Five points given by A,B,C,D and E are in a plane. Three forces AC,AD and AE act at A annd three forces CB,DB and EB act B. then, their resultant is

Options

A. 2AC
B. 3AB
C. 3DB
D. 2BC

Explanation

To solve the problem, we need to find the resultant of the forces acting at points A and B. The forces acting at A are AC, AD, and AE, and the forces acting at B are CB, DB, and EB.

### Step-by-Step Solution:

1. **Identify the Forces:**
- At point A, the forces are:
- \( \vec{F}_{AC} \) (Force from A to C)
- \( \vec{F}_{AD} \) (Force from A to D)
- \( \vec{F}_{AE} \) (Force from A to E)
- At point B, the forces are:
- \( \vec{F}_{CB} \) (Force from C to B)
- \( \vec{F}_{DB} \) (Force from D to B)
- \( \vec{F}_{EB} \) (Force from E to B)

2. **Write the Resultant Forces:**
- The resultant force at point A can be expressed as:
\[ \vec{R}_A = \vec{F}_{AC} + \vec{F}_{AD} + \vec{F}_{AE} \]
- The resultant force at point B can be expressed as:
\[ \vec{R}_B = \vec{F}_{CB} + \vec{F}_{DB} + \vec{F}_{EB} \]

3. **Group the Forces:**
- We can group the forces acting at A and B:
\[ \vec{R} = (\vec{F}_{AC} + \vec{F}_{CB}) + (\vec{F}_{AD} + \vec{F}_{DB}) + (\vec{F}_{AE} + \vec{F}_{EB}) \]

4. **Simplify the Grouping:**
- Notice that:
- \( \vec{F}_{AC} + \vec{F}_{CB} \) can be considered as a single resultant force.
- \( \vec{F}_{AD} + \vec{F}_{DB} \) can be considered as another resultant force.
- \( \vec{F}_{AE} + \vec{F}_{EB} \) can be considered as the third resultant force.
- Therefore, we can express the resultant as:
\[ \vec{R} = \vec{R}_{ACB} + \vec{R}_{ADB} + \vec{R}_{AEB} \]

5. **Final Resultant:**
- Since each of the pairs results in a force acting in the same direction, we can conclude that the overall resultant force is:
\[ \vec{R} = 3 \vec{R}_{AB} \]
- This indicates that the resultant is three times the vector from A to B.

### Conclusion:
The resultant of the forces acting at points A and B is \( 3 \vec{R}_{AB} \).