The unit vector `veca` and `vecb` are perpendicular, and the unit vector `vecc` is inclined at an angle `theta` to both `veca` and `vecb`. If `vecc=alpha veca+betavecb+gamma(vecaxxvecb)`, then which one of the following is incorrect?

Options

A. `alpha!=beta`
B. `gamma^(2)=1-2alpha^(2)`
C. `gamma^(2)=-cos 2theta`
D. `beta^(2)=(1+cos 2 theta)/2`

Explanation

To solve the problem, we need to analyze the given information about the unit vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- We have two unit vectors \(\vec{a}\) and \(\vec{b}\) that are perpendicular to each other. Therefore, we know:
\[ |\vec{a}| = 1, \quad |\vec{b}| = 1, \quad \vec{a} \cdot \vec{b} = 0 \]

2. **Unit Vector \(\vec{c}\)**:
- The unit vector \(\vec{c}\) is inclined at an angle \(\theta\) to both \(\vec{a}\) and \(\vec{b}\). Thus, we can express the dot products:
\[ \vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \theta = 1 \cdot 1 \cdot \cos \theta = \cos \theta \]
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta = 1 \cdot 1 \cdot \cos \theta = \cos \theta \]

3. **Expression for \(\vec{c}\)**:
- The vector \(\vec{c}\) is given as:
\[ \vec{c} = \alpha \vec{a} + \beta \vec{b} + \gamma (\vec{a} \times \vec{b}) \]
- Here, \(\alpha\), \(\beta\), and \(\gamma\) are coefficients we need to determine.

4. **Dot Product with \(\vec{a}\)**:
- Taking the dot product of \(\vec{c}\) with \(\vec{a}\):
\[ \vec{c} \cdot \vec{a} = (\alpha \vec{a} + \beta \vec{b} + \gamma (\vec{a} \times \vec{b})) \cdot \vec{a} \]
- This simplifies to:
\[ \alpha (\vec{a} \cdot \vec{a}) + \beta (\vec{b} \cdot \vec{a}) + \gamma ((\vec{a} \times \vec{b}) \cdot \vec{a}) \]
- Since \(\vec{a} \cdot \vec{a} = 1\), \(\vec{b} \cdot \vec{a} = 0\), and \((\vec{a} \times \vec{b}) \cdot \vec{a} = 0\), we have:
\[ \vec{c} \cdot \vec{a} = \alpha \]
- Therefore, we find that:
\[ \alpha = \cos \theta \]

5. **Dot Product with \(\vec{b}\)**:
- Similarly, taking the dot product of \(\vec{c}\) with \(\vec{b}\):
\[ \vec{c} \cdot \vec{b} = \alpha (\vec{a} \cdot \vec{b}) + \beta (\vec{b} \cdot \vec{b}) + \gamma ((\vec{a} \times \vec{b}) \cdot \vec{b}) \]
- This simplifies to:
\[ 0 + \beta (1) + 0 = \beta \]
- Therefore, we find that:
\[ \beta = \cos \theta \]

6. **Magnitude of \(\vec{c}\)**:
- Since \(\vec{c}\) is a unit vector, we have:
\[ |\vec{c}|^2 = 1 \]
- Expanding the magnitude:
\[ |\vec{c}|^2 = \alpha^2 + \beta^2 + \gamma^2 \]
- Substituting \(\alpha\) and \(\beta\):
\[ 1 = \cos^2 \theta + \cos^2 \theta + \gamma^2 \]
\[ 1 = 2\cos^2 \theta + \gamma^2 \]
- Rearranging gives:
\[ \gamma^2 = 1 - 2\cos^2 \theta \]

7. **Identifying Incorrect Option**:
- From our calculations, we have determined that \(\alpha = \beta = \cos \theta\).
- Therefore, any statement asserting that \(\alpha \neq \beta\) is incorrect.

### Conclusion:
The incorrect statement among the options provided is:
- **Option 1: \(\alpha \neq \beta\)** (This is incorrect since \(\alpha = \beta = \cos \theta\)).