Write down a unit vector in XY plane, making an angle of `30^(@)` with positive direction of x-axis.

Answer

To find a unit vector in the XY plane that makes an angle of \(30^\circ\) with the positive direction of the x-axis, we can follow these steps:

### Step 1: Understand the Definition of a Unit Vector
A unit vector is a vector with a magnitude of 1. In the XY plane, we can express a unit vector \( \mathbf{A} \) in terms of its components along the x-axis and y-axis:
\[ \mathbf{A} = x \mathbf{i} + y \mathbf{j} \]

### Step 2: Use the Angle Information
Since the vector makes an angle of \(30^\circ\) with the positive x-axis, we can relate the components \(x\) and \(y\) using trigonometric functions:
- The x-component can be expressed as:
\[ x = \cos(30^\circ) \]
- The y-component can be expressed as:
\[ y = \sin(30^\circ) \]

### Step 3: Calculate the Values of Cosine and Sine
Using the known values of cosine and sine for \(30^\circ\):
- \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)
- \(\sin(30^\circ) = \frac{1}{2}\)

Thus, we can substitute these values into our expressions for \(x\) and \(y\):
\[ x = \frac{\sqrt{3}}{2}, \quad y = \frac{1}{2} \]

### Step 4: Write the Unit Vector
Now that we have the components, we can write the unit vector \( \mathbf{A} \):
\[ \mathbf{A} = \frac{\sqrt{3}}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} \]

### Step 5: Verify the Magnitude
To ensure that \( \mathbf{A} \) is indeed a unit vector, we can check its magnitude:
\[ |\mathbf{A}| = \sqrt{x^2 + y^2} = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \]

### Final Answer
Thus, the unit vector in the XY plane making an angle of \(30^\circ\) with the positive x-axis is:
\[ \mathbf{A} = \frac{\sqrt{3}}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} \]