Unit vectors `veca and vecb` ar perpendicular , and unit vector `vecc` is inclined at an angle `theta` to both `veca and vecb . If alpha veca + beta vecb + gamma (veca xx vecb)` then.

Options

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

Explanation

To solve the given problem step by step, we will analyze the relationships between the vectors and their properties.

### Step 1: Understand the given vectors
We have three unit vectors: \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \). It is given that \( \vec{a} \) and \( \vec{b} \) are perpendicular to each other. This means:
\[ \vec{a} \cdot \vec{b} = 0 \]
Also, since they are unit vectors, we have:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]

### Step 2: Analyze the vector \( \vec{c} \)
The vector \( \vec{c} \) is inclined at an angle \( \theta \) to both \( \vec{a} \) and \( \vec{b} \). Therefore, we can express the dot products:
\[ \vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \theta = 1 \cdot 1 \cdot \cos \theta = \cos \theta \]
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta = 1 \cdot 1 \cdot \cos \theta = \cos \theta \]

### Step 3: Set up the equation involving \( \vec{c} \)
We are given that:
\[ \vec{c} = \alpha \vec{a} + \beta \vec{b} + \gamma (\vec{a} \times \vec{b}) \]
We need to find \( \alpha \), \( \beta \), and \( \gamma \).

### Step 4: Calculate \( \vec{a} \cdot \vec{c} \)
Taking the dot product of \( \vec{c} \) with \( \vec{a} \):
\[ \vec{a} \cdot \vec{c} = \vec{a} \cdot (\alpha \vec{a} + \beta \vec{b} + \gamma (\vec{a} \times \vec{b})) \]
Using the properties of dot products:
\[ \vec{a} \cdot \vec{c} = \alpha (\vec{a} \cdot \vec{a}) + \beta (\vec{a} \cdot \vec{b}) + \gamma (\vec{a} \cdot (\vec{a} \times \vec{b})) \]
Since \( \vec{a} \cdot \vec{a} = 1 \), \( \vec{a} \cdot \vec{b} = 0 \), and \( \vec{a} \cdot (\vec{a} \times \vec{b}) = 0 \):
\[ \vec{a} \cdot \vec{c} = \alpha \]

### Step 5: Relate \( \alpha \) to \( \cos \theta \)
From Step 2, we know that:
\[ \vec{a} \cdot \vec{c} = \cos \theta \]
Thus, we have:
\[ \alpha = \cos \theta \]

### Step 6: Calculate \( \vec{b} \cdot \vec{c} \)
Now, taking the dot product of \( \vec{c} \) with \( \vec{b} \):
\[ \vec{b} \cdot \vec{c} = \vec{b} \cdot (\alpha \vec{a} + \beta \vec{b} + \gamma (\vec{a} \times \vec{b})) \]
Using the properties of dot products:
\[ \vec{b} \cdot \vec{c} = \alpha (\vec{b} \cdot \vec{a}) + \beta (\vec{b} \cdot \vec{b}) + \gamma (\vec{b} \cdot (\vec{a} \times \vec{b})) \]
Again, since \( \vec{b} \cdot \vec{a} = 0 \), \( \vec{b} \cdot \vec{b} = 1 \), and \( \vec{b} \cdot (\vec{a} \times \vec{b}) = 0 \):
\[ \vec{b} \cdot \vec{c} = \beta \]

### Step 7: Relate \( \beta \) to \( \cos \theta \)
From Step 2, we also know that:
\[ \vec{b} \cdot \vec{c} = \cos \theta \]
Thus, we have:
\[ \beta = \cos \theta \]

### Step 8: Conclusion
Since both \( \alpha \) and \( \beta \) are equal to \( \cos \theta \), we conclude:
\[ \alpha = \beta \]

### Final Answer
The correct answer is:
\[ \alpha = \beta \]