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
Best Answer
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}$.