ABCDEF is a regular hexagon in the x-y plance with vertices in the anticlockwise direction. If `vecAB=2hati`, then `vecCD` is

Options

A. `hati+sqrt(3)hatj`
B. `hati-sqrt(3)hatj`
C. `-hati+sqrt(3)hatj`
D. `sqrt(3)hati-hatj`

Explanation

To find the vector \( \vec{CD} \) in the regular hexagon \( ABCDEF \) where \( \vec{AB} = 2\hat{i} \), we can follow these steps:

### Step 1: Understand the Geometry of the Hexagon
A regular hexagon can be divided into 6 equilateral triangles. The internal angles of a regular hexagon are \(120^\circ\). The vertices \( A, B, C, D, E, F \) can be represented in the Cartesian plane.

### Step 2: Place the Hexagon in the Coordinate System
Let's place vertex \( A \) at the origin \( (0, 0) \) and vertex \( B \) at \( (2, 0) \) since \( \vec{AB} = 2\hat{i} \).

### Step 3: Determine the Coordinates of Other Vertices
Using the properties of the hexagon:
- The coordinates of \( C \) can be found by rotating \( B \) by \( 60^\circ \) counterclockwise around \( A \):
\[ C = \left(2\cos(60^\circ), 2\sin(60^\circ)\right) = \left(2 \cdot \frac{1}{2}, 2 \cdot \frac{\sqrt{3}}{2}\right) = (1, \sqrt{3}) \]

- The coordinates of \( D \) can be found by rotating \( C \) by \( 60^\circ \) counterclockwise:
\[ D = \left(2\cos(120^\circ), 2\sin(120^\circ)\right) = \left(2 \cdot -\frac{1}{2}, 2 \cdot \frac{\sqrt{3}}{2}\right) = (-1, \sqrt{3}) \]

### Step 4: Calculate the Vector \( \vec{CD} \)
The vector \( \vec{CD} \) can be calculated as:
\[ \vec{CD} = \vec{D} - \vec{C} = (-1, \sqrt{3}) - (1, \sqrt{3}) = (-1 - 1, \sqrt{3} - \sqrt{3}) = (-2, 0) \]

### Step 5: Write the Result in Vector Notation
Thus, the vector \( \vec{CD} \) is:
\[ \vec{CD} = -2\hat{i} \]

### Conclusion
The final answer is:
\[ \vec{CD} = -2\hat{i} \]

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