(a) Derive an expression for unit vector along reflected ray `(hat r)` if unit vectors `hat i` and `hat n` represents unit vectors along incident light ray and normal (at point of reflection and outward from surface) respectively.
(b) If vector along the incident ray on a mirror is `-2 hat i+ 3 hat j + 4 hat k`. Considering the x-axis to be along the normal. Then, find the unit vector along the reflected ray.

Explanation

To solve the problem, we will break it down into two parts as specified in the question.

### Part (a): Deriving the expression for the unit vector along the reflected ray \(\hat{r}\)

1. **Understanding the Geometry**:
- Let \(\hat{i}\) be the unit vector along the incident ray.
- Let \(\hat{n}\) be the unit vector normal to the surface at the point of reflection.
- The angle of incidence \(\theta\) is equal to the angle of reflection \(\theta'\).

2. **Using the Reflection Law**:
- According to the law of reflection, the angle of incidence is equal to the angle of reflection:
\[ \theta = \theta' \]

3. **Dot Product Relations**:
- The dot product of the incident ray vector \(\hat{i}\) with the normal vector \(\hat{n}\) gives:
\[ \hat{i} \cdot \hat{n} = \cos(\theta) \]
- For the reflected ray vector \(\hat{r}\), the dot product with the normal vector gives:
\[ \hat{r} \cdot \hat{n} = -\cos(\theta) \]

4. **Setting Up the Equation**:
- From the above two equations, we can express the relationship between the incident and reflected rays:
\[ \hat{r} \cdot \hat{n} = -\hat{i} \cdot \hat{n} \]

5. **Deriving the Expression**:
- We can express \(\hat{r}\) in terms of \(\hat{i}\) and \(\hat{n}\):
\[ \hat{r} = \hat{i} - 2(\hat{i} \cdot \hat{n})\hat{n} \]
- This equation shows that the reflected ray \(\hat{r}\) can be derived from the incident ray \(\hat{i}\) by subtracting twice the projection of \(\hat{i}\) onto \(\hat{n}\).

### Part (b): Finding the unit vector along the reflected ray

1. **Given Incident Ray Vector**:
- The incident ray vector is given as:
\[ \vec{E_i} = -2\hat{i} + 3\hat{j} + 4\hat{k} \]

2. **Calculating the Magnitude of the Incident Vector**:
- The magnitude of \(\vec{E_i}\) is:
\[ |\vec{E_i}| = \sqrt{(-2)^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

3. **Finding the Unit Vector Along the Incident Ray**:
- The unit vector along the incident ray \(\hat{i}\) is:
\[ \hat{i} = \frac{\vec{E_i}}{|\vec{E_i}|} = \frac{-2\hat{i} + 3\hat{j} + 4\hat{k}}{\sqrt{29}} \]

4. **Normal Vector**:
- Since the x-axis is along the normal, the normal vector is:
\[ \hat{n} = \hat{i} \]

5. **Calculating the Dot Product**:
- Now, we calculate \(\hat{i} \cdot \hat{n}\):
\[ \hat{i} \cdot \hat{n} = \frac{-2}{\sqrt{29}} \]

6. **Finding the Reflected Ray Vector**:
- Using the derived expression for \(\hat{r}\):
\[ \hat{r} = \hat{i} - 2(\hat{i} \cdot \hat{n})\hat{n} \]
- Substitute \(\hat{i} \cdot \hat{n}\):
\[ \hat{r} = \hat{i} - 2\left(-\frac{2}{\sqrt{29}}\right)\hat{n} \]
- Since \(\hat{n} = \hat{i}\):
\[ \hat{r} = \hat{i} + \frac{4}{\sqrt{29}}\hat{i} = \left(1 + \frac{4}{\sqrt{29}}\right)\hat{i} \]

7. **Final Unit Vector Along the Reflected Ray**:
- The unit vector along the reflected ray is:
\[ \hat{r} = \frac{(1 + \frac{4}{\sqrt{29}})\hat{i}}{|\hat{r}|} \]