# Finding vectors in terms of other vectors

Suppose that O, A and B are three non-collinear points in a plane. Let $\vec {OC}:=\vec {OB}-\vec{2OA}$, and $\vec {OE} :=\vec {-OA}$

Express $\vec {OM}$ in terms of the vectors $\vec {OA}$ and $\vec {OB}$ where M is the point of intersection of the line through O and C and the line through B and E.

I dont know how to come up with 2 equations for OM to solve. I would need a step by step answer if that is possible, any takers?

• Could you post the diagram you drew? – Carser Jun 3 '14 at 4:01
• Assigning O as origin might simplify the problem – GTX OC Jun 3 '14 at 4:01
• math.stackexchange.com/questions/818899/… – Gerry Myerson Jun 3 '14 at 7:23
• Hi sorry, I tried to add a comment but could not, then I changed from google chrome to internet explorer and it worked, very sorry. – user148615 Jun 3 '14 at 10:04

Since M belongs to $(OC)$ and $(BE)$, then you have two ways to consider $\vec{OM}$: $$\vec{OM}=\alpha\vec{OC}=\alpha(\vec{OB}-2\vec{OA})$$ $$\vec{OM}=\vec{OE}+\vec{EM}=\vec{OE}+\beta\vec{EB}=\vec{OE}+\beta(\vec{EO}+\vec{OB})=-\vec{OA}+\beta(\vec{OA}+\vec{OB})$$
so $$\alpha(\vec{OB}-2\vec{OA})=-\vec{OA}+\beta(\vec{OA}+\vec{OB})$$ so $$(\alpha-\beta)\vec{OB} + (-2\alpha+1-\beta)\vec{OA}=\vec{0}$$ so $$\begin{cases} \alpha-\beta=0 \\ -2\alpha+1-\beta=0 \end{cases}$$ so $$\begin{cases} \alpha=\beta \\ \alpha=\frac13 \end{cases}$$
so $$\vec{OM}=\frac13(\vec{OB}-2\vec{OA})$$
• @user148615, consider $\vec{EB}=\vec{EO}+\vec{OB}$ – Fabien Jun 3 '14 at 23:48