The resultant of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\overrightarrow{c}$. If $\overrightarrow{c}$ trisects the angle between $\overrightarrow{a}$ and $|\overrightarrow{b},|\overrightarrow{a}|=6,|\overrightarrow{b}\mid =4$ then $\left|\overrightarrow{c}\right|$ is

Let $\overrightarrow{c}$ be inclined at angles $\theta$ with $\overrightarrow{a}$ and 2$\theta$ with $\overrightarrow{b}$

$\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}\Rightarrow \overrightarrow{a}\cdot \overrightarrow{a}+\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{a}\cdot \overrightarrow{c}$ and $\overrightarrow{b}\cdot \overrightarrow{a}+\overrightarrow{b}\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{c}$

$\therefore$$6+4\cos 3\theta +4=c \cos \theta$       …(i)

$6+4\cos 3\theta +4=c \cos 2\theta$           …(ii)

 Eliminating $c,(6+4\cos 3\theta )\cos 2\theta =(6\cos 3\theta +4)\cos \theta$

$\therefore$$4(\cos 3\theta \cos 2\theta -\cos \theta )=6(\cos 3\theta \cos \theta -\cos 2\theta )$

$\begin{array}{l}\Rightarrow 2(\cos 3\theta \cos 2\theta -\cos (3\theta -2\theta \left)\right)\\ =3(\cos 3\theta \cos \theta -\cos (3\theta -\theta \left)\right)\\ \therefore 2\sin 2\theta =3\sin \theta \Rightarrow \cos \theta =\frac{3}{4}\end{array}$

$\begin{array}{l}\left(\mathrm{i}\right)\Rightarrow c=\frac{6+4\cos 3\theta }{\cos \theta }=\frac{8}{3}(3+2\cos 3\theta )\\ =\frac{8}{3}\left(3+2\left(4{\cos }^3\theta -3\cos \theta \right)\right)=\frac{8}{3}\left[3+2\left(\frac{27}{16}-\frac{9}{4}\right)\right]=5\end{array}$