Let a vector r make angles `60^(@), 30^(@)` with X and Y -axis, respectively.
What are the direction cosines of r ?

Options

A. `((1)/(2), (sqrt3)/(2), 0)`
B. `((1)/(2), - (sqrt3)/(2), 0)`
C. `((1)/(sqrt2), (1)/(sqrt2), 0)`
D. `(-(1)/(2), (sqrt3)/(2), 0)`

Answer

To find the direction cosines of the vector \( \mathbf{r} \) that makes angles of \( 60^\circ \) and \( 30^\circ \) with the X and Y axes respectively, we can follow these steps:

### Step 1: Understand Direction Cosines
The direction cosines of a vector are the cosines of the angles that the vector makes with the coordinate axes. If a vector \( \mathbf{r} \) makes angles \( \alpha \), \( \beta \), and \( \gamma \) with the X, Y, and Z axes respectively, then the direction cosines \( l \), \( m \), and \( n \) are given by:
- \( l = \cos(\alpha) \)
- \( m = \cos(\beta) \)
- \( n = \cos(\gamma) \)

### Step 2: Write Down Known Angles
From the problem:
- \( \alpha = 60^\circ \)
- \( \beta = 30^\circ \)

We need to find \( \gamma \) (the angle with the Z-axis).

### Step 3: Use the Property of Direction Cosines
The sum of the squares of the direction cosines is equal to 1:
\[ l^2 + m^2 + n^2 = 1 \]

### Step 4: Calculate \( l \) and \( m \)
Now we can calculate \( l \) and \( m \):
- \( l = \cos(60^\circ) = \frac{1}{2} \)
- \( m = \cos(30^\circ) = \frac{\sqrt{3}}{2} \)

### Step 5: Substitute into the Equation
Substituting \( l \) and \( m \) into the equation:
\[ \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 + n^2 = 1 \]
Calculating the squares:
\[ \frac{1}{4} + \frac{3}{4} + n^2 = 1 \]
\[ 1 + n^2 = 1 \]

### Step 6: Solve for \( n^2 \)
From the equation:
\[ n^2 = 1 - 1 = 0 \]
Thus:
\[ n = 0 \]

### Step 7: Conclusion
The direction cosines of the vector \( \mathbf{r} \) are:
\[ l = \frac{1}{2}, \quad m = \frac{\sqrt{3}}{2}, \quad n = 0 \]
Therefore, the direction cosines of vector \( \mathbf{r} \) are:
\[ \left( \frac{1}{2}, \frac{\sqrt{3}}{2}, 0 \right) \]