`veca and vecb` are two unit vectors that are mutually perpendicular. A unit vector that if equally inclined to `veca, vecb and veca xxvecb` is equal to

Options

A. `1/sqrt2(veca+vecb+vecaxxvecb)`
B. `1/2(vecaxxvecb+veca+vecb)`
C. `1/sqrt3(veca+vecb+vecaxxvecb)`
D. `1/3(veca+vecb+vecaxxvecb)`

Explanation

To solve the problem step by step, we will find a unit vector that is equally inclined to the two unit vectors \(\vec{a}\) and \(\vec{b}\), which are mutually perpendicular, as well as the vector \(\vec{a} \times \vec{b}\).

### Step 1: Define the Required Vector
Let the required unit vector be represented as:
\[ \vec{r} = x_1 \vec{a} + x_2 \vec{b} + x_3 (\vec{a} \times \vec{b}) \]

### Step 2: Set Up the Equations for Equal Inclination
Since \(\vec{r}\) is equally inclined to \(\vec{a}\), \(\vec{b}\), and \(\vec{a} \times \vec{b}\), we can write the following equations based on the dot product:
\[ \vec{r} \cdot \vec{a} = \vec{r} \cdot \vec{b} = \vec{r} \cdot (\vec{a} \times \vec{b}) \]

### Step 3: Express the Dot Products
From the definition of \(\vec{r}\):
- \(\vec{r} \cdot \vec{a} = x_1\)
- \(\vec{r} \cdot \vec{b} = x_2\)
- \(\vec{r} \cdot (\vec{a} \times \vec{b}) = x_3\)

Thus, we have:
\[ x_1 = x_2 = x_3 \]

### Step 4: Substitute the Values
Let \(x_1 = x_2 = x_3 = \lambda\). Therefore, we can rewrite \(\vec{r}\) as:
\[ \vec{r} = \lambda \vec{a} + \lambda \vec{b} + \lambda (\vec{a} \times \vec{b}) = \lambda (\vec{a} + \vec{b} + \vec{a} \times \vec{b}) \]

### Step 5: Normalize the Vector
Since \(\vec{r}\) is a unit vector, we need to satisfy the condition:
\[ |\vec{r}| = 1 \]
Thus, we have:
\[ |\lambda (\vec{a} + \vec{b} + \vec{a} \times \vec{b})| = 1 \]
This implies:
\[ \lambda^2 |\vec{a} + \vec{b} + \vec{a} \times \vec{b}|^2 = 1 \]

### Step 6: Calculate the Magnitude
To find \(|\vec{a} + \vec{b} + \vec{a} \times \vec{b}|\), we need to compute:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{a} \times \vec{b}|^2 \]
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors:
\[ |\vec{a}|^2 = 1, \quad |\vec{b}|^2 = 1, \quad |\vec{a} \times \vec{b}| = 1 \]
Thus:
\[ |\vec{a} + \vec{b} + \vec{a} \times \vec{b}|^2 = 1 + 1 + 1 = 3 \]

### Step 7: Solve for \(\lambda\)
Substituting back, we have:
\[ \lambda^2 \cdot 3 = 1 \implies \lambda^2 = \frac{1}{3} \implies \lambda = \pm \frac{1}{\sqrt{3}} \]

### Step 8: Final Expression for \(\vec{r}\)
Thus, the required unit vector \(\vec{r}\) is:
\[ \vec{r} = \pm \frac{1}{\sqrt{3}} (\vec{a} + \vec{b} + \vec{a} \times \vec{b}) \]

### Final Answer
The required unit vector that is equally inclined to \(\vec{a}\), \(\vec{b}\), and \(\vec{a} \times \vec{b}\) is:
\[ \vec{r} = \pm \frac{1}{\sqrt{3}} (\vec{a} + \vec{b} + \vec{a} \times \vec{b}) \]