Forces `3 O vec A , 5 O vec B ` act along OA and OB. If their resultant passes through C on AB, then

Options

A. C is a mid-point of AB
B. C divides AB in the ratio `2:1`
C. `3 AC = 5 CB`
D. `2 AC = 3 CB`

Answer

To solve the problem, we need to analyze the forces acting along the lines OA and OB and find the conditions under which their resultant passes through point C on line AB.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- We have two forces: \( \vec{F_1} = 3 \vec{A} \) and \( \vec{F_2} = 5 \vec{B} \).
- These forces act along the lines OA and OB, respectively.

2. **Resultant Force Calculation**:
- The resultant force \( \vec{R} \) can be expressed as:
\[ \vec{R} = \vec{F_1} + \vec{F_2} = 3 \vec{A} + 5 \vec{B} \]

3. **Position of Point C**:
- Point C lies on line AB. We can express the position of C in terms of the segments AC and CB.
- Let \( OC = x \) and \( CA = y \). Thus, \( OA = OC + CA = x + y \).

4. **Using the Section Formula**:
- Since C divides AB in the ratio \( k:1 \) (where \( k \) is the ratio of AC to CB), we can express the position vector of C as:
\[ \vec{C} = \frac{k \vec{B} + \vec{A}}{k + 1} \]

5. **Equating Resultant with C**:
- For the resultant to pass through C, the resultant vector \( \vec{R} \) must be proportional to the vector \( \vec{C} \):
\[ \vec{R} = \lambda \vec{C} \]
- Substituting the expressions for \( \vec{R} \) and \( \vec{C} \):
\[ 3 \vec{A} + 5 \vec{B} = \lambda \left( \frac{k \vec{B} + \vec{A}}{k + 1} \right) \]

6. **Finding the Ratios**:
- By equating coefficients of \( \vec{A} \) and \( \vec{B} \):
- For \( \vec{A} \):
\[ 3 = \frac{\lambda}{k + 1} \]
- For \( \vec{B} \):
\[ 5 = \frac{k \lambda}{k + 1} \]

7. **Solving the Equations**:
- From the first equation, we can express \( \lambda \):
\[ \lambda = 3(k + 1) \]
- Substituting \( \lambda \) into the second equation:
\[ 5 = \frac{k \cdot 3(k + 1)}{k + 1} \]
- Simplifying gives:
\[ 5 = 3k \]
- Thus, \( k = \frac{5}{3} \).

8. **Conclusion**:
- The ratio \( AC:CB \) is \( \frac{5}{3} \), which means the conditions are satisfied for the resultant to pass through point C.