Find Value of Eight Cos Squared Alpha for Unit Vectors Linear Combination

Given that $\vec{a}$ and $\vec{b}$ are two unit vectors such that $\vec{a} \cdot \vec{b} = 0$, we have: $|\vec{a}| = 1, \quad |\vec{b}| = 1, \quad \text{and} \quad \vec{a} \cdot \vec{b} = 0$

The magnitude of the cross product $\vec{a} \times \vec{b}$ is given by: $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\left(\frac{\pi}{2}\right) = 1$

Furthermore, the vectors $\vec{a}$, $\vec{b}$, and $\vec{a} \times \vec{b}$ are mutually orthogonal unit vectors, forming an orthonormal basis.

We are given the vector $\vec{c}$: $\vec{c} = x\vec{a} + y\vec{b} + (\vec{a} \times \vec{b})$

Taking the magnitude squared on both sides: $|\vec{c}|^2 = x^2 |\vec{a}|^2 + y^2 |\vec{b}|^2 + |\vec{a} \times \vec{b}|^2$

Given that $|\vec{c}| = 2$, we substitute the magnitudes: $2^2 = x^2(1) + y^2(1) + 1^2$ $4 = x^2 + y^2 + 1 \implies x^2 + y^2 = 3 \quad \text{--- (1)}$

Next, taking the dot product of $\vec{c}$ with $\vec{a}$: $\vec{c} \cdot \vec{a} = (x\vec{a} + y\vec{b} + \vec{a} \times \vec{b}) \cdot \vec{a} = x(\vec{a} \cdot \vec{a}) + y(\vec{b} \cdot \vec{a}) + (\vec{a} \times \vec{b}) \cdot \vec{a} = x$

Since the vector $\vec{c}$ makes an angle $\alpha$ with $\vec{a}$: $\vec{c} \cdot \vec{a} = |\vec{c}||\vec{a}|\cos \alpha = 2 \cdot 1 \cdot \cos \alpha = 2 \cos \alpha$

Equating the two expressions for $\vec{c} \cdot \vec{a}$: $x = 2 \cos \alpha$

Similarly, taking the dot product of $\vec{c}$ with $\vec{b}$: $\vec{c} \cdot \vec{b} = (x\vec{a} + y\vec{b} + \vec{a} \times \vec{b}) \cdot \vec{b} = y$

Since $\vec{c}$ also makes an angle $\alpha$ with $\vec{b}$: $\vec{c} \cdot \vec{b} = |\vec{c}||\vec{b}|\cos \alpha = 2 \cos \alpha$

Thus, we have: $y = 2 \cos \alpha$

Now, substituting $x = 2 \cos \alpha$ and $y = 2 \cos \alpha$ into equation (1): $(2\cos \alpha)^2 + (2\cos \alpha)^2 = 3$ $4\cos^2 \alpha + 4\cos^2 \alpha = 3$ $8\cos^2 \alpha = 3$