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JEE Main 2017

JEE Main 2017 Mathematics question on Vector Algebra

Let OO be the origin and let PQRPQR be an arbitrary triangle. The point SS is such that
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 the triangle PQRPQR has SS as its
  1. A.
    centroid
  2. B.
    orthocentre
  3. C.
    incentre
  4. D.
    circumcentre
Show answer and solution

Correct option: B

Solution

From the first equality: OP·OQ + OR·OS = OR·OP + OQ·OS

Rearranging: OP·OQ - OR·OP = OQ·OS - OR·OS

OP·(OQ - OR) = OS·(OQ - OR)

(OP - OS)·(OQ - OR) = 0

This means SP·QR = 0, i.e., SP ⊥ QR.

From the second equality: OR·OP + OQ·OS = OQ·OR + OP·OS

OR·(OP - OQ) = OS·(OP - OQ)

(OR - OS)·(OP - OQ) = 0

This means SR·PQ = 0, i.e., SR ⊥ PQ.

So S is a point from which SP ⊥ QR and SR ⊥ PQ — meaning S lies on two altitudes of triangle PQR simultaneously, making S the orthocentre.

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