A vector ` vec O P` is inclined to `O X at 45^0 and O Y at 60^0` . Find the angle at which ` vec O P` is inclined to `O Zdot`

Options

A. `75^(@)`
B. `60^(@)` or `120^(@)`
C. `75^(@)` or `105^(@)`
D. `255^(@)`

Explanation

To solve the problem of finding the angle at which vector \( \vec{OP} \) is inclined to the \( OZ \) axis, given that it is inclined to the \( OX \) axis at \( 45^\circ \) and to the \( OY \) axis at \( 60^\circ \), we can follow these steps:

### Step-by-Step Solution:

1. **Define the Angles**:
Let:
- \( \alpha = 45^\circ \) (angle with the \( OX \) axis)
- \( \beta = 60^\circ \) (angle with the \( OY \) axis)
- \( \gamma \) = angle with the \( OZ \) axis (unknown)

2. **Use the Cosine Relation**:
We know from the properties of angles in three-dimensional space that:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

3. **Substitute Known Values**:
Substitute the values of \( \alpha \) and \( \beta \):
\[ \cos^2(45^\circ) + \cos^2(60^\circ) + \cos^2(\gamma) = 1 \]

4. **Calculate Cosine Values**:
- \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \cos(60^\circ) = \frac{1}{2} \)

Therefore:
\[ \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2(\gamma) = 1 \]
Simplifying gives:
\[ \frac{1}{2} + \frac{1}{4} + \cos^2(\gamma) = 1 \]

5. **Combine the Terms**:
Convert \( \frac{1}{2} \) to have a common denominator of 4:
\[ \frac{2}{4} + \frac{1}{4} + \cos^2(\gamma) = 1 \]
This simplifies to:
\[ \frac{3}{4} + \cos^2(\gamma) = 1 \]

6. **Isolate \( \cos^2(\gamma) \)**:
\[ \cos^2(\gamma) = 1 - \frac{3}{4} = \frac{1}{4} \]

7. **Take the Square Root**:
\[ \cos(\gamma) = \pm \frac{1}{2} \]

8. **Find Possible Angles**:
The angles corresponding to \( \cos(\gamma) = \frac{1}{2} \) are:
- \( \gamma = 60^\circ \)

The angles corresponding to \( \cos(\gamma) = -\frac{1}{2} \) are:
- \( \gamma = 120^\circ \)

### Conclusion:
Thus, the angle \( \gamma \) at which vector \( \vec{OP} \) is inclined to the \( OZ \) axis can be either \( 60^\circ \) or \( 120^\circ \).