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Question

Let ABC be a triangle and a, b and c be the position vectors of the point. A,Band C, respectively. External bisectors of B and\angleCmeetatPwiththesidesofthetriangleas\vec a, \vec b and\vec cthepositionvectorsofP$ becomes
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A
(b)b+(c)c(b+c)
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B
aa+(b)b+(c)c(abc)
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C
(a+b+c3)(abc)
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D
aa+bb+cc(a+b+c)
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Solution

The correct option is C aa+(b)b+(c)c(abc)
Vectors along AB, BC and CA are b - c, c - b and a - c, respectively. Hence, the bisectors of the angular B and C, respectively are
r=b+μ(bac+cba)
and r=c+t(bca+cab)
So that for the point P,
b+μ(bac+cba)=c+t(bca+cab)
t=bμc and μ=acb+ca
So, the position vector of P is bb+ccaab+ca

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