A line lies in ZX plane and makes `60^(@)` with x - axis ,then direction cosines of the line are :

Options

A. ` 1/2,0,(sqrt(3))/2`
B. `(sqrt(3))/2,0,1/2`
C. `0,1/2,(sqrt(3))/2`
D. `1/2 , sqrt(3),0`

Answer

To find the direction cosines of a line that lies in the ZX plane and makes an angle of \(60^\circ\) with the x-axis, we can follow these steps:

### Step 1: Understand the Direction Cosines
Direction cosines are defined as the cosines of the angles that a line makes with the coordinate axes. If a line makes angles \(\alpha\), \(\beta\), and \(\gamma\) with the x, y, and z axes respectively, then the direction cosines are given by:
- \(l = \cos \alpha\)
- \(m = \cos \beta\)
- \(n = \cos \gamma\)

### Step 2: Identify Given Angles
From the problem, we know:
- The line makes an angle of \(60^\circ\) with the x-axis, so \(\alpha = 60^\circ\).
- Since the line lies in the ZX plane, it makes an angle of \(90^\circ\) with the y-axis, so \(\beta = 90^\circ\).

### Step 3: Calculate the Direction Cosines
Now we can calculate the direction cosines using the angles:
1. Calculate \(l = \cos \alpha = \cos 60^\circ = \frac{1}{2}\).
2. Calculate \(m = \cos \beta = \cos 90^\circ = 0\).
3. We need to find \(n = \cos \gamma\), where \(\gamma\) is the angle with the z-axis.

### Step 4: Use the Relation of Direction Cosines
The relation for direction cosines is given by:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]
Substituting the known values:
\[ \left(\frac{1}{2}\right)^2 + \cos^2 90^\circ + \cos^2 \gamma = 1 \]
This simplifies to:
\[ \frac{1}{4} + 0 + \cos^2 \gamma = 1 \]
\[ \cos^2 \gamma = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 5: Solve for \(\cos \gamma\)
Taking the square root, we find:
\[ \cos \gamma = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2} \]

### Step 6: Write the Direction Cosines
Now we can summarize the direction cosines:
- \(l = \cos \alpha = \frac{1}{2}\)
- \(m = \cos \beta = 0\)
- \(n = \cos \gamma = \pm \frac{\sqrt{3}}{2}\)

Thus, the direction cosines of the line are:
\[ \left(\frac{1}{2}, 0, \pm \frac{\sqrt{3}}{2}\right) \]