Discussion Forum

1)

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

Answer:

Option B

Explanation:

OP.OQ+OR.OS

=OR.OP+OQ.OS

$\Rightarrow$ OP(OQ-OR)+OS(OR-OQ)=0

$\Rightarrow$  (OP-OS)(OQ-OR)=0

$\Rightarrow$ SP.RQ=0

Similarly SR.PQ=0

and SQ.PR=0

 So   S is orthocentre