Verified Solution • 2025–26 Session moderate

QJEEAdv17Maf0bc: JEE Advanced 2017 Physics - Class 11 JEEAdv17Maf0bc Solution: Let O be the origin and let PQR be an arbitrary triangle. The…

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that

OP→\overrightarrow{OP} . OQ→\overrightarrow{OQ} + OR→\overrightarrow{OR} . OS→\overrightarrow{OS} = OR→\overrightarrow{OR} . OP→\overrightarrow{OP} + OQ→\overrightarrow{OQ} . OS→\overrightarrow{OS} = OQ→\overrightarrow{OQ} . OR→\overrightarrow{OR} + OP→\overrightarrow{OP} . OS→\overrightarrow{OS}

Then the triangle PQR has S as its

Multiple Choice Options

  • A centroid
  • B orthocentre
  • C incentre
  • D circumcentre

Step-by-Step Verified Solution

OP→\overrightarrow{OP} . OQ→\overrightarrow{OQ} + OR→\overrightarrow{OR} . OS→\overrightarrow{OS} = OR→\overrightarrow{OR} . OP→\overrightarrow{OP} + OQ→\overrightarrow{OQ} . OS→\overrightarrow{OS}

⇒\Rightarrow OP→\overrightarrow{OP} ( OQ→\overrightarrow{OQ} −- OR→\overrightarrow{OR} ) + OS→\overrightarrow{OS} ( OR→\overrightarrow{OR} −- OQ→\overrightarrow{OQ} ) = 0

⇒\Rightarrow ( OP→\overrightarrow{OP} −- OS→\overrightarrow{OS} )( OQ→\overrightarrow{OQ} −- OR→\overrightarrow{OR} ) = 0

⇒\Rightarrow SP→\overrightarrow{SP} . RQ→\overrightarrow{RQ} = 0

Similarly SR→\overrightarrow{SR} . PQ→\overrightarrow{PQ} = 0 and SQ→\overrightarrow{SQ} . PR→\overrightarrow{PR} = 0

∴S→\therefore \overrightarrow{S} is orthocentre.

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