A unit vector `hata` makes an angle `pi/4` with z-axis, `if hata+hati+hatj` is a unit vector then `hata` is equal to
(A) `hati+hatj+hatk/2` (B) `hati/2+hatj/2-hatk/sqrt(2)` (C) `-hati/2-hat/2+hatk/sqrt(2)` (D) `hati/2-hatj/2-hatk/sqrt(2)`

Options

A. A. `(hati)/(2)+(hatj)/(2)+(hatk)/(sqrt(2))`
B. B. `(hati)/(2)+(hatj)/(2)-(hatk)/(sqrt(2))`
C. C. `-(hati)/(2)-(hatj)/(2)+(hatk)/(sqrt(2))`
D. D. none of these

Explanation

To solve the problem, we need to find the unit vector \(\hat{a}\) that makes an angle \(\frac{\pi}{4}\) with the z-axis, given that \(\hat{a} + \hat{i} + \hat{j}\) is also a unit vector.

### Step-by-step Solution:

1. **Define the Unit Vector**:
Let \(\hat{a} = x \hat{i} + y \hat{j} + z \hat{k}\). Since \(\hat{a}\) is a unit vector, we have:
\[ \|\hat{a}\| = 1 \implies x^2 + y^2 + z^2 = 1 \tag{1} \]

2. **Angle with the z-axis**:
The angle \(\theta\) that \(\hat{a}\) makes with the z-axis is given as \(\frac{\pi}{4}\). The cosine of this angle is:
\[ \cos \theta = \frac{z}{\|\hat{a}\|} = z \quad (\text{since } \|\hat{a}\| = 1) \]
Therefore, we have:
\[ z = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \tag{2} \]

3. **Substituting z in Equation (1)**:
Substitute \(z\) from equation (2) into equation (1):
\[ x^2 + y^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = 1 \]
This simplifies to:
\[ x^2 + y^2 + \frac{1}{2} = 1 \implies x^2 + y^2 = \frac{1}{2} \tag{3} \]

4. **Condition for \(\hat{a} + \hat{i} + \hat{j}\)**:
We know that \(\hat{a} + \hat{i} + \hat{j}\) is a unit vector. Hence:
\[ \|\hat{a} + \hat{i} + \hat{j}\| = 1 \]
This can be expressed as:
\[ \|(x + 1) \hat{i} + (y + 1) \hat{j} + z \hat{k}\| = 1 \]
Therefore:
\[ (x + 1)^2 + (y + 1)^2 + z^2 = 1 \]
Substituting \(z\) from equation (2):
\[ (x + 1)^2 + (y + 1)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = 1 \]
This simplifies to:
\[ (x + 1)^2 + (y + 1)^2 + \frac{1}{2} = 1 \]
Thus:
\[ (x + 1)^2 + (y + 1)^2 = \frac{1}{2} \tag{4} \]

5. **Expanding Equation (4)**:
Expanding equation (4):
\[ (x^2 + 2x + 1) + (y^2 + 2y + 1) = \frac{1}{2} \]
Substituting \(x^2 + y^2 = \frac{1}{2}\) from equation (3):
\[ \frac{1}{2} + 2x + 1 + 1 = \frac{1}{2} \]
This simplifies to:
\[ 2x + 2 = 0 \implies x = -1 \]

6. **Finding y**:
Substitute \(x = -1\) into equation (3):
\[ (-1)^2 + y^2 = \frac{1}{2} \implies 1 + y^2 = \frac{1}{2} \implies y^2 = -\frac{1}{2} \]
Since this is not possible, we need to check our calculations.

Let's consider the equations again. We can also use the equations derived from \(x + y = -1\) and \(x^2 + y^2 = \frac{1}{2}\) to find \(x\) and \(y\).

7. **Solving for x and y**:
From \(y = -1 - x\), substitute into \(x^2 + y^2 = \frac{1}{2}\):
\[ x^2 + (-1 - x)^2 = \frac{1}{2} \]
Expanding:
\[ x^2 + (1 + 2x + x^2) = \frac{1}{2} \implies 2x^2 + 2x + 1 = \frac{1}{2} \]
Rearranging gives:
\[ 2x^2 + 2x + \frac{1}{2} = 0 \implies 4x^2 + 4x + 1 = 0 \]
Solving this quadratic equation using the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{16 - 16}}{8} = -\frac{1}{2} \]

8. **Finding y and z**:
Now substituting \(x = -\frac{1}{2}\) into \(y = -1 - x\):
\[ y = -1 + \frac{1}{2} = -\frac{1}{2} \]
We have \(z = \frac{1}{\sqrt{2}}\).

9. **Final Result**:
Thus, the vector \(\hat{a}\) is:
\[ \hat{a} = -\frac{1}{2} \hat{i} - \frac{1}{2} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \]

### Conclusion:
The correct answer is:
\[ \hat{a} = -\frac{1}{2} \hat{i} - \frac{1}{2} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \quad \text{(Option C)} \]