If ABCDEF is a regular hexagon , then `A vec D + E vec B + F vec C ` equals

Options

A. ` 2 A vec B `
B. `vec 0`
C. `3 A vec B `
D. `4 A vec B `

Explanation

To solve the problem, we need to find the vector sum \( \vec{AD} + \vec{EB} + \vec{FC} \) for the regular hexagon ABCDEF.

### Step-by-Step Solution:

1. **Understanding the Regular Hexagon**:
A regular hexagon can be inscribed in a circle. Let's place the hexagon in a coordinate system. We can assign coordinates to each vertex:
- \( A(1, 0) \)
- \( B\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \)
- \( C\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \)
- \( D(-1, 0) \)
- \( E\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \)
- \( F\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \)

2. **Finding the Vectors**:
We need to find the vectors \( \vec{AD} \), \( \vec{EB} \), and \( \vec{FC} \).
- **Vector \( \vec{AD} \)**:
\[ \vec{AD} = D - A = (-1, 0) - (1, 0) = (-2, 0) \]
- **Vector \( \vec{EB} \)**:
\[ \vec{EB} = B - E = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) = \left(1, \sqrt{3}\right) \]
- **Vector \( \vec{FC} \)**:
\[ \vec{FC} = C - F = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) = (-1, \sqrt{3}) \]

3. **Adding the Vectors**:
Now we add the vectors:
\[ \vec{AD} + \vec{EB} + \vec{FC} = (-2, 0) + (1, \sqrt{3}) + (-1, \sqrt{3}) \]
Breaking it down into components:
- **X-component**:
\[ -2 + 1 - 1 = -2 \]
- **Y-component**:
\[ 0 + \sqrt{3} + \sqrt{3} = 2\sqrt{3} \]
Therefore, the resultant vector is:
\[ \vec{AD} + \vec{EB} + \vec{FC} = (-2, 2\sqrt{3}) \]

4. **Final Result**:
The final answer for \( \vec{AD} + \vec{EB} + \vec{FC} \) is:
\[ \vec{AD} + \vec{EB} + \vec{FC} = (-2, 2\sqrt{3}) \]