Let a vector r make angles `60^(@), 30^(@)` with X and Y -axis, respectively.
What angle does r mark the Z-axis?

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Answer

To find the angle that the vector \( \mathbf{r} \) makes with the Z-axis, we can use the property of direction cosines. The angles made by the vector with the X, Y, and Z axes are denoted as \( \alpha \), \( \beta \), and \( \gamma \) respectively.

### Step-by-Step Solution:

1. **Identify the angles**:
- Given \( \alpha = 60^\circ \) (angle with the X-axis)
- Given \( \beta = 30^\circ \) (angle with the Y-axis)
- We need to find \( \gamma \) (angle with the Z-axis).

2. **Use the property of direction cosines**:
The relationship between the angles is given by the equation:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

3. **Calculate \( \cos \alpha \) and \( \cos \beta \)**:
- \( \cos 60^\circ = \frac{1}{2} \)
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)

4. **Substitute the values into the equation**:
\[ \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 + \cos^2 \gamma = 1 \]
This simplifies to:
\[ \frac{1}{4} + \frac{3}{4} + \cos^2 \gamma = 1 \]

5. **Combine the terms**:
\[ 1 + \cos^2 \gamma = 1 \]

6. **Solve for \( \cos^2 \gamma \)**:
\[ \cos^2 \gamma = 1 - 1 = 0 \]

7. **Find \( \cos \gamma \)**:
\[ \cos \gamma = 0 \]

8. **Determine \( \gamma \)**:
The angle \( \gamma \) for which \( \cos \gamma = 0 \) is:
\[ \gamma = 90^\circ \]

### Final Answer:
The angle that the vector \( \mathbf{r} \) makes with the Z-axis is \( 90^\circ \).

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