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Diagram

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.

1 Answers 1

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At some point you said $ 2(1-\ell)- 4k = 1-2 \ell $ But you should've noticed that you could say that because $ 2(1-\ell) - 4k = 0 = 1-2 \ell. $ I think that gives you $\ell = 1/2$.

Hope that helps,

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    Thanks again for your help! It really helped me understand this better.2011-07-23