Q.

Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that QAAR=RBBP=PCCQ=12. Then Area(∆PQR)Area(∆ABC) is equal to              [2023]

1 4  
2 3  
3 2  
4 5/2  

Ans.

(2)

Let the position vectors of P,Q,R be 0→, a→, b→.

⇒ P.V. of A=2a→+b→3, P.V. of B=2b→3

and P.V. of C=a→3

∴  AB→=2b→3-(2a→+b→3)=b→3-2a→3=b→-2a→3

CA→=2a→+b→3-a→3=a→+b→3

Area of ∆PQR=12|PQ→×PR→|=12|a→×b→|

Area of ∆ABC=12|CA→×AB→|

=12|(a→+b→3)×(b→-2a→3)|=12|a→×b→3|

∴ Area(∆PQR)Area(∆ABC)=12|a→×b→|12|a→×b→3|=3