Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two vectors such that $|2\overrightarrow{a}+3\overrightarrow{b}|=|3\overrightarrow{a}+\overrightarrow{b}|$ and the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is $60°.$ If $\frac{1}{8}\overrightarrow{a}$ is a unit vector, then $\left|\overrightarrow{b}\right|$ is equal to : 

Given that 

$\frac{1}{8}\overrightarrow{a}$ is a unit vector.

$\left|\overrightarrow{a}\right|=8$
$|2\overrightarrow{a}+3\overrightarrow{b}{|}^2=|3\overrightarrow{a}+\overrightarrow{b}{|}^2$

We know that$\overrightarrow{a} · \overrightarrow{b} = \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right| \cos \theta ,$
$\Rightarrow 4|\overrightarrow{a}{|}^2+9|\overrightarrow{b}{|}^2+12\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos 60°=9|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+6\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos 60°$
$\Rightarrow 5|\overrightarrow{a}{|}^2=8|\overrightarrow{b}{|}^2+6\left|\overrightarrow{a}\right|\left|\right|\overrightarrow{b}\mid \cos 60°$
$\Rightarrow 5\left(64\right)=8|\overrightarrow{b}{|}^2+6\left(8\right)·|\overrightarrow{b}|\frac{1}{2}$
$\left.\Rightarrow |\overrightarrow{b}{|}^2+3|\overrightarrow{b}|-40=0\Rightarrow \left(\left|\overrightarrow{b}\right|+8\right)|\overrightarrow{b}|-5\right)=0$
$\Rightarrow \left|\overrightarrow{b}\right|=5$