The unit vector in `XOZ` plane and making angles `45^@` and `60^@` respectively with `vec(a)=2i+2j-k` and `vecb=0i+j-k`, is

Options

A. `(1)/(sqrt(2))(-hati+hatk)`
B. `(1)/(sqrt(2))(hati-hatk)`
C. `(sqrt(3))/(2)(hati+hatk)`
D. none of these

Answer

To solve the problem, we need to find a unit vector \( \vec{L} \) in the \( XOZ \) plane that makes angles of \( 45^\circ \) and \( 60^\circ \) with the vectors \( \vec{a} = 2\hat{i} + 2\hat{j} - \hat{k} \) and \( \vec{b} = 0\hat{i} + \hat{j} - \hat{k} \) respectively.

### Step-by-Step Solution:

1. **Define the Unit Vector in the XOZ Plane:**
Since the unit vector \( \vec{L} \) lies in the \( XOZ \) plane, we can express it as:
\[ \vec{L} = a\hat{i} + b\hat{k} \]
where \( a \) and \( b \) are the components along the \( x \) and \( z \) axes respectively.

2. **Magnitude of the Unit Vector:**
The magnitude of \( \vec{L} \) must equal 1 because it is a unit vector:
\[ \sqrt{a^2 + b^2} = 1 \implies a^2 + b^2 = 1 \quad \text{(Equation 1)} \]

3. **Dot Product with Vector \( \vec{a} \):**
The unit vector \( \vec{L} \) makes an angle of \( 45^\circ \) with \( \vec{a} \). Using the dot product formula:
\[ \vec{L} \cdot \vec{a} = |\vec{L}| |\vec{a}| \cos(45^\circ) \]
Calculating \( |\vec{a}| \):
\[ |\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]
Thus, we have:
\[ \vec{L} \cdot \vec{a} = (a\hat{i} + b\hat{k}) \cdot (2\hat{i} + 2\hat{j} - \hat{k}) = 2a + 0 - b = 2a - b \]
Therefore, we can write:
\[ 2a - b = 3 \cdot \frac{1}{\sqrt{2}} \quad \text{(Equation 2)} \]

4. **Dot Product with Vector \( \vec{b} \):**
The unit vector \( \vec{L} \) also makes an angle of \( 60^\circ \) with \( \vec{b} \):
\[ \vec{L} \cdot \vec{b} = |\vec{L}| |\vec{b}| \cos(60^\circ) \]
Calculating \( |\vec{b}| \):
\[ |\vec{b}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]
Thus, we have:
\[ \vec{L} \cdot \vec{b} = (a\hat{i} + b\hat{k}) \cdot (0\hat{i} + \hat{j} - \hat{k}) = 0 + 0 - b = -b \]
Therefore, we can write:
\[ -b = 1 \cdot \frac{1}{2} \implies b = -\frac{1}{2} \quad \text{(Equation 3)} \]

5. **Substituting \( b \) into Equation 1:**
Substitute \( b = -\frac{1}{2} \) into Equation 1:
\[ a^2 + \left(-\frac{1}{2}\right)^2 = 1 \implies a^2 + \frac{1}{4} = 1 \implies a^2 = 1 - \frac{1}{4} = \frac{3}{4} \implies a = \pm \frac{\sqrt{3}}{2} \]

6. **Finding the Unit Vector \( \vec{L} \):**
We can take \( a = \frac{\sqrt{3}}{2} \) (the positive root) for the unit vector:
\[ \vec{L} = \frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{k} \]

7. **Final Form of the Unit Vector:**
To express \( \vec{L} \) in a more standard form, we can factor out \( \frac{1}{2} \):
\[ \vec{L} = \frac{1}{2} \left( \sqrt{3}\hat{i} - \hat{k} \right) \]

### Conclusion:
The unit vector \( \vec{L} \) in the \( XOZ \) plane making angles of \( 45^\circ \) and \( 60^\circ \) with the given vectors is:
\[ \vec{L} = \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{k} \]