Unit vector `vecc` is inclined at an angle `theta` to unit vectors `veca and vecb` which are perpendicular.
If `vecc=lambda(veca+vecb)+mu(veca xx vecb), lambda, mu` real, then `theta` belongs to:

Options

A. `(-(pi)/(4),0)`
B. `[0,(pi)/(4))`
C. `[(3pi)/(4),pi)`
D. `[(pi)/(4),(3pi)/(4)]`

Explanation

To solve the problem, we need to analyze the given information and derive the necessary relationships step by step.

### Step 1: Understand the vectors involved
We have three unit vectors: \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). The vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular to each other. Therefore, we know that:
\[ \vec{a} \cdot \vec{b} = 0 \]

### Step 2: Express \(\vec{c}\) in terms of \(\vec{a}\) and \(\vec{b}\)
The vector \(\vec{c}\) is given by:
\[ \vec{c} = \lambda \vec{a} + \mu (\vec{a} \times \vec{b}) \]
where \(\lambda\) and \(\mu\) are real numbers.

### Step 3: Calculate the dot product of \(\vec{c}\) with \(\vec{a}\)
Taking the dot product of \(\vec{c}\) with \(\vec{a}\):
\[ \vec{c} \cdot \vec{a} = (\lambda \vec{a} + \mu (\vec{a} \times \vec{b})) \cdot \vec{a} \]
This simplifies to:
\[ \vec{c} \cdot \vec{a} = \lambda (\vec{a} \cdot \vec{a}) + \mu ((\vec{a} \times \vec{b}) \cdot \vec{a}) \]
Since \(\vec{a} \cdot \vec{a} = 1\) and \((\vec{a} \times \vec{b}) \cdot \vec{a} = 0\) (because the cross product is perpendicular to both \(\vec{a}\) and \(\vec{b}\)), we have:
\[ \vec{c} \cdot \vec{a} = \lambda \]

### Step 4: Calculate the dot product of \(\vec{c}\) with \(\vec{b}\)
Now, taking the dot product of \(\vec{c}\) with \(\vec{b}\):
\[ \vec{c} \cdot \vec{b} = (\lambda \vec{a} + \mu (\vec{a} \times \vec{b})) \cdot \vec{b} \]
This simplifies to:
\[ \vec{c} \cdot \vec{b} = \lambda (\vec{a} \cdot \vec{b}) + \mu ((\vec{a} \times \vec{b}) \cdot \vec{b}) \]
Again, since \(\vec{a} \cdot \vec{b} = 0\) and \((\vec{a} \times \vec{b}) \cdot \vec{b} = 0\), we find:
\[ \vec{c} \cdot \vec{b} = 0 \]

### Step 5: Calculate the magnitude of \(\vec{c}\)
Since \(\vec{c}\) is a unit vector, we have:
\[ \|\vec{c}\|^2 = 1 \]
Calculating the magnitude:
\[ \|\vec{c}\|^2 = (\lambda \vec{a} + \mu (\vec{a} \times \vec{b})) \cdot (\lambda \vec{a} + \mu (\vec{a} \times \vec{b})) \]
This expands to:
\[ \|\vec{c}\|^2 = \lambda^2 (\vec{a} \cdot \vec{a}) + \mu^2 ((\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b})) + 2\lambda\mu (\vec{a} \cdot (\vec{a} \times \vec{b})) \]
Since \(\vec{a} \cdot \vec{a} = 1\) and \((\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b}) = 1\) (because \(\|\vec{a} \times \vec{b}\| = 1\)), and the last term is zero:
\[ 1 = \lambda^2 + \mu^2 \]

### Step 6: Relate \(\lambda\) and \(\mu\) to \(\theta\)
From the earlier steps, we have:
\[ \lambda = \cos \theta \]
Substituting this into the equation:
\[ 1 = \cos^2 \theta + \mu^2 \]
This gives:
\[ \mu^2 = 1 - \cos^2 \theta = \sin^2 \theta \]
Thus, \(\mu = \sin \theta\) or \(\mu = -\sin \theta\).

### Step 7: Determine the range of \(\theta\)
Since \(\mu^2 \geq 0\), we have:
\[ 1 - \cos^2 \theta \geq 0 \implies \cos^2 \theta \leq 1 \]
This means:
\[ \cos \theta \in [-1, 1] \]
From the earlier derived relationship, we have:
\[ \cos \theta \leq \frac{1}{\sqrt{2}} \implies \theta \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right] \]

### Final Answer
Thus, the angle \(\theta\) belongs to:
\[ \theta \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right] \]