A particle is moving in xy-plane in a circular path with centre at origin. If at an instant the position of particle is given by `(1)/(sqrt(2)) ( hat(i) + hat(j))`, then velocity of particle is along

Options

A. `(1)/(sqrt(2)) (hat(i) - hat(j))`
B. `(1)/(sqrt(2))( hat(j) - hat(i))`
C. `(1)/(sqrt(2))( hat(i) + hat(j))`
D. Either (1) or (2)

Answer

To solve the problem step by step, we need to find the direction of the velocity of a particle moving in a circular path in the xy-plane, given its position vector.

### Step 1: Identify the position vector
The position vector of the particle is given as:
\[ \mathbf{r} = \frac{1}{\sqrt{2}} \hat{i} + \hat{j} \]

### Step 2: Determine the angular velocity vector
Since the particle is moving in a circular path in the xy-plane, the angular velocity vector \(\mathbf{\omega}\) will be along the z-axis. It can be either in the positive z-direction (\(\hat{k}\)) or negative z-direction (\(-\hat{k}\)).

### Step 3: Calculate the velocity vector using the cross product
The velocity \(\mathbf{v}\) of the particle can be calculated using the formula:
\[ \mathbf{v} = \mathbf{r} \times \mathbf{\omega} \]
Substituting the position vector and the angular velocity vector, we have:
\[ \mathbf{v} = \left(\frac{1}{\sqrt{2}} \hat{i} + \hat{j}\right) \times \mathbf{\omega} \]

### Step 4: Consider both cases for \(\mathbf{\omega}\)
1. **Case 1**: If \(\mathbf{\omega} = \hat{k}\):
\[ \mathbf{v} = \left(\frac{1}{\sqrt{2}} \hat{i} + \hat{j}\right) \times \hat{k} \]
Using the right-hand rule for the cross product:
\[ \hat{i} \times \hat{k} = -\hat{j}, \quad \hat{j} \times \hat{k} = \hat{i} \]
Therefore,
\[ \mathbf{v} = \frac{1}{\sqrt{2}}(-\hat{j}) + \hat{i} = \frac{1}{\sqrt{2}} \hat{i} - \hat{j} \]

2. **Case 2**: If \(\mathbf{\omega} = -\hat{k}\):
\[ \mathbf{v} = \left(\frac{1}{\sqrt{2}} \hat{i} + \hat{j}\right) \times (-\hat{k}) \]
This gives:
\[ \mathbf{v} = -\left(\frac{1}{\sqrt{2}}(-\hat{j}) + \hat{i}\right) = \frac{1}{\sqrt{2}} \hat{j} - \hat{i} \]

### Step 5: Conclusion
Thus, the velocity of the particle can either be:
\[ \mathbf{v} = \frac{1}{\sqrt{2}} \hat{i} - \hat{j} \quad \text{or} \quad \mathbf{v} = \frac{1}{\sqrt{2}} \hat{j} - \hat{i} \]
This indicates that the velocity vector is along the directions of these two vectors.

### Final Answer
The velocity of the particle is along either:
1. \(\frac{1}{\sqrt{2}} \hat{i} - \hat{j}\)
2. \(\frac{1}{\sqrt{2}} \hat{j} - \hat{i}\)