In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by k and `2hat(i) - 2hat(j)` , respectively. What is the unit vector along direction of propagation of the wave.

Options

A. `1/(sqrt(2)) (hat(i)+hat(j))`
B. `1/(sqrt(5))(2hat(i)+hat(j))`
C. `1/(sqrt(5)) (2hat(i)+2hat(j))`
D. `1/(sqrt(2)) (hat(j)+hat(k))`

Explanation

To solve the problem, we need to find the unit vector along the direction of propagation of a plane electromagnetic wave, given the directions of the electric field and magnetic field.

### Step-by-Step Solution:

1. **Identify the Electric Field (E) and Magnetic Field (B) Vectors:**
- The electric field is represented by the unit vector \( \hat{k} \) (which we can assume is in the z-direction).
- The magnetic field is given as \( \vec{B} = 2\hat{i} - 2\hat{j} \).

2. **Determine the Direction of Propagation:**
- In an electromagnetic wave, the electric field \( \vec{E} \), magnetic field \( \vec{B} \), and the direction of propagation \( \vec{k} \) are mutually perpendicular.
- We can find the direction of propagation by taking the cross product of the electric field and magnetic field vectors: \( \vec{k} = \vec{E} \times \vec{B} \).

3. **Set Up the Cross Product:**
- Let's represent the electric field vector \( \vec{E} \) as \( \vec{E} = 0\hat{i} + 0\hat{j} + E\hat{k} \) (where \( E \) is the magnitude of the electric field).
- The magnetic field vector is \( \vec{B} = 2\hat{i} - 2\hat{j} + 0\hat{k} \).

4. **Calculate the Cross Product:**
- The cross product \( \vec{E} \times \vec{B} \) can be calculated using the determinant of a matrix:
\[ \vec{E} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & E \\ 2 & -2 & 0 \end{vmatrix} \]

- Expanding this determinant:
\[ = \hat{i} \left( 0 \cdot 0 - E \cdot (-2) \right) - \hat{j} \left( 0 \cdot 0 - E \cdot 2 \right) + \hat{k} \left( 0 \cdot (-2) - 0 \cdot 2 \right) \]
\[ = \hat{i} (2E) + \hat{j} (2E) + 0\hat{k} \]
\[ = 2E\hat{i} + 2E\hat{j} \]

5. **Find the Unit Vector:**
- The resulting vector \( \vec{k} = 2E\hat{i} + 2E\hat{j} \).
- To find the unit vector in the direction of propagation, we need to divide by its magnitude:
\[ |\vec{k}| = \sqrt{(2E)^2 + (2E)^2} = \sqrt{4E^2 + 4E^2} = \sqrt{8E^2} = 2E\sqrt{2} \]
- Therefore, the unit vector \( \hat{k} \) is:
\[ \hat{k} = \frac{2E\hat{i} + 2E\hat{j}}{2E\sqrt{2}} = \frac{\hat{i} + \hat{j}}{\sqrt{2}} \]

### Final Answer:
The unit vector along the direction of propagation of the wave is:
\[ \hat{k} = \frac{\hat{i} + \hat{j}}{\sqrt{2}} \]