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Vector Algebra question

2017 · Shift 2 · Q24
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  5. /2017 · Shift 2 · Q24

Vector Algebra question

2017 · Shift 2 · Q24

JEE AdvancedMathematicsVector AlgebraMCQ+3 / −1
Let O be the origin and let PQR be an arbitrary triangle. The point S is such that OP→\overrightarrow{OP}OP. OQ→\overrightarrow{OQ}OQ​+OR→\overrightarrow{OR}OR. OS→\overrightarrow{OS}OS=OR→\overrightarrow{OR}OR. OP→\overrightarrow{OP}OP+OQ→\overrightarrow{OQ}OQ​. OS→\overrightarrow{OS}OS=OQ→\overrightarrow{OQ}OQ​. OR→\overrightarrow{OR}OR+OP→\overrightarrow{OP}OP. OS→\overrightarrow{OS}OS Then the triangle PQR has S as its
  1. A
    centroid
  2. B
    orthocentre
  3. C
    incentre
  4. D
    circumcentre
View written solutionFree

Correct answer: B

Let p⃗=OP→,q⃗=OQ→,r⃗=OR→,s⃗=OS→.\vec p=\overrightarrow{OP},\quad \vec q=\overrightarrow{OQ},\quad \vec r=\overrightarrow{OR},\quad \vec s=\overrightarrow{OS}.p​=OP,q​=OQ​,r=OR,s=OS.

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.p​⋅q​+r⋅s=r⋅p​+q​⋅s=q​⋅r+p​⋅s.

We must identify what special point SSS is for triangle PQRPQRPQR.

1. Equate the first two expressions

Since 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,p​⋅q​+r⋅s=r⋅p​+q​⋅s, we get p⃗⋅q⃗−r⃗⋅p⃗=q⃗⋅s⃗−r⃗⋅s⃗.\vec p\cdot \vec q-\vec r\cdot \vec p=\vec q\cdot \vec s-\vec r\cdot \vec s.p​⋅q​−r⋅p​=q​⋅s−r⋅s. Factor both sides: p⃗⋅(q⃗−r⃗)=s⃗⋅(q⃗−r⃗).\vec p\cdot(\vec q-\vec r)=\vec s\cdot(\vec q-\vec r).p​⋅(q​−r)=s⋅(q​−r). Hence, (p⃗−s⃗)⋅(q⃗−r⃗)=0.(\vec p-\vec s)\cdot(\vec q-\vec r)=0.(p​−s)⋅(q​−r)=0.

But p⃗−s⃗=SP→,q⃗−r⃗=RQ→.\vec p-\vec s=\overrightarrow{SP},\qquad \vec q-\vec r=\overrightarrow{RQ}.p​−s=SP,q​−r=RQ​. So, SP→⋅RQ→=0.\overrightarrow{SP}\cdot \overrightarrow{RQ}=0.SP⋅RQ​=0. Thus, SP⊥QR.SP\perp QR.SP⊥QR.

2. Equate the second and third expressions

Now, 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.r⋅p​+q​⋅s=q​⋅r+p​⋅s. So, r⃗⋅p⃗−q⃗⋅r⃗=p⃗⋅s⃗−q⃗⋅s⃗.\vec r\cdot \vec p-\vec q\cdot \vec r=\vec p\cdot \vec s-\vec q\cdot \vec s.r⋅p​−q​⋅r=p​⋅s−q​⋅s. Factor: r⃗⋅(p⃗−q⃗)=s⃗⋅(p⃗−q⃗).\vec r\cdot(\vec p-\vec q)=\vec s\cdot(\vec p-\vec q).r⋅(p​−q​)=s⋅(p​−q​). Hence, (r⃗−s⃗)⋅(p⃗−q⃗)=0.(\vec r-\vec s)\cdot(\vec p-\vec q)=0.(r−s)⋅(p​−q​)=0. That is, SR→⋅QP→=0.\overrightarrow{SR}\cdot \overrightarrow{QP}=0.SR⋅QP​=0. Therefore, SR⊥PQ.SR\perp PQ.SR⊥PQ.

3. Equate the first and third expressions

Also, p⃗⋅q⃗+r⃗⋅s⃗=q⃗⋅r⃗+p⃗⋅s⃗.\vec p\cdot \vec q+\vec r\cdot \vec s=\vec q\cdot \vec r+\vec p\cdot \vec s.p​⋅q​+r⋅s=q​⋅r+p​⋅s. Thus, p⃗⋅q⃗−q⃗⋅r⃗=p⃗⋅s⃗−r⃗⋅s⃗.\vec p\cdot \vec q-\vec q\cdot \vec r=\vec p\cdot \vec s-\vec r\cdot \vec s.p​⋅q​−q​⋅r=p​⋅s−r⋅s. Factor: q⃗⋅(p⃗−r⃗)=s⃗⋅(p⃗−r⃗).\vec q\cdot(\vec p-\vec r)=\vec s\cdot(\vec p-\vec r).q​⋅(p​−r)=s⋅(p​−r). Hence, (q⃗−s⃗)⋅(p⃗−r⃗)=0.(\vec q-\vec s)\cdot(\vec p-\vec r)=0.(q​−s)⋅(p​−r)=0. So, SQ→⋅PR→=0,\overrightarrow{SQ}\cdot \overrightarrow{PR}=0,SQ​⋅PR=0, which gives SQ⊥PR.SQ\perp PR.SQ⊥PR.

4. Geometrical interpretation

We have shown:

  • SP⊥QRSP\perp QRSP⊥QR
  • SQ⊥PRSQ\perp PRSQ⊥PR
  • SR⊥PQSR\perp PQSR⊥PQ

So SSS lies on all three altitudes of triangle PQRPQRPQR.

Therefore, SSS is the orthocentre of triangle PQRPQRPQR.

5. Checking options

  • A: centroid — false
  • B: orthocentre — true
  • C: incentre — false
  • D: circumcentre — false

Hence the correct option is: B: orthocentre\boxed{\text{B: orthocentre}}B: orthocentre​

6. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So the answer agrees with the stored correct answer.

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