Let `vec(a)` and `vec(b)` be two unit vectors such that `vec(a).vec(b)=0` For some `x,y in R`, let `vec(c)=xvec(a)+yvec(b)+(vec(a)xxvec(b))` If `|vec(c)|=2` and the vector `vec(c)` is inclined at same angle `alpha` to both `vec(a)` and `vec(b)` then the value of `8cos^2alpha` is

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have two unit vectors \(\vec{a}\) and \(\vec{b}\) such that \(\vec{a} \cdot \vec{b} = 0\). This means that \(\vec{a}\) and \(\vec{b}\) are perpendicular to each other. We also have a vector \(\vec{c} = x\vec{a} + y\vec{b} + \vec{a} \times \vec{b}\).

### Step 2: Find the magnitude of \(\vec{c}\)
We know that \(|\vec{c}| = 2\). Therefore, we can write:
\[ |\vec{c}|^2 = (x\vec{a} + y\vec{b} + \vec{a} \times \vec{b}) \cdot (x\vec{a} + y\vec{b} + \vec{a} \times \vec{b}) = 4 \]

### Step 3: Expand the dot product
Using the properties of dot products:
\[ |\vec{c}|^2 = x^2 |\vec{a}|^2 + y^2 |\vec{b}|^2 + |\vec{a} \times \vec{b}|^2 + 2xy (\vec{a} \cdot \vec{b}) + 2x(\vec{a} \cdot (\vec{a} \times \vec{b})) + 2y(\vec{b} \cdot (\vec{a} \times \vec{b})) \]
Since \(\vec{a} \cdot \vec{b} = 0\) and \(\vec{a} \cdot (\vec{a} \times \vec{b}) = 0\) and \(\vec{b} \cdot (\vec{a} \times \vec{b}) = 0\), we simplify this to:
\[ |\vec{c}|^2 = x^2 + y^2 + |\vec{a} \times \vec{b}|^2 \]

### Step 4: Calculate \(|\vec{a} \times \vec{b}|\)
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors and perpendicular, we have:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(90^\circ) = 1 \]
Thus, \(|\vec{a} \times \vec{b}|^2 = 1\).

### Step 5: Substitute back into the equation
Now substituting back, we get:
\[ x^2 + y^2 + 1 = 4 \implies x^2 + y^2 = 3 \]

### Step 6: Relate \(x\) and \(y\) to \(\alpha\)
Since \(\vec{c}\) is inclined at the same angle \(\alpha\) to both \(\vec{a}\) and \(\vec{b}\), we have:
\[ \vec{a} \cdot \vec{c} = |\vec{c}| \cos(\alpha) \quad \text{and} \quad \vec{b} \cdot \vec{c} = |\vec{c}| \cos(\alpha) \]
Thus:
\[ x = 2 \cos(\alpha) \quad \text{and} \quad y = 2 \cos(\alpha) \]

### Step 7: Substitute \(x\) and \(y\) into the equation
Substituting \(x\) and \(y\) into \(x^2 + y^2 = 3\):
\[ (2 \cos(\alpha))^2 + (2 \cos(\alpha))^2 = 3 \]
\[ 4 \cos^2(\alpha) + 4 \cos^2(\alpha) = 3 \implies 8 \cos^2(\alpha) = 3 \]

### Step 8: Conclusion
Thus, the value of \(8 \cos^2(\alpha)\) is:
\[ \boxed{3} \]