Can a line make angle `45^(@), 60^(@), 120^(@)` with `x-,y-` and z-axes respectively.

Answer

To determine if a line can make angles of \(45^\circ\), \(60^\circ\), and \(120^\circ\) with the \(x\)-, \(y\)-, and \(z\)-axes respectively, we can use the relationship between the angles and the direction cosines of the line.

### Step-by-Step Solution:

1. **Understand the Relationship**:
The direction cosines of a line making angles \(\alpha\), \(\beta\), and \(\gamma\) with the \(x\)-, \(y\)-, and \(z\)-axes are given by:
\[ \cos \alpha, \cos \beta, \cos \gamma \]
The condition that must be satisfied for a line to exist is:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

2. **Substitute the Angles**:
Given:
- \(\alpha = 45^\circ\)
- \(\beta = 60^\circ\)
- \(\gamma = 120^\circ\)

We can calculate the cosines:
\[ \cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 60^\circ = \frac{1}{2}, \quad \cos 120^\circ = -\frac{1}{2} \]

3. **Calculate the Squares of the Cosines**:
Now, we compute:
\[ \cos^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
\[ \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
\[ \cos^2 120^\circ = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \]

4. **Add the Squares Together**:
Now, we sum these values:
\[ \cos^2 45^\circ + \cos^2 60^\circ + \cos^2 120^\circ = \frac{1}{2} + \frac{1}{4} + \frac{1}{4} \]

5. **Find a Common Denominator**:
The common denominator for the fractions is 4:
\[ \frac{1}{2} = \frac{2}{4} \]
Thus:
\[ \frac{2}{4} + \frac{1}{4} + \frac{1}{4} = \frac{2 + 1 + 1}{4} = \frac{4}{4} = 1 \]

6. **Conclusion**:
Since the sum of the squares of the cosines equals 1, it confirms that a line can indeed make angles of \(45^\circ\), \(60^\circ\), and \(120^\circ\) with the \(x\)-, \(y\)-, and \(z\)-axes respectively.

### Final Answer:
Yes, a line can make angles of \(45^\circ\), \(60^\circ\), and \(120^\circ\) with the \(x\)-, \(y\)-, and \(z\)-axes respectively.