If a vector `vecr` is equall inclined with the vectors `veca=costhetahati+sinthetahatj, vecb=-sinthetahati+costhetahatj` and `vecc=hatk`, then the angle between `vecr` and `veca` is

Options

A. `cos^(-1)(1/sqrt(2))`
B. `cos^(-1)(1/sqrt(3))`
C. `cos^(-1)(1/3)`
D. `cos^(-1)(1/2)`

Explanation

To find the angle between the vector \(\vec{r}\) and the vector \(\vec{a}\), we follow these steps:

### Step 1: Understand the vectors
Given:
- \(\vec{a} = \cos \theta \hat{i} + \sin \theta \hat{j}\)
- \(\vec{b} = -\sin \theta \hat{i} + \cos \theta \hat{j}\)
- \(\vec{c} = \hat{k}\)

The vector \(\vec{r}\) is equally inclined to \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

### Step 2: Set \(\theta = 0\)
For simplicity, let's evaluate the vectors at \(\theta = 0\):
- \(\vec{a} = \cos(0) \hat{i} + \sin(0) \hat{j} = 1 \hat{i} + 0 \hat{j} = \hat{i}\)
- \(\vec{b} = -\sin(0) \hat{i} + \cos(0) \hat{j} = 0 \hat{i} + 1 \hat{j} = \hat{j}\)
- \(\vec{c} = \hat{k}\)

### Step 3: Determine the vector \(\vec{r}\)
Since \(\vec{r}\) is equally inclined to \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), it can be represented as:
\[ \vec{r} = k(\hat{i} + \hat{j} + \hat{k}) \]
where \(k\) is a scalar.

### Step 4: Normalize \(\vec{r}\)
To find the unit vector, we can take \(k = \frac{1}{\sqrt{3}}\) (to ensure that the magnitude is 1):
\[ \vec{r} = \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) \]

### Step 5: Calculate the angle between \(\vec{r}\) and \(\vec{a}\)
The formula for the angle \(\theta\) between two vectors \(\vec{u}\) and \(\vec{v}\) is given by:
\[ \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} \]

Here, \(\vec{u} = \vec{r}\) and \(\vec{v} = \vec{a}\):
- \(\vec{r} \cdot \vec{a} = \left(\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})\right) \cdot \hat{i} = \frac{1}{\sqrt{3}} (1 + 0 + 0) = \frac{1}{\sqrt{3}}\)
- The magnitude of \(\vec{r}\) is 1 (as we normalized it).
- The magnitude of \(\vec{a}\) is also 1.

### Step 6: Substitute into the formula
Substituting into the cosine formula:
\[ \cos \theta = \frac{\frac{1}{\sqrt{3}}}{1 \cdot 1} = \frac{1}{\sqrt{3}} \]

### Step 7: Find the angle \(\theta\)
To find \(\theta\):
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Final Answer
Thus, the angle between \(\vec{r}\) and \(\vec{a}\) is:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]