If `A B C D E F` is a regular hexagon, them ` vec (A D)+ vec (E B)+ vec (F C)` equals

Options

A. a) `2 vec (A B)`
B. b) ` vec0`
C. c) `3 vec (A B)`
D. d) `4 vec (A B)`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Geometry of the Hexagon:**
- A regular hexagon has equal sides and symmetrical properties. The vertices are labeled as \( A, B, C, D, E, F \).

2. **Identify the Vectors:**
- The vectors we need to analyze are:
- \( \vec{AD} \) (from point A to point D)
- \( \vec{EB} \) (from point E to point B)
- \( \vec{FC} \) (from point F to point C)

3. **Express Each Vector in Terms of Side Length:**
- In a regular hexagon, the diagonal \( \vec{AD} \) can be expressed in terms of the side length \( s \):
\[ \vec{AD} = 2 \vec{AB} \]
- For \( \vec{EB} \):
\[ \vec{EB} = 2 \vec{FA} \]
- For \( \vec{FC} \):
\[ \vec{FC} = 2 \vec{AB} \]

4. **Combine the Vectors:**
- Now, we can substitute the expressions for \( \vec{AD} \), \( \vec{EB} \), and \( \vec{FC} \):
\[ \vec{AD} + \vec{EB} + \vec{FC} = 2 \vec{AB} + 2 \vec{FA} + 2 \vec{AB} \]
- Since \( \vec{FA} \) is equal to \( \vec{AB} \) (because they are both sides of the hexagon):
\[ \vec{EB} = 2 \vec{AB} \]
- Thus, we can rewrite:
\[ \vec{AD} + \vec{EB} + \vec{FC} = 2 \vec{AB} + 2 \vec{AB} + 2 \vec{AB} = 4 \vec{AB} \]

5. **Final Result:**
- Therefore, the final result is:
\[ \vec{AD} + \vec{EB} + \vec{FC} = 4 \vec{AB} \]