Let a vector `bar(r)` make angle `60^(@), 30^(@)` with x and y-axes respectively.
What angle does `bar(r)` make with z-axis ?

Options

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

Explanation

To find the angle that the vector \(\bar{r}\) makes with the z-axis, we can use the relationship between the angles made with the x, y, and z axes. The angles made with the axes are related to the direction cosines of the vector.

### Step-by-Step Solution:

1. **Define the Angles**:
Let the angle that the vector \(\bar{r}\) makes with the x-axis be \( \alpha = 60^\circ \) and with the y-axis be \( \beta = 30^\circ \). We need to find the angle \( \theta \) that \(\bar{r}\) makes with the z-axis.

2. **Use Direction Cosines**:
The direction cosines are given by:
- \( l = \cos(\alpha) = \cos(60^\circ) = \frac{1}{2} \)
- \( m = \cos(\beta) = \cos(30^\circ) = \frac{\sqrt{3}}{2} \)
- \( n = \cos(\theta) \) (the unknown we want to find)

3. **Apply the Direction Cosine Condition**:
The sum of the squares of the direction cosines is equal to 1:
\[ l^2 + m^2 + n^2 = 1 \]
Substituting the known values:
\[ \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 + n^2 = 1 \]

4. **Calculate the Squares**:
- \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
- \( \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \)

5. **Substitute and Simplify**:
\[ \frac{1}{4} + \frac{3}{4} + n^2 = 1 \]
\[ 1 + n^2 = 1 \]

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

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

8. **Determine the Angle \( \theta \)**:
Since \( n = \cos(\theta) = 0 \), this implies:
\[ \theta = 90^\circ \]

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