Position vector ` hat k` is rotated about the origin by angle `135^0` in such a way that the plane made by it bisects the angel between ` hat ia n d hatjdot` Then its new position is `+-( hat i)/(sqrt(2))+-( hat j)/(sqrt(2))` b. `+-( hat i)/2+-( hat j)/2-( hat k)/(sqrt(2))` c. `( hat i)/(sqrt(2))-( hat k)/(sqrt(2))` d. none of these

Options

A. `+-hati/sqrt2+-hatj/sqrt2`
B. `+-hati/2+-hatj/2-hatk/sqrt2`
C. `hati/sqrt2-hatk/sqrt2`
D. none of these

Explanation

To solve the problem step by step, we will analyze the rotation of the position vector \( \hat{k} \) about the origin by an angle of \( 135^\circ \) and find its new position vector.

### Step 1: Understand the Rotation
The position vector \( \hat{k} \) is rotated about the origin by \( 135^\circ \). The angle \( 135^\circ \) can be expressed in radians as \( \frac{3\pi}{4} \).

### Step 2: Calculate the Cosine of the Angle
We know that:
\[ \cos(135^\circ) = -\frac{1}{\sqrt{2}} \]

### Step 3: Define the New Position Vector
Let the new position vector after rotation be represented as \( \mathbf{r} \). We can express \( \mathbf{r} \) in terms of its components along \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \):
\[ \mathbf{r} = \lambda \hat{k} + \mu_1 \hat{i} + \mu_2 \hat{j} \]
where \( \lambda \) is the component along \( \hat{k} \), and \( \mu_1 \) and \( \mu_2 \) are the components along \( \hat{i} \) and \( \hat{j} \) respectively.

### Step 4: Use the Dot Product
From the rotation, we have:
\[ \mathbf{r} \cdot \hat{k} = \lambda \]
Given that the rotation is about \( \hat{k} \), we also have:
\[ \mathbf{r} \cdot \hat{k} = -\frac{1}{\sqrt{2}} \]
Thus, we find:
\[ \lambda = -\frac{1}{\sqrt{2}} \]

### Step 5: Find the Magnitude of the New Position Vector
The magnitude of \( \mathbf{r} \) should remain 1 (since we are rotating a unit vector):
\[ |\mathbf{r}| = \sqrt{\lambda^2 + \mu_1^2 + \mu_2^2} = 1 \]
Substituting \( \lambda \):
\[ \sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 + \mu_1^2 + \mu_2^2} = 1 \]
This simplifies to:
\[ \sqrt{\frac{1}{2} + \mu_1^2 + \mu_2^2} = 1 \]
Squaring both sides gives:
\[ \frac{1}{2} + \mu_1^2 + \mu_2^2 = 1 \]
Thus:
\[ \mu_1^2 + \mu_2^2 = \frac{1}{2} \]

### Step 6: Choose Components for \( \mu_1 \) and \( \mu_2 \)
Assuming \( \mu_1 = \mu_2 = \mu \):
\[ 2\mu^2 = \frac{1}{2} \implies \mu^2 = \frac{1}{4} \implies \mu = \pm \frac{1}{2} \]

### Step 7: Substitute Back into the Position Vector
Substituting \( \lambda \) and \( \mu \) back into the expression for \( \mathbf{r} \):
\[ \mathbf{r} = -\frac{1}{\sqrt{2}} \hat{k} + \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} \]
This can be expressed as:
\[ \mathbf{r} = \pm \frac{1}{2} \hat{i} \pm \frac{1}{2} \hat{j} - \frac{1}{\sqrt{2}} \hat{k} \]

### Final Answer
Thus, the new position vector after rotation is:
\[ \mathbf{r} = \pm \frac{1}{2} \hat{i} \pm \frac{1}{2} \hat{j} - \frac{1}{\sqrt{2}} \hat{k} \]

### Conclusion
The correct option is:
**b. \( \pm \frac{\hat{i}}{2} \pm \frac{\hat{j}}{2} - \frac{\hat{k}}{\sqrt{2}} \)**