Let `hati+yhatj+zhatk` and `xhati-hatj+hatk` are parallel then unit vector parallel to `xhati+yhatj+zhatk`

Options

A. `i/sqrt2(hati-hatj)`
B. `i/sqrt2(hati+hatj)`
C. `i/sqrt3(hati-hatj+hatk)`
D. `i/sqrt3(hati+hatj+hatk)`

Answer

To solve the problem, we need to determine a unit vector that is parallel to the vector \( \mathbf{x} \hat{i} + \mathbf{y} \hat{j} + \mathbf{z} \hat{k} \) given that the vectors \( \mathbf{a_1} = \mathbf{x} \hat{i} - \hat{j} + \hat{k} \) and \( \mathbf{a_2} = \hat{i} + \mathbf{y} \hat{j} + \mathbf{z} \hat{k} \) are parallel (collinear).

### Step-by-Step Solution:

1. **Identify the Vectors**:
- Let \( \mathbf{a_1} = \mathbf{x} \hat{i} - \hat{j} + \hat{k} \)
- Let \( \mathbf{a_2} = \hat{i} + \mathbf{y} \hat{j} + \mathbf{z} \hat{k} \)

2. **Set Up the Condition for Collinearity**:
- For two vectors to be collinear, the ratios of their corresponding components must be equal. Thus, we can write:
\[ \frac{x}{1} = \frac{-1}{y} = \frac{1}{z} = \lambda \]
- Here, \( \lambda \) is a scalar.

3. **Express \( x, y, z \) in terms of \( \lambda \)**:
- From the equations, we can express \( x, y, z \) as:
\[ x = \lambda \]
\[ y = -\frac{1}{\lambda} \]
\[ z = \frac{1}{\lambda} \]

4. **Form the Vector \( \mathbf{v} \)**:
- Now, substitute \( x, y, z \) into the vector \( \mathbf{v} = x \hat{i} + y \hat{j} + z \hat{k} \):
\[ \mathbf{v} = \lambda \hat{i} - \frac{1}{\lambda} \hat{j} + \frac{1}{\lambda} \hat{k} \]

5. **Calculate the Magnitude of \( \mathbf{v} \)**:
- The magnitude \( |\mathbf{v}| \) is given by:
\[ |\mathbf{v}| = \sqrt{\lambda^2 + \left(-\frac{1}{\lambda}\right)^2 + \left(\frac{1}{\lambda}\right)^2} \]
\[ = \sqrt{\lambda^2 + \frac{1}{\lambda^2} + \frac{1}{\lambda^2}} = \sqrt{\lambda^2 + \frac{2}{\lambda^2}} \]

6. **Simplify the Magnitude**:
- Combine the terms:
\[ |\mathbf{v}| = \sqrt{\lambda^2 + \frac{2}{\lambda^2}} = \sqrt{\frac{\lambda^4 + 2}{\lambda^2}} = \frac{\sqrt{\lambda^4 + 2}}{\lambda} \]

7. **Find the Unit Vector**:
- The unit vector \( \mathbf{u} \) parallel to \( \mathbf{v} \) is given by:
\[ \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{\lambda \hat{i} - \frac{1}{\lambda} \hat{j} + \frac{1}{\lambda} \hat{k}}{\frac{\sqrt{\lambda^4 + 2}}{\lambda}} = \frac{\lambda^2 \hat{i} - \hat{j} + \hat{k}}{\sqrt{\lambda^4 + 2}} \]

8. **Choose \( \lambda = 1 \) for Simplicity**:
- If we set \( \lambda = 1 \):
\[ \mathbf{u} = \frac{1 \hat{i} - 1 \hat{j} + 1 \hat{k}}{\sqrt{1^4 + 2}} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} \]

### Final Answer:
The unit vector parallel to \( \mathbf{x} \hat{i} + \mathbf{y} \hat{j} + \mathbf{z} \hat{k} \) is:
\[ \mathbf{u} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} \]