# Problem Study : Geometric Vector

The diagram shows triangle $\displaystyle OAB$ in which $\displaystyle \overrightarrow{{OA}}=\vec{a}$ and $\displaystyle \overrightarrow{{OB}}=\vec{b}$. The point $\displaystyle C$ is the midpoint of $\displaystyle OA$ and the point $\displaystyle D$ is the midpoint of $\displaystyle DC$.
(a) Express $\overrightarrow{{OD}}$ in terms of $\vec{a}$ and $\vec{b}$.
(b) If the point E lies on AB, then show that $\overrightarrow{{OE}}$ can be written in the form $\displaystyle \vec{a}+k\left( {\vec{b}-\vec{a}} \right)$, where k is a constant.
Given also that $\displaystyle OB$ produced meets $\displaystyle AB$ at $\displaystyle E$,
(c) find $\overrightarrow{{OE}}$,
(d) $\displaystyle AB:EB=2:1$
Solution

$\displaystyle \text{Since}$ $\displaystyle D$ $\displaystyle \text{is the midpoint of}$ $\displaystyle BC$, $\displaystyle \text{by midpoint formula,}$

$\displaystyle \begin{array}{l}\overrightarrow{{OD}}=\frac{1}{2}\left( {\overrightarrow{{OC}}+\overrightarrow{{OB}}} \right)\\\\\ \ \ \ \ \ =\frac{1}{2}\left( {\frac{1}{2}\vec{a}+\vec{b}} \right)\\\\\ \ \ \ \ \ =\frac{1}{4}\vec{a}+\frac{1}{2}\vec{b}\end{array}$

$\displaystyle \text{Let}$ $\displaystyle E$ $\displaystyle \text{be a point on}$ $\displaystyle AB$ $\displaystyle \text{such that}$ $\displaystyle \overrightarrow{{AE}}=k\overrightarrow{{AB}}.$

$\displaystyle \begin{array}{l}\therefore \overrightarrow{{OE}}=\overrightarrow{{OA}}+\overrightarrow{{AE}}\\\\\ \ \ \ \ \ \ \ =\overrightarrow{{OA}}+k\overrightarrow{{AB}}\\\\\ \ \ \ \ \ \ \ =\overrightarrow{{OA}}+k\left( {\overrightarrow{{OB}}-\overrightarrow{{OA}}} \right)\\\\\ \ \ \ \ \ \ \ =\vec{a}+k\left( {\vec{b}-\vec{a}} \right)\end{array}$

$\displaystyle \text{Since}$ $\displaystyle OD$ $\displaystyle \text{produced meets}$ $\displaystyle AB$ $\displaystyle \text{at}$ $\displaystyle E,$ $\displaystyle \text{let}$ $\displaystyle \overrightarrow{{OE}}=h\overrightarrow{{OD}}.$

$\displaystyle \therefore \left( {1-k} \right)\vec{a}+k\vec{b}=\frac{h}{4}\vec{a}+\frac{h}{2}\vec{b}$

$\displaystyle \text{Since}$ $\displaystyle \vec{a}$ $\displaystyle \text{and}$ $\displaystyle \vec{b}$ $\displaystyle \text{are not parallel and}$ $\displaystyle \vec{a}\ne \vec{0},\vec{b}\ne\vec{0},$

$\displaystyle \begin{array}{l}\ \ \ k=\frac{h}{2}\Rightarrow h=2k\\\\\ \ \ 1-k=\frac{h}{4}\Rightarrow h=4-4k\\\\\therefore 2k=4-4k\Rightarrow k=\frac{2}{3}\end{array}$

$\displaystyle \text{Since }\overrightarrow{{AE}}=k\overrightarrow{{AB}},$

$\displaystyle \begin{array}{l}\overrightarrow{{AE}}=\frac{2}{3}\overrightarrow{{AB}}\Rightarrow AE=\frac{2}{3}AB\\\\\therefore \overrightarrow{{EB}}=\frac{1}{3}\overrightarrow{{AB}}\Rightarrow EB=\frac{1}{3}AB\\\\\therefore AE:EB=\frac{2}{3}AB:\frac{1}{3}AB\\\\\therefore AE:EB=2:1\end{array}$