A ray of light is incident on a plane mirror along a vector `hat i+hat j- hat k.`
The normal on incidence point is along `hat i+ hat j `.Find a unit vector along the
reflected ray.

Explanation

To find the unit vector along the reflected ray when a ray of light is incident on a plane mirror, we can follow these steps:

### Step 1: Identify the incident ray and the normal vector
The incident ray is given as the vector \( \hat{i} + \hat{j} - \hat{k} \) and the normal vector at the point of incidence is given as \( \hat{i} + \hat{j} \).

### Step 2: Normalize the normal vector
To work with unit vectors, we need to normalize the normal vector. The magnitude of the normal vector \( \hat{n} = \hat{i} + \hat{j} \) is calculated as follows:
\[ |\hat{n}| = \sqrt{(1^2 + 1^2)} = \sqrt{2} \]
Thus, the unit normal vector \( \hat{n}_{unit} \) is:
\[ \hat{n}_{unit} = \frac{\hat{n}}{|\hat{n}|} = \frac{\hat{i} + \hat{j}}{\sqrt{2}} \]

### Step 3: Find the component of the incident ray along the normal
To find the component of the incident ray along the normal, we use the formula:
\[ \text{Component along } \hat{n} = (\vec{I} \cdot \hat{n}_{unit}) \hat{n}_{unit} \]
Calculating the dot product:
\[ \vec{I} \cdot \hat{n}_{unit} = (\hat{i} + \hat{j} - \hat{k}) \cdot \left(\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right) = \frac{1 + 1 + 0}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]
Thus, the component along the normal is:
\[ \text{Component along } \hat{n} = \sqrt{2} \cdot \frac{\hat{i} + \hat{j}}{\sqrt{2}} = \hat{i} + \hat{j} \]

### Step 4: Find the component of the incident ray along the plane
The component of the incident ray along the plane (perpendicular to the normal) can be found by subtracting the normal component from the incident ray:
\[ \text{Component along plane} = \vec{I} - \text{Component along } \hat{n} = (\hat{i} + \hat{j} - \hat{k}) - (\hat{i} + \hat{j}) = -\hat{k} \]

### Step 5: Determine the reflected ray
According to the law of reflection, the component along the normal will invert its direction, while the component along the plane remains unchanged. Therefore, the reflected ray \( \vec{R} \) can be expressed as:
\[ \vec{R} = \text{Component along plane} + (-\text{Component along } \hat{n}) = (-\hat{k}) + (-(\hat{i} + \hat{j})) = -\hat{i} - \hat{j} + \hat{k} \]

### Step 6: Normalize the reflected ray to find the unit vector
Now, we need to find the unit vector along the reflected ray:
\[ \vec{R} = -\hat{i} - \hat{j} + \hat{k} \]
Calculating the magnitude of \( \vec{R} \):
\[ |\vec{R}| = \sqrt{(-1)^2 + (-1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]
Thus, the unit vector along the reflected ray \( \hat{R}_{unit} \) is:
\[ \hat{R}_{unit} = \frac{\vec{R}}{|\vec{R}|} = \frac{-\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} = -\frac{1}{\sqrt{3}} \hat{i} - \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \]

### Final Answer
The unit vector along the reflected ray is:
\[ \hat{R}_{unit} = -\frac{1}{\sqrt{3}} \hat{i} - \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \]