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MathematicsSingle correct+3 / -1Question 38 of 54

JEE Advanced 2017 · Paper 2, Mathematics question 2

Let OO be the origin and let PQRPQR be an arbitrary triangle. The point SS is such that

OP→⋅OQ→+OR→⋅OS→=OR→⋅OP→+OQ→⋅OS→=OQ→⋅OR→+OP→⋅OS→\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 PQRPQR has SS as its

Choose one option
Answer and worked solution

Official answer: D

Rearrange. Write p⃗,q⃗,r⃗,s⃗\vec p,\vec q,\vec r,\vec s for the position vectors. The first equality gives
p⃗⋅q⃗+r⃗⋅s⃗=r⃗⋅p⃗+q⃗⋅s⃗  ⟹  (p⃗−s⃗)⋅(q⃗−r⃗)=0.\vec p\cdot\vec q+\vec r\cdot\vec s=\vec r\cdot\vec p+\vec q\cdot\vec s\implies(\vec p-\vec s)\cdot(\vec q-\vec r)=0.

So SP⊥QRSP\perp QR. The other equalities give, in the same way, SQ⊥RPSQ\perp RP and SR⊥PQSR\perp PQ.

SS lies on all three altitudes: it is the orthocentre.

(D)

Topics:Dot product
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