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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?

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    $\begingroup$ Could you post the diagram you drew? $\endgroup$ – Carser Jun 3 '14 at 4:01
  • $\begingroup$ Assigning O as origin might simplify the problem $\endgroup$ – GTX OC Jun 3 '14 at 4:01
  • $\begingroup$ math.stackexchange.com/questions/818899/… $\endgroup$ – Gerry Myerson Jun 3 '14 at 7:23
  • $\begingroup$ 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. $\endgroup$ – user148615 Jun 3 '14 at 10:04
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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 $$\vec{OM}=\frac13(\vec{OB}-2\vec{OA})$$

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  • $\begingroup$ Hi Fabien, Thank you for answering, just wondering how you managed to deduce the first 2 equations? Also I am unsure what is meant by the first and second projection and also why is it required and how you came about finding the alpha values? $\endgroup$ – user148615 Jun 3 '14 at 10:04
  • $\begingroup$ @user148615, I've edited. Feel free to post another comment if it's not clear enough ;) $\endgroup$ – Fabien Jun 3 '14 at 12:05
  • $\begingroup$ Hi again Fabien, thank you for the reply, just wondering how you managed to conclude that, on the second line of equations, the vector EB is equal to OA+OB? Is it because the two vectors are parallel? $\endgroup$ – user148615 Jun 3 '14 at 23:45
  • $\begingroup$ @user148615, consider $\vec{EB}=\vec{EO}+\vec{OB}$ $\endgroup$ – Fabien Jun 3 '14 at 23:48
  • $\begingroup$ Ah sorry! I have always learnt vectors to be in the form EB = OB - OE but I realised that E is actually in front of the O which makes EO=OA :) Thankyou Fabien! $\endgroup$ – user148615 Jun 3 '14 at 23:56

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