JEE Challenger
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MathMedAdv 2017 • Paper 2

Point S Relation in Vector Product with Origin

Let OO represent the origin and PQRPQR be any triangle. Suppose a point SS satisfies OP⃗⋅OQ⃗+OR⃗⋅OS⃗=OR⃗⋅OP⃗+OQ⃗⋅OS⃗=OQ⃗⋅OR⃗+OP⃗⋅OS⃗\vec{OP} \cdot \vec{OQ} + \vec{OR} \cdot \vec{OS} = \vec{OR} \cdot \vec{OP} + \vec{OQ} \cdot \vec{OS} = \vec{OQ} \cdot \vec{OR} + \vec{OP} \cdot \vec{OS} Then for the triangle PQRPQR, the point SS serves as its

Options

A

centroid

B

circumcentre

C

incentre

D

orthocenter

Correct

Topics & Concepts

Vector AlgebraDot and Cross Product

Step-by-Step Solution

To determine the nature of the point SS relative to the triangle PQRPQR, let us denote the position vectors of the points P,Q,R,P, Q, R, and SS with respect to the origin OO as p⃗,q⃗,r⃗,\vec{p}, \vec{q}, \vec{r}, and s⃗\vec{s} respectively.

That is: OP⃗=p⃗,OQ⃗=q⃗,OR⃗=r⃗,andOS⃗=s⃗\vec{OP} = \vec{p}, \quad \vec{OQ} = \vec{q}, \quad \vec{OR} = \vec{r}, \quad \text{and} \quad \vec{OS} = \vec{s}

The given condition is: p⃗⋅q⃗+r⃗⋅s⃗=r⃗⋅p⃗+q⃗⋅s⃗=q⃗⋅r⃗+p⃗⋅s⃗\vec{p} \cdot \vec{q} + \vec{r} \cdot \vec{s} = \vec{r} \cdot \vec{p} + \vec{q} \cdot \vec{s} = \vec{q} \cdot \vec{r} + \vec{p} \cdot \vec{s}

Step 1: Equating the first and second terms

p⃗⋅q⃗+r⃗⋅s⃗=r⃗⋅p⃗+q⃗⋅s⃗\vec{p} \cdot \vec{q} + \vec{r} \cdot \vec{s} = \vec{r} \cdot \vec{p} + \vec{q} \cdot \vec{s}

Rearranging the terms: p⃗⋅q⃗−r⃗⋅p⃗+r⃗⋅s⃗−q⃗⋅s⃗=0\vec{p} \cdot \vec{q} - \vec{r} \cdot \vec{p} + \vec{r} \cdot \vec{s} - \vec{q} \cdot \vec{s} = 0 p⃗⋅(q⃗−r⃗)−s⃗⋅(q⃗−r⃗)=0\vec{p} \cdot (\vec{q} - \vec{r}) - \vec{s} \cdot (\vec{q} - \vec{r}) = 0 (p⃗−s⃗)⋅(q⃗−r⃗)=0(\vec{p} - \vec{s}) \cdot (\vec{q} - \vec{r}) = 0

Since p⃗−s⃗=SP⃗\vec{p} - \vec{s} = \vec{SP} and q⃗−r⃗=RQ⃗\vec{q} - \vec{r} = \vec{RQ}, this yields: SP⃗⋅RQ⃗=0  ⟹  SP⊥QR\vec{SP} \cdot \vec{RQ} = 0 \implies SP \perp QR

This means that SPSP is perpendicular to the side QRQR. Thus, SPSP lies along the altitude from vertex PP to side QRQR.

Step 2: Equating the second and third terms

r⃗⋅p⃗+q⃗⋅s⃗=q⃗⋅r⃗+p⃗⋅s⃗\vec{r} \cdot \vec{p} + \vec{q} \cdot \vec{s} = \vec{q} \cdot \vec{r} + \vec{p} \cdot \vec{s}

Rearranging the terms: r⃗⋅p⃗−q⃗⋅r⃗+q⃗⋅s⃗−p⃗⋅s⃗=0\vec{r} \cdot \vec{p} - \vec{q} \cdot \vec{r} + \vec{q} \cdot \vec{s} - \vec{p} \cdot \vec{s} = 0 r⃗⋅(p⃗−q⃗)−s⃗⋅(p⃗−q⃗)=0\vec{r} \cdot (\vec{p} - \vec{q}) - \vec{s} \cdot (\vec{p} - \vec{q}) = 0 (r⃗−s⃗)⋅(p⃗−q⃗)=0(\vec{r} - \vec{s}) \cdot (\vec{p} - \vec{q}) = 0

Since r⃗−s⃗=SR⃗\vec{r} - \vec{s} = \vec{SR} and p⃗−q⃗=QP⃗\vec{p} - \vec{q} = \vec{QP}, this yields: SR⃗⋅QP⃗=0  ⟹  SR⊥PQ\vec{SR} \cdot \vec{QP} = 0 \implies SR \perp PQ

This means that SRSR lies along the altitude from vertex RR to side PQPQ.

Step 3: Equating the first and third terms

Similarly, equating the first and third terms yields: (q⃗−s⃗)⋅(p⃗−r⃗)=0  ⟹  SQ⃗⋅RP⃗=0  ⟹  SQ⊥PR(\vec{q} - \vec{s}) \cdot (\vec{p} - \vec{r}) = 0 \implies \vec{SQ} \cdot \vec{RP} = 0 \implies SQ \perp PR

Conclusion

Since the line segments SP,SQ,SP, SQ, and SRSR are perpendicular to the opposite sides QR,PR,QR, PR, and PQPQ respectively, the point SS is the point of concurrence of the altitudes of △PQR\triangle PQR.

Therefore, SS serves as the orthocenter of triangle PQRPQR.

Correct Answer: [D] orthocenter

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