Calculate the bisection vector OM, by using the vectors that form an angle with it, $OA_1$ and $OA_2$
I was thinking of using the Angle bisector theorem, but i can't seem to get that end formula look.
Calculate the bisection vector OM
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$\begingroup$
geometry
vectors
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0Do you mean $M$ as the midpoint, or that the *angle* gets bisected? – 2017-02-27
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0From the "answer" it seems actually the question must have been about locating $M$ so that it cuts $A_1A_2$ in the ratio $K.$ – 2017-02-27
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0M is not the midpoint by default, but if you let K = 1, it is. And the O angle gets bisected, yes. – 2017-02-27
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0If $A_1,A_2$ have different lengths and $K=1$ it will not give the bisection of the angle, only the midpoint (not the same points). – 2017-02-27
1 Answers
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$$\begin{cases} \vec{OM}=\vec{OA_1}+\vec{A_1M}\\ \vec{OM}=\vec{OA_2}+\vec{A_2M} \end{cases}$$ $$\begin{cases} \vec{OM}=\vec{OA_1}+k\vec{MA_2}\\ \vec{OM}=\vec{OA_2}-\vec{MA_2} \end{cases}$$ $$\begin{cases} \vec{OM}=\vec{OA_1}+k\vec{MA_2}\\ k\vec{OM}=k\vec{OA_2}-k\vec{MA_2} \end{cases}$$
Now , add two equations to obtain :
$$(1+k)\vec{OM}=\vec{OA_1}+k\vec{OA_2}$$
from which we have :
$$\vec{OM}=\frac{\vec{OA_1}+k\vec{OA_2}}{1+k}$$
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0Thanks, looks like i didn't know you could inverse the letters... – 2017-02-27