Question

Let $ABC$ be a triangle and $\overline{a},\overline{b},\overline{c}$ be the position vectors of the vertices $A,B,C$ respectively, the internal bisectors of angle $C$ and the external bisectors of $\angle A$ and $\angle B$ meet at a point $P$ if$a,b,c$ be the lengths of the sides $BC,CA$ and $AB$ respectively then position vector of point $P$ if vertex $C$ referred to origin is

• A. $\frac{a\overline{a}-b\overline{b}-c\overline{c}}{a-b-c}$
• B. $\frac{a\overline{a}-b\overline{b}-c\overline{c}}{a+b+c}$
• C. $\frac{\overline{a}\times \overline{b}+\overline{b}\times \overline{c}+\overline{c}\times \overline{a}}{a^2+b^2+c^2}$
• D. $\frac{a\overline{a}+b\overline{b}}{a+b-c}$

Answer 1

The correct option is D $\frac{a\overline{a}+b\overline{b}}{a+b-c}$

As vertex' $C$ referred as origin $\overline{CA}=\overline{a},\overline{CB}=\overline{b}$
$\therefore$ vector equation of internal bisector of $C$ is given by
$\overline{r}={\lambda }_1(\frac{\overline{CA}}{CA}+\frac{\overline{CB}}{CB})$
$\Rightarrow \overline{r}={\lambda }_1(\frac{\overline{a}}{b}+\frac{\overline{b}}{a})={\lambda }_1\left(\frac{a\overline{a}+b\overline{b}}{ab}\right)=\frac{{\lambda }_1}{ab}(a\overline{a}+b\overline{b})$ ...............(i)
Now external bisector of $A$ is the internal bisectors of the
angle between the vectors $\overline{CA}$ and $\overline{AB}$ ' Therefore vector equation of the external bisectors of $A$ is
$\overline{r}=\overline{a}+{\lambda }_2\{\frac{\overline{CA}}{CA}+\frac{\overline{AB}}{AB}\}={\lambda }_2\{\frac{\overline{a}}{b}+\frac{\overline{b}-\overline{a}}{c}\}+\overline{a}$ ..........(ii)
Similarly the external vector of $B$ is the internal bisectors of the angle between the vectors $\overline{CB}$ & $\overline{BA}$ Therefore, the vector equation ofthe external bisector of $B$ is given by
$\overline{r}=\overline{b}+{\lambda }_3\{\frac{\overline{CB}}{CB}+\frac{\overline{BA}}{BA}\}=b+{\lambda }_3\{\frac{\overline{b}}{a}+\frac{\overline{a}-\overline{b}}{c}\}$ ................(iii)
Now Equating (ii) and (iii)
$\overline{a}+{\lambda }_2\{\frac{\overline{a}}{b}+\frac{\overline{b}-\overline{a}}{c}\}=b+{\lambda }_3\{\frac{\overline{b}}{a}+\frac{\overline{a}-\overline{b}}{c}\}$
$\Rightarrow \overline{a}\{1+\frac{{\lambda }_2}{b}-\frac{{\lambda }_2}{c}\}+\frac{{\lambda }_2}{c}\overline{b}=\frac{{\lambda }_3}{c}\overline{a}+\overline{b}\{1+\frac{{\lambda }_3}{a}-\frac{{\lambda }_3}{c}\}.$
$\Rightarrow \{\begin{array}{c}1+\frac{{\lambda }_2}{b}-\frac{{\lambda }_2}{c}=\frac{{\lambda }_3}{c}\\ 1+\frac{{\lambda }_3}{a}-\frac{{\lambda }_3}{c}=\frac{{\lambda }_2}{c}\end{array}$
(on equating the coefficient of $\overline{a}$ & $\overline{b}$)
$\Rightarrow \{\begin{array}{c}\frac{{\lambda }_3}{c}=1+\frac{{\lambda }_2}{bc}(c-b)........(\ast )\\ \frac{{\lambda }_2}{c}=1+\frac{{\lambda }_3}{ac}(c-a).....(\ast \ast )\end{array}$
$\Rightarrow \frac{{\lambda }_2}{c}=1+\frac{1}{a}(c-a)(1+\frac{{\lambda }_2}{bc}(c-b))$
by putting $(\ast )$ in $(\ast \ast )$
$\Rightarrow \frac{{\lambda }_2}{c}=\frac{c}{a}+\frac{{\lambda }_2}{abc}(c-a)(c-b)$
$\Rightarrow \frac{{\lambda }_2}{abc}(ab+bc+ac-ab-c^2)=\frac{c}{a}$
$\Rightarrow \frac{{\lambda }_2}{ab}(a+b-c)=\frac{c}{a}\Rightarrow {\lambda }_2=\frac{bc}{a+b-c}$
On putting the value of ${\lambda }_2$ in (*) we get
$\frac{{\lambda }_3}{c}=1+\frac{c-b}{a+b-c}or{\lambda }_3=\frac{ac}{a+b-c}$
Now for position vector of point $P$ Putting ${\lambda }_2$ in (ii) or that of ${\lambda }_3$ in (iii) we have
$\overline{r}=\overline{a}+\frac{bc}{a+b-c}\{\frac{\overline{a}}{b}+\frac{\overline{b}-\overline{a}}{c}\}$
$\Rightarrow \overline{r}=\frac{\overline{a}(a+b-c)+c\overline{a}+b(\overline{b}-\overline{a})}{a+b-c}\Rightarrow \overline{r}=\frac{a\overline{a}+b\overline{b}}{a+b-c}$
Hence (d) is correct answer