Let the position vector of Aand C be $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{c}}$ respectively.
Therefore,
Position vector of $\mathrm{B}=\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}$   --------- (i)

Also, E divides BC in ratio of 2:2

∴ Position vector of $\mathrm{E}=\frac{\overrightarrow{\mathrm{b}}+2\overrightarrow{\mathrm{c}}}{3}$⟶ (ii)

∴ Position vector of $\mathrm{E}=\frac{\overrightarrow{\mathrm{b}}+2\overrightarrow{\mathrm{c}}}{3}$ (from (1))

∴  Position vector of $\mathrm{E}=\frac{\overrightarrow{\mathrm{a}}+3\overrightarrow{\mathrm{c}}}{3}$

The vector equation of the bisector of ∠AOC is

 $\overrightarrow{\mathrm{\gamma }}=\mathrm{\lambda }\left\{\frac{\overrightarrow{\mathrm{a}}}{\left|\overrightarrow{\mathrm{a}}\right|}+\frac{\overrightarrow{\mathrm{c}}}{\left|\overrightarrow{\mathrm{c}}\right|}\right\}$   ------ (iii)

The vector equation of AE is

$\overrightarrow{\mathrm{\gamma }}=\overrightarrow{\mathrm{a}}+\mathrm{\mu }\left\{\frac{\overrightarrow{\mathrm{a}}+3\overrightarrow{\mathrm{c}}}{3}-\overrightarrow{\mathrm{a}}\right\}$   ----- (iv)

Lines (3) and (4) intersects at p.
∴ for point P we must have

$\mathrm{\lambda }\left\{\frac{\overrightarrow{\mathrm{a}}}{\left|\overrightarrow{\mathrm{a}}\right|}+1\frac{\overrightarrow{\mathrm{c}}}{\left|\overrightarrow{\mathrm{c}}\right|}\right\}\overrightarrow{\mathrm{a}}+\mathrm{\mu }\left\{\frac{3\overrightarrow{\mathrm{c}}-2\overrightarrow{\mathrm{a}}}{3}\right\}$ 

$\Rightarrow \left\{\frac{\mathrm{\lambda }}{\left|\overrightarrow{\mathrm{a}}\right|}-1+\frac{2\mathrm{\mu }}{3}\right\}\overrightarrow{\mathrm{a}}+\left\{\frac{\mathrm{\lambda }}{\left|\overrightarrow{\mathrm{c}}\right|}-\mathrm{\mu }\right\}\overrightarrow{\mathrm{c}}=\overrightarrow{0}$ 

$\Rightarrow \frac{\mathrm{\lambda }}{\left|\overrightarrow{\mathrm{a}}\right|}-1+\frac{2\mathrm{\mu }}{3}=0\text{ and }\frac{\mathrm{\lambda }}{\left|\overrightarrow{\mathrm{c}}\right|}-\mathrm{\mu }=0$ 

( $\because \overrightarrow{\mathrm{a}}\text{ and }\overrightarrow{\mathrm{c}}$ are non- colinear)

$\Rightarrow \mathrm{\lambda }=\frac{3\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{c}}\right|}{3\left|\overrightarrow{\mathrm{c}}\right|+2\left|\overrightarrow{\mathrm{a}}\right|}\text{ and }\mathrm{\mu }=\frac{3\left|\overrightarrow{\mathrm{a}}\right|}{3\left|\overrightarrow{\mathrm{c}}\right|+2\left|\overrightarrow{\mathrm{a}}\right|}$

Putting the value of λ in (iii) or that of μ in (iv) we obtain the positive vector of p as 

${\overrightarrow{\mathrm{\gamma }}}_1=\frac{3\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{c}}\right|}{3\left|\overrightarrow{\mathrm{c}}\right|+2\left|\overrightarrow{\mathrm{a}}\right|}\left\{\frac{\overrightarrow{\mathrm{a}}}{\left|\overrightarrow{\mathrm{a}}\right|}+\frac{\overrightarrow{\mathrm{c}}}{\left|\overrightarrow{\mathrm{c}}\right|}\right\}$

Now, vector eqaution of line cp:

${\overrightarrow{\mathrm{\gamma }}}_1=\overrightarrow{\mathrm{c}}+{\mathrm{\lambda }}_1\left\{\frac{3\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{c}}\right|}{3\left|\overrightarrow{\mathrm{c}}\right|+2\left|\overrightarrow{\mathrm{a}}\right|}\left\{\frac{\overrightarrow{\mathrm{a}}}{\left|\overrightarrow{\mathrm{a}}\right|}+\frac{\overrightarrow{\mathrm{c}}}{\left|\overrightarrow{\mathrm{c}}\right|}\right\}-\overrightarrow{\mathrm{c}}\right\}$   ------ (v)

vector eqaution of line AB is $\overrightarrow{\mathrm{\gamma }}=\overrightarrow{\mathrm{a}}+{\mathrm{\lambda }}_2\overrightarrow{\mathrm{c}}$  --- (vi)

These two line intersets at F, we must have for f

$\overrightarrow{\mathrm{c}}+{\mathrm{\lambda }}_1\left[\frac{3\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{c}}\right|}{2\left|\overrightarrow{\mathrm{a}}\right|+3\left|\overrightarrow{\mathrm{c}}\right|}\left\{\frac{\overrightarrow{\mathrm{a}}}{\left|\overrightarrow{\mathrm{a}}\right|}+\frac{\overrightarrow{\mathrm{c}}}{\left|\overrightarrow{\mathrm{c}}\right|}\right\}-\overrightarrow{\mathrm{c}}\right]=\overrightarrow{\mathrm{a}}+{\mathrm{\lambda }}_2\overrightarrow{\mathrm{c}}$

$\Rightarrow \frac{3\left|\overrightarrow{\mathrm{c}}\right|{\mathrm{\lambda }}_1}{3\left|\overrightarrow{\mathrm{c}}\right|+2\left|\overrightarrow{\mathrm{a}}\right|}-1=0\text{ and }1+\frac{3{\mathrm{\lambda }}_1\left|\overrightarrow{\mathrm{a}}\right|}{3\left|\overrightarrow{\mathrm{c}}\right|+2\left|\overrightarrow{\mathrm{a}}\right|}-{\mathrm{\lambda }}_1-{\mathrm{\lambda }}_2=0$

($\because \overrightarrow{\mathrm{a}}\text{ and }\overrightarrow{\mathrm{c}}$ are non- colinear)

$\Rightarrow {\mathrm{\lambda }}_1=\frac{3\left|\overrightarrow{\mathrm{c}}\right|+2\left|\overrightarrow{\mathrm{a}}\right|}{3\left|\overrightarrow{\mathrm{c}}\right|}\text{ and }{\mathrm{\lambda }}_2=\frac{1}{3}\frac{\left|\overrightarrow{\mathrm{a}}\right|}{\left|\overrightarrow{\mathrm{c}}\right|}$

putting values of λ₂ in (vi) we get

${\overrightarrow{\mathrm{r}}}_4=\overrightarrow{\mathrm{a}}+\frac{1}{3}\frac{\left|\overrightarrow{\mathrm{a}}\right|}{\left|\overrightarrow{\mathrm{l}}\right|}\overrightarrow{\mathrm{c}}$

$\therefore \overrightarrow{\mathrm{AF}}=\overrightarrow{{\mathrm{r}}_4}-\overrightarrow{\mathrm{a}}$