Angle bisector between two vectors, which are expressed by unit non-orthogonal vectors Given vectors $\,a=2m-2n\,$ and $\,b=3m+6n\,$, where $\ \left\lvert m \right \rvert =\left\lvert n \right \rvert =1\,$ and $\,\angle\left(m,n\right)=\dfrac{2\pi}{3},\,$ find vector of angle bisector of the angle $\angle\left(a,b\right)$, which intensity is $10\sqrt{3}$.
Advice on how to start?
EDIT: intensity=magnitude
 A: First, compute angle $\theta := \angle (a,b)$ between $a$ and $b$ using cosine law. 
$$
\left\langle a, b \right\rangle =  \left\| a\right\| \left\| b\right\|  \cos \theta
\implies 
\cos \theta = \frac{\left\langle a, b \right\rangle}{\left\| a\right\| \left\| b\right\| },
$$
where $\left\langle \,\cdot\,, \cdot \,\right\rangle$ is inner product of vectors $a$ and $b$.
Note that
$$
\left\| a \right\| = \sqrt{\left\langle a, a \right\rangle }  
= \sqrt{\left\langle 2m - 2n , 2m - 2n  \right\rangle } 
= 2 \sqrt{\left\langle m - n , m - n \right\rangle } 
= 2 \sqrt{ \left\langle m,m \right\rangle   - 2  \left\langle m, n \right\rangle + \left\langle m,m \right\rangle }
= 2\sqrt{\left\|m\right\|^2 - 2 \left\langle m, n \right\rangle + \left\|n\right\|^2}
$$
Since $\left\langle m, n \right\rangle = \left\|m\right\| \left\|n\right\|\cos\angle\left(m,n \right) = 1\cdot 1\cdot \cos \frac{2\pi}{3} = -\frac{1}{2}$, 
we have 
$$
\begin{aligned}
\left\| a \right\| 
&= 2\sqrt{\left\|m\right\|^2 - 2 \left\langle m, n \right\rangle + \left\|n\right\|^2} 
= 2\sqrt{1^2 + \frac{2}{2} + 1^2}
= 2\sqrt{3}
\\
\left\| b \right\|  
&= \sqrt{\left\langle 3m+6n,3m+6n\right\rangle}
=3\sqrt{\left\|m\right\|^2 + 4\left\langle m,n\right\rangle + 4\left\|n\right\|^2} 
=3\sqrt{1 - \frac{4}{2} + 4 }
= 3\sqrt{3}
\\
\left\langle a,b\right\rangle 
&= \left\langle 2m-2n,3m+6n\right\rangle
= 6 \left\| m\right\| ^2 + 6 \left\langle m,n\right\rangle - 12 \left\| n\right\|^2 
= 6 - \frac{6}{2} - 12 = -9
\end{aligned}
$$
And so
$$
\cos \theta 
= \frac{\left\langle a, b \right\rangle}{\left\| a\right\| \left\| b\right\| }
= \frac{-9}{2\sqrt{3} \cdot 3\sqrt{3}}
= -\frac{1}{2}
$$

You are looking for a vector $v$ such that $\angle\left(a,v\right)  = \theta/2 $ and $\left\| v\right\| = 10\sqrt 3$. 
Assume $x,y$ are coefficients of decomposition of $v$ in terms of $m$ and $n$, i.e. $v = x m + y n$, then we write
$$
v = x m + yn, \quad 
\left\| v\right\| = 10\sqrt 3, \quad 
\cos \frac{\theta}{2} = \sqrt{\frac{1+ \cos \theta}{2}} 
= \sqrt{\frac{1-\frac{1}{2}}{2}} 
= \pm \frac{1}{2}
$$
Let us write out what we know about $v$:
$$
\begin{aligned}
\left\langle v,v\right\rangle  
& = \left\| v \right\|^2 = 300
\\
\left\langle v,a\right\rangle  
& = \left\| v \right\| \left\| a \right\| \cos \frac{\theta}{2}
=  10\sqrt{3}\cdot 2\sqrt{3}\cdot \frac{\pm 1}{2}= \pm 30
\\
\left\langle v,b\right\rangle  
& = \left\| v \right\| \left\| b \right\| \cos \frac{\theta}{2}
=  10\sqrt{3}\cdot3\sqrt{3}\cdot\frac{\pm 1}{2}=\pm 45
\end{aligned}
$$
On the other hand,
$$
\begin{aligned}
30 = \left\langle v,a\right\rangle &= \left\langle xm+yn,2m-2n\right\rangle
= 2\left(x \left\| m\right\| ^2 + (y - x) \left\langle m,n\right\rangle - y \left\| n\right\|^2 \right) = 
\\  & 
=  2\left(x -\frac{y - x}{2}  - y  \right) = 3x-3y
\\
45 =  \left\langle v,b\right\rangle &= \left\langle xm+yn,3m+6n\right\rangle
= 3\left(x \left\| m\right\| ^2 + (y + 2x) \left\langle m,n\right\rangle + 2 y \left\| n\right\|^2 \right)= 
\\  & 
= 3\left(x -\frac{y + 2 x}{2} + y  \right) =  \frac{9}{2}y
\end{aligned}
$$
Thus we have a linear system
$$
\begin{cases}
\displaystyle 3x - 3y = 30 \\
\displaystyle \dfrac{9}{2}y = 45
\end{cases}
\implies 
\begin{cases}
\displaystyle x = 10  + y\\
\displaystyle y = 10
\end{cases}
\implies 
\begin{cases}
\displaystyle x = 20\\
\displaystyle y = 10
\end{cases}
$$
Finally,  we write the answer
$$
\bbox[5pt, border: 2pt solid #FF0000]{v = 20 m + 10 n}
$$
One can check that indeed $ \left\| v\right\| = 10\sqrt{3}$.
