A ray of light in incident along a vector `hati + hatj - hatk` on a plane mirror lying in y-z plane. The unit vector along the reflected ray can be

Options

A. `(hati + hatj -hatk)/(sqrt(3))`
B. `(hati - hatj +hatk)/(sqrt(3))`
C. `(-hati + hatj -hatk)/(sqrt(3))`
D. `(3hati + hatj -hatk)/(sqrt(3))`

Explanation

To solve the problem of finding 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 Vector
The incident ray is given as a vector:
\[ \mathbf{e} = \hat{i} + \hat{j} - \hat{k} \]

### Step 2: Normalize the Incident Vector
To find the unit vector of the incident ray, we need to normalize it:
\[ |\mathbf{e}| = \sqrt{(1^2 + 1^2 + (-1)^2)} = \sqrt{3} \]
Thus, the unit vector along the incident ray is:
\[ \mathbf{e}_{\text{unit}} = \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}} \]

### Step 3: Determine the Normal Vector
Since the mirror lies in the y-z plane, the normal vector to the mirror can be taken as:
\[ \mathbf{n} = \hat{i} \]

### Step 4: Apply the Law of Reflection
According to the law of reflection, the reflected vector \(\mathbf{r}\) can be calculated using the formula:
\[ \mathbf{r} = \mathbf{e} - 2(\mathbf{e} \cdot \mathbf{n})\mathbf{n} \]
First, we need to calculate the dot product \(\mathbf{e} \cdot \mathbf{n}\):
\[ \mathbf{e} \cdot \mathbf{n} = \left(\hat{i} + \hat{j} - \hat{k}\right) \cdot \hat{i} = 1 \]

### Step 5: Substitute into the Reflection Formula
Now substituting the values into the reflection formula:
\[ \mathbf{r} = \left(\hat{i} + \hat{j} - \hat{k}\right) - 2(1)\hat{i} \]
\[ \mathbf{r} = \hat{i} + \hat{j} - \hat{k} - 2\hat{i} = -\hat{i} + \hat{j} - \hat{k} \]

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

### Step 7: Check for Additional Reflected Direction
Since the incident ray can reflect in two directions, we also consider the reflection in the opposite direction:
\[ \mathbf{r'} = \mathbf{e} + 2(\mathbf{e} \cdot \mathbf{n})\mathbf{n} \]
This gives:
\[ \mathbf{r'} = \left(\hat{i} + \hat{j} - \hat{k}\right) + 2\hat{i} = 3\hat{i} + \hat{j} - \hat{k} \]
And normalizing this:
\[ \mathbf{r'}_{\text{unit}} = \frac{3\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}} \]

### Final Answer
Thus, the unit vectors along the reflected rays are:
1. \(\frac{-\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}}\)
2. \(\frac{3\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}}\)