(A) If a line makes angle `90^(@), 135^(@), 45^(@)` with the x , y and z respectively, find its direction-cosines.
(b) If a line has direction-ratio `lt ` 2, -1, -2, `gt` , determine its direction-cosines.

Answer

To solve the given problem, we will break it down into two parts as specified in the question.

### Part (A): Finding Direction Cosines

Given that a line makes angles of \(90^\circ\), \(135^\circ\), and \(45^\circ\) with the x, y, and z axes respectively, we need to find its direction cosines.

1. **Identify the angles**:
- Let \( \alpha = 90^\circ \) (angle with x-axis)
- Let \( \beta = 135^\circ \) (angle with y-axis)
- Let \( \gamma = 45^\circ \) (angle with z-axis)

2. **Use the formula for direction cosines**:
- The direction cosines \( l, m, n \) are given by:
\[ l = \cos \alpha, \quad m = \cos \beta, \quad n = \cos \gamma \]

3. **Calculate \( l \)**:
- \( l = \cos(90^\circ) = 0 \)

4. **Calculate \( m \)**:
- \( m = \cos(135^\circ) = \cos(90^\circ + 45^\circ) = -\sin(45^\circ) = -\frac{1}{\sqrt{2}} \)

5. **Calculate \( n \)**:
- \( n = \cos(45^\circ) = \frac{1}{\sqrt{2}} \)

6. **Direction cosines**:
- Therefore, the direction cosines of the line are:
\[ (l, m, n) = \left(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \]

### Part (B): Finding Direction Cosines from Direction Ratios

Given the direction ratios \( (2, -1, -2) \), we need to find the direction cosines.

1. **Identify the direction ratios**:
- Let the direction ratios be \( a = 2, b = -1, c = -2 \).

2. **Calculate the magnitude of the direction ratios**:
- The magnitude \( |d| \) is given by:
\[ |d| = \sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

3. **Calculate the direction cosines**:
- The direction cosines are given by:
\[ l = \frac{a}{|d|}, \quad m = \frac{b}{|d|}, \quad n = \frac{c}{|d|} \]
- So,
\[ l = \frac{2}{3}, \quad m = \frac{-1}{3}, \quad n = \frac{-2}{3} \]

4. **Direction cosines**:
- Therefore, the direction cosines corresponding to the direction ratios are:
\[ (l, m, n) = \left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right) \]

### Final Answers:
- Part (A): Direction cosines are \( \left(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \)
- Part (B): Direction cosines are \( \left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right) \)