Let`veca` and `vecb` be two unit vectors such that `veca.vecb=0`. For some `x,y in R`, let `vecc = xveca + yvecb + (veca xx vecb)` . If `|vecc|=2` and the vector `vecc` is inclined at the same angle `angle` to both `veca` and `vecb`, then the value of `8cos^(2)alpha` is________________

Answer

To solve the problem step by step, we will analyze the given information and derive the required value systematically.

### 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 implies that \(\vec{a}\) and \(\vec{b}\) are perpendicular to each other.

### Step 2: Define the Vector \(\vec{c}\)
The vector \(\vec{c}\) is defined as:
\[ \vec{c} = x\vec{a} + y\vec{b} + (\vec{a} \times \vec{b}) \]
where \(x\) and \(y\) are real numbers.

### Step 3: Calculate the Magnitude of \(\vec{c}\)
We know that \(|\vec{c}| = 2\). To find the magnitude of \(\vec{c}\), we will use the property that the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{a} \times \vec{b}\) are mutually perpendicular. Thus, we can write:
\[ |\vec{c}|^2 = |x\vec{a}|^2 + |y\vec{b}|^2 + |\vec{a} \times \vec{b}|^2 \]
Since \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\), we have:
\[ |\vec{c}|^2 = x^2 + y^2 + |\vec{a} \times \vec{b}|^2 \]
The magnitude of \(\vec{a} \times \vec{b}\) is given by:
\[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(\theta) = 1 \cdot 1 \cdot \sin(90^\circ) = 1 \]
Thus, we can substitute this into our equation:
\[ |\vec{c}|^2 = x^2 + y^2 + 1 \]
Given that \(|\vec{c}| = 2\), we have:
\[ 4 = x^2 + y^2 + 1 \]
This simplifies to:
\[ x^2 + y^2 = 3 \]

### Step 4: Find the Angles
Since \(\vec{c}\) is inclined at the same angle \(\alpha\) to both \(\vec{a}\) and \(\vec{b}\), we can write:
\[ \vec{c} \cdot \vec{a} = |\vec{c}||\vec{a}|\cos(\alpha) = 2 \cdot 1 \cdot \cos(\alpha) = 2\cos(\alpha) \]
Calculating \(\vec{c} \cdot \vec{a}\):
\[ \vec{c} \cdot \vec{a} = (x\vec{a} + y\vec{b} + \vec{a} \times \vec{b}) \cdot \vec{a} = x(\vec{a} \cdot \vec{a}) + y(\vec{b} \cdot \vec{a}) + (\vec{a} \times \vec{b}) \cdot \vec{a} \]
Since \(\vec{b} \cdot \vec{a} = 0\) and \((\vec{a} \times \vec{b}) \cdot \vec{a} = 0\):
\[ \vec{c} \cdot \vec{a} = x \]
Thus, we have:
\[ x = 2\cos(\alpha) \]

Similarly, for \(\vec{c} \cdot \vec{b}\):
\[ \vec{c} \cdot \vec{b} = (x\vec{a} + y\vec{b} + \vec{a} \times \vec{b}) \cdot \vec{b} = x(\vec{a} \cdot \vec{b}) + y(\vec{b} \cdot \vec{b}) + (\vec{a} \times \vec{b}) \cdot \vec{b} \]
Again, since \(\vec{a} \cdot \vec{b} = 0\) and \((\vec{a} \times \vec{b}) \cdot \vec{b} = 0\):
\[ \vec{c} \cdot \vec{b} = y \]
Thus, we have:
\[ y = 2\cos(\alpha) \]

### Step 5: Substitute \(x\) and \(y\) into the Equation
Now substituting \(x\) and \(y\) into the equation \(x^2 + y^2 = 3\):
\[ (2\cos(\alpha))^2 + (2\cos(\alpha))^2 = 3 \]
This simplifies to:
\[ 4\cos^2(\alpha) + 4\cos^2(\alpha) = 3 \]
\[ 8\cos^2(\alpha) = 3 \]

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