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Relationship between Unequal Sides of Triangle and the Angles Opposite to It.

Let ABC be a ...

Question

Let ABC be a triangle and $\to a$ , $\to b$ and $\to c$ be the position vectors of the point. $A, B$ and $C,$ respectively. External bisectors of $∠$ B $a n d$ \angle $C m e e t a t P w i t h t h e s i d e s o f t h e t r i a n g l e a s$ \vec a, \vec b $a n d$ \vec c $t h e p o s i t i o n v e c t o r s o f$ P$ becomes

A

\frac{(- b) b + (- c) c}{(b + c)}

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B

\frac{a a + (- b) b + (- c) c}{(a - b - c)}

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C

(\frac{a + b + c}{3}) (a b c)

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D

\frac{a a + b b + c c}{(a + b + c)}

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Solution

The correct option is C $\frac{a a + (- b) b + (- c) c}{(a - b - c)}$
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
$r = b + μ (\frac{b - a}{c} + \frac{c - b}{a})$
and $r = c + t (\frac{b - c}{a} + \frac{c - a}{b})$
So that for the point P,
$b + μ (\frac{b - a}{c} + \frac{c - b}{a}) = c + t (\frac{b - c}{a} + \frac{c - a}{b})$
$\Rightarrow t = \frac{b μ}{c}$ and $μ = \frac{a c}{b + c - a}$
So, the position vector of P is $\frac{b b + c c - a a}{b + c - a}$

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