If the sides `vec(AB)` of an equilateral triangle ABC lying in the xy-plane is `3hati` then the side `vec(CB)` can be (A) `-3/2(hati-sqrt(3))` (B) `3/2(hati-sqrt(3))` (C) `-3/2(hati+sqrt(3))` (D) `3/2(hati+sqrt(3))`

Options

A. `-3/2(hati- sqrt3hatj)`
B. `-3/2(hati- sqrt3hatj)`
C. `-3/2(hati + sqrt3hatj)`
D. `3/2(hati + sqrt3hatj)`

Explanation

To solve the problem, we need to find the vector \(\vec{CB}\) of an equilateral triangle \(ABC\) where the side \(\vec{AB} = 3\hat{i}\).

### Step-by-Step Solution:

1. **Understanding the Triangle**:
- Given that \(AB\) is a side of an equilateral triangle and is represented as \(\vec{AB} = 3\hat{i}\), we can denote the coordinates of points \(A\) and \(B\).
- Let \(A = (0, 0)\) and \(B = (3, 0)\).

2. **Finding Point C**:
- In an equilateral triangle, the angle between any two sides is \(60^\circ\).
- The coordinates of point \(C\) can be found using the properties of rotation.
- The point \(C\) can be obtained by rotating point \(B\) around point \(A\) by \(60^\circ\) counterclockwise.

3. **Using Rotation Matrix**:
- The rotation matrix for an angle \(\theta\) is given by:
\[ R(\theta) = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} \]
- For \(\theta = 60^\circ\):
\[ R(60^\circ) = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \]

4. **Applying the Rotation**:
- The coordinates of point \(B\) are \((3, 0)\). We can apply the rotation matrix to find the coordinates of point \(C\):
\[ C = R(60^\circ) \cdot \begin{pmatrix} 3 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \cdot 3 + 0 \cdot (-\frac{\sqrt{3}}{2}) \\ \frac{\sqrt{3}}{2} \cdot 3 + 0 \cdot \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{3}{2} \\ \frac{3\sqrt{3}}{2} \end{pmatrix} \]

5. **Finding Vector CB**:
- Now, we can find the vector \(\vec{CB}\) which is given by:
\[ \vec{CB} = \vec{B} - \vec{C} = (3, 0) - \left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right) = \left(3 - \frac{3}{2}, 0 - \frac{3\sqrt{3}}{2}\right) = \left(\frac{3}{2}, -\frac{3\sqrt{3}}{2}\right) \]

6. **Expressing in Vector Form**:
- Thus, the vector \(\vec{CB}\) can be expressed as:
\[ \vec{CB} = \frac{3}{2} \hat{i} - \frac{3\sqrt{3}}{2} \hat{j} \]

7. **Comparing with Options**:
- Now we can compare \(\vec{CB}\) with the given options:
- (A) \(-\frac{3}{2}(\hat{i} - \sqrt{3})\)
- (B) \(\frac{3}{2}(\hat{i} - \sqrt{3})\)
- (C) \(-\frac{3}{2}(\hat{i} + \sqrt{3})\)
- (D) \(\frac{3}{2}(\hat{i} + \sqrt{3})\)

- After simplifying the options, we find that option (B) matches our derived vector.

### Conclusion:
The correct answer is **(B) \(\frac{3}{2}(\hat{i} - \sqrt{3})\)**.