ABCDEF si a regular hexagon with centre at the origin such that `A vec D + E vec B + F vec C = lambda E vec D . " Then, " lambda ` equals

Options

A. 2
B. 4
C. 6
D. 3

Explanation

To solve the problem, we need to analyze the vectors associated with the regular hexagon ABCDEF centered at the origin. We are given the equation:

\[ \vec{AD} + \vec{EB} + \vec{FC} = \lambda \vec{ED} \]

### Step-by-Step Solution:

1. **Understanding the Hexagon**:
- A regular hexagon has six vertices (A, B, C, D, E, F) and is symmetric about its center (O). The coordinates of the vertices can be represented in terms of angles. For a hexagon centered at the origin, we can assign coordinates as follows:
- \( A(1, 0) \)
- \( B\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \)
- \( C\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \)
- \( D(-1, 0) \)
- \( E\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \)
- \( F\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \)

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

3. **Summing the Vectors**:
- Now, we sum these vectors:
\[ \vec{AD} + \vec{EB} + \vec{FC} = (-2, 0) + (1, \sqrt{3}) + (-1, \sqrt{3}) \]
- Simplifying this gives:
\[ (-2 + 1 - 1, 0 + \sqrt{3} + \sqrt{3}) = (-2, 2\sqrt{3}) \]

4. **Expressing \( \vec{ED} \)**:
- Now we need to express \( \vec{ED} \):
- \( \vec{ED} = \vec{D} - \vec{E} = (-1, 0) - \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \)

5. **Finding \( \lambda \)**:
- We need to find \( \lambda \) such that:
\[ (-2, 2\sqrt{3}) = \lambda \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \]
- This gives us two equations:
- For the x-component: \( -2 = -\frac{\lambda}{2} \) → \( \lambda = 4 \)
- For the y-component: \( 2\sqrt{3} = \frac{\lambda \sqrt{3}}{2} \) → \( \lambda = 4 \)

Thus, we find that \( \lambda = 4 \).

### Final Answer:
\[ \lambda = 4 \]