Point $A$ and $B$ have position vectors $\vec a$ and $\vec b$ respectively relative to an orgin $O$.
The point $D$ is such that $\overrightarrow{OD} = k\overrightarrow{OA}$ and the point $E$ is such that $\overrightarrow{AE} = l\overrightarrow{AB}$.
The line segments $BD$ and $OE$ intersect at $X$.
If $\overrightarrow{OX} = \frac{2}{5}\overrightarrow{OE}$ and $\overrightarrow{XB} = \dfrac{4}{5}\overrightarrow{DB}$.
Express $\overrightarrow{OX}$ and $XB$ in terms of $\vec a, \vec b, k, l$ and hence evaluate $k$ and $l$.
I have worked out most of the problem but can't figure out how to evaluate $l$.
Using ratio theorem, I got $\overrightarrow{OX}$ as,
$ \overrightarrow{OX} = \dfrac{2}{5}\Big[(1-l)a + lb\Big] $
And similarly, $\overrightarrow{XB}$
$ \overrightarrow{XB} = \dfrac{4}{5}(b - ka) $
Then using, $ \begin{align} \overrightarrow{OX} + \overrightarrow{XB} &= \overrightarrow{OB} \\ \dfrac{2}{5}\Big[(1-l)a + lb\Big] + \dfrac{4}{5}(b - ka) &= b\\ \\ \text{...}\\ \\ [2(1-l) - 4k]a &= (1-2l)b\\ 2(1-l)- 4k &= 1 - 2l & \text{(a and b are non-zero and non-parallel)}\\ -4k &= -1 \\ k &= \dfrac{1}{4}\\ \end{align} $
I can't seem to figure out how to get $l$. I tried $\overrightarrow{DA} + \overrightarrow{AE} + \overrightarrow{E} = \overrightarrow{DX}$ and $\overrightarrow{OA} + \overrightarrow{AE} = \overrightarrow{OE}$, these just give an equality statement.
How do I evaluate $l$?
Thanks for your help.