OS ⋅( OQ - OR ) ⇒ ( OP - OS ) ⋅( RQ )=0 ( SP ) ⋅( RQ )=0 ⇒ SP perp overline RQ Similarly SR perp QP and SQ perp PR Hence, S is orthocentre

Q. Let $O$ be the origin and let $PQR$ be an arbitrary triangle. The point $S$ is such that
$OP \cdot OQ + OR \cdot OS = OR \cdot OP + OQ \cdot OS = OQ \cdot OR + OP \cdot OS$
Then the triangle $PQR$ has $S$ as its

centroid

circumcentre

incentre

orthocenter

$OP \cdot OQ + OR \cdot OS = OR \cdot OP + OQ \cdot OS$
$\Rightarrow OP \cdot (OQ - OR) = OS \cdot (OQ - OR)$
$\Rightarrow (OP - OS) \cdot (RQ) = 0$
$(SP) \cdot (RQ) = 0$
$\Rightarrow SP ⊥ \overline{RQ}$
Similarly $SR ⊥ QP$ and $SQ ⊥ PR$
Hence, $S$ is orthocentre

Q. Let O be the origin and let PQR be an arbitrary triangle. The point S is such that OP⋅OQ​+OR⋅OS=OR⋅OP+OQ​⋅OS=OQ​⋅OR+OP⋅OS Then the triangle PQR has S as its