Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two unit vectors such that $\overrightarrow{a}·\overrightarrow{b}=0$.

For some $x,y\in R$, let $\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})$. If $\left|\overrightarrow{c}\right|=2$ and the 

vector $\overrightarrow{c}$ is inclined to the same angle  $\alpha$ to both $\overrightarrow{a}$ and $\overrightarrow{b}$, then the value of 

$8{\cos }^2\alpha ,$ is

It is given that $\overrightarrow{c}$ is inclined to the same angle $\alpha$ to both $\overrightarrow{a}$ and $\overrightarrow{b}$.

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overrightarrow{c}\cdot \overrightarrow{a}=\left|\overrightarrow{c}\right|\left|\overrightarrow{a}\right|\cos \alpha ,\overrightarrow{c}\cdot \overrightarrow{b}=\left|\overrightarrow{c}\right|\left|\overrightarrow{b}\right|\cos \alpha \\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\to \overrightarrow{c}\cdot \overrightarrow{a}=2\cos \alpha ,\overrightarrow{c}\cdot \overrightarrow{b}=2\cos \alpha [\because |\overrightarrow{c}|=2,|\overrightarrow{a}|=|\overrightarrow{b}|=1]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overrightarrow{c}\cdot \overrightarrow{a}=x\left(\overrightarrow{a}\cdot \overrightarrow{a}\right)+y(\overrightarrow{b}\cdot \overrightarrow{a})+(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{a}=x+y\left(0\right)+0\\ \Rightarrow x=2\cos \alpha \end{array}$

and, $x(\overrightarrow{a}\cdot \overrightarrow{b})+y(\overrightarrow{b}\cdot \overrightarrow{b})+(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{b}=2\cos \alpha \Rightarrow y=2\cos \alpha$

$\Rightarrow x=2\cos \alpha ,y=2\cos \alpha$

Now, $\left|\overrightarrow{c}\right|=2$

$\begin{array}{l}\Rightarrow x^2|\overrightarrow{a}{|}^2+y^2|\overrightarrow{b}{|}^2+|\overrightarrow{a}\times \overrightarrow{b}{|}^2+2xy(\overrightarrow{a}\cdot \overrightarrow{b})+2y\overrightarrow{b}\cdot (\overrightarrow{a}\times \overrightarrow{b})\\ +2x\overrightarrow{a}\cdot (\overrightarrow{a}\times \overrightarrow{b})=4\\ \Rightarrow x^2+y^2+\left|\overrightarrow{a}{|}^2\right|\overrightarrow{b}{|}^2-(\overrightarrow{a}\cdot \overrightarrow{b}{)}^2+2xy(\overrightarrow{a}\cdot \overrightarrow{b})\\ \Rightarrow x^2+y^2+1-0+2xy\times 0=4+2y\times 0+2x\times 0=4\\ \Rightarrow x^2+y^2=3\Rightarrow 8{\cos }^2\alpha =3\end{array}$