Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that $\vec{a}\cdot\vec{b}=0$. For some $x,y\in\mathbb{R}$, let $\vec{c}=x\vec{a}+y\vec{b}+(\vec{a}\times\vec{b})$. If $|\vec{c}|=2$ and the vector $\vec{c}$ is inclined at the same angle $\alpha$ to both $\vec{a}$ and $\vec{b}$, then the value of $8\cos^{2}\alpha$ is ________.

$\text{Given: } |\vec{a}|=|\vec{b}|=1,\quad \vec{a}\cdot\vec{b}=0$

$\vec{c}=x\vec{a}+y\vec{b}+(\vec{a}\times\vec{b}),\quad x,y\in\mathbb{R}$

$|\vec{c}|=2$

$\text{To find: } 8\cos^{2}\alpha$

$\angle(\vec{c},\vec{a})=\angle(\vec{c},\vec{b})=\alpha$

$\vec{n}=\vec{a}\times\vec{b},\quad |\vec{n}|=1,\quad \vec{n}\perp\vec{a},\ \vec{n}\perp\vec{b}$

$\text{Plan: (i) write } \vec{c} \text{ in the } \{\vec{a},\vec{b},\vec{n}\} \text{ frame}$

$\text{(ii) } |\vec{c}|=2 \Rightarrow x^2+y^2=3;\quad \text{(iii) equal angles} \Rightarrow x=y$

$\text{(iv) find } 8\cos^2\alpha$

$\vec{c}=x\vec{a}+y\vec{b}+\vec{n}=x\vec{a}+y\vec{b}+(\vec{a}\times\vec{b})$

$\vec{c}=(x,\ y,\ 1) \quad \text{in the frame } \{\vec{a},\vec{b},\vec{n}\}$

$|\vec{c}|^2 = x^2+y^2+1$

$(\text{cross terms vanish;} \ |\vec{a}|=|\vec{b}|=|\vec{n}|=1)$

$|\vec{c}|=2 \Rightarrow x^2+y^2+1=4$

$x^2+y^2=3$

$\cos\alpha = \dfrac{\vec{c}\cdot\vec{a}}{|\vec{c}||\vec{a}|} = \dfrac{x}{2}$

$\cos\alpha = \dfrac{\vec{c}\cdot\vec{b}}{|\vec{c}||\vec{b}|} = \dfrac{y}{2}$

$\dfrac{x}{2} = \dfrac{y}{2} \Rightarrow x=y$

$\vec{c} = x(\vec{a}+\vec{b}) + \vec{n}$

$\Rightarrow \text{projection of } \vec{c} \text{ on the } \vec{a}\vec{b}\text{-plane is along } \vec{a}+\vec{b}$

$x^2+y^2=3,\quad x=y \Rightarrow 2x^2=3$

$x^2 = \dfrac{3}{2}$

$\cos^2\alpha = \left(\dfrac{x}{2}\right)^2 = \dfrac{x^2}{4}$

$= \dfrac{3/2}{4} = \dfrac{3}{8}$

$8\cos^2\alpha = 8 \times \dfrac{3}{8} = 3$