A unit vector a makes an angle `pi//4` with z-axis and if `a + i+j` is a unit vector,k then a =

Options

A. `((1)/(2), (1)/(2),(1)/(sqrt2))`
B. `(1)/(2), (1)/(2),(1)/(sqrt2)`
C. `(-(1)/(2),-(1)/(2),(1)/(sqrt2))`
D. none

Answer

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

### Step-by-Step Solution:

1. **Define the Unit Vector**:
Since \( \mathbf{a} \) is a unit vector making an angle \( \frac{\pi}{4} \) with the z-axis, we can express it in terms of its components:
\[ \mathbf{a} = l \mathbf{i} + m \mathbf{j} + n \mathbf{k} \]
where \( l, m, n \) are the components along the x, y, and z axes respectively.

2. **Use the Angle Condition**:
The angle \( \frac{\pi}{4} \) with the z-axis gives us:
\[ n = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

3. **Magnitude of the Unit Vector**:
Since \( \mathbf{a} \) is a unit vector, its magnitude must equal 1:
\[ \sqrt{l^2 + m^2 + n^2} = 1 \]
Squaring both sides, we get:
\[ l^2 + m^2 + n^2 = 1 \]

4. **Substituting for \( n \)**:
Substitute \( n = \frac{1}{\sqrt{2}} \) into the magnitude equation:
\[ l^2 + m^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = 1 \]
This simplifies to:
\[ l^2 + m^2 + \frac{1}{2} = 1 \]
Therefore,
\[ l^2 + m^2 = 1 - \frac{1}{2} = \frac{1}{2} \]

5. **Condition for the Vector \( \mathbf{a} + \mathbf{i} + \mathbf{j} \)**:
We know that \( \mathbf{a} + \mathbf{i} + \mathbf{j} \) is also a unit vector:
\[ \mathbf{a} + \mathbf{i} + \mathbf{j} = (l + 1) \mathbf{i} + (m + 1) \mathbf{j} + n \mathbf{k} \]
The magnitude of this vector must also equal 1:
\[ \sqrt{(l + 1)^2 + (m + 1)^2 + n^2} = 1 \]
Squaring both sides gives:
\[ (l + 1)^2 + (m + 1)^2 + n^2 = 1 \]

6. **Expanding the Equation**:
Expanding the left-hand side:
\[ (l^2 + 2l + 1) + (m^2 + 2m + 1) + n^2 = 1 \]
Substituting \( n^2 = \frac{1}{2} \) and \( l^2 + m^2 = \frac{1}{2} \):
\[ \left(\frac{1}{2} + 2l + 1\right) + \left(\frac{1}{2} + 2m + 1\right) + \frac{1}{2} = 1 \]
Simplifying gives:
\[ 2l + 2m + 2 = 1 \]
Thus,
\[ l + m = -\frac{1}{2} \]

7. **Solving the System of Equations**:
We now have two equations:
- \( l^2 + m^2 = \frac{1}{2} \) (Equation 1)
- \( l + m = -\frac{1}{2} \) (Equation 2)

From Equation 2, we can express \( m \) in terms of \( l \):
\[ m = -\frac{1}{2} - l \]
Substituting into Equation 1:
\[ l^2 + \left(-\frac{1}{2} - l\right)^2 = \frac{1}{2} \]
Expanding and simplifying:
\[ l^2 + \left(\frac{1}{4} + l + l^2\right) = \frac{1}{2} \]
This leads to:
\[ 2l^2 + l + \frac{1}{4} - \frac{1}{2} = 0 \]
Simplifying gives:
\[ 2l^2 + l - \frac{1}{4} = 0 \]
Multiplying through by 4:
\[ 8l^2 + 4l - 1 = 0 \]
Using the quadratic formula:
\[ l = \frac{-4 \pm \sqrt{16 + 32}}{16} = \frac{-4 \pm 8}{16} \]
This gives:
\[ l = \frac{1}{2} \quad \text{or} \quad l = -\frac{3}{4} \]

8. **Finding \( m \)**:
Using \( l = -\frac{3}{4} \):
\[ m = -\frac{1}{2} - \left(-\frac{3}{4}\right) = \frac{1}{4} \]
Using \( l = \frac{1}{2} \):
\[ m = -\frac{1}{2} - \frac{1}{2} = -1 \]

9. **Final Values**:
Thus, we have two potential vectors:
- For \( l = -\frac{3}{4}, m = \frac{1}{4}, n = \frac{1}{\sqrt{2}} \)
- For \( l = \frac{1}{2}, m = -1, n = \frac{1}{\sqrt{2}} \)

Therefore, the unit vector \( \mathbf{a} \) can be expressed as:
\[ \mathbf{a} = -\frac{3}{4} \mathbf{i} + \frac{1}{4} \mathbf{j} + \frac{1}{\sqrt{2}} \mathbf{k} \quad \text{or} \quad \mathbf{a} = \frac{1}{2} \mathbf{i} - \mathbf{j} + \frac{1}{\sqrt{2}} \mathbf{k} \]