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To solve for the ratio \( \frac{OR}{RB} \), let's break down the given information from the diagram and problem statement:

### Given:
- Triangle \( OAB \) is described, where:
  - \( \vec{OA} = \mathbf{a} \)
  - \( \vec{OB} = \mathbf{b} \)
  
- Point \( P \) divides line segment \( AB \) such that:
  \[
  \frac{AP}{PB} = 2:1
  \]

- Point \( Q \) divides line segment \( OP \) such that:
  \[
  \frac{OQ}{QP} = 1:3
  \]

- \( R \) is a point on \( OB \) such that \( RQA \) is a straight line.

### Step-by-Step Solution:

#### Step 1: Coordinates of Points
We can express the coordinates of points \( P \), \( Q \), and \( R \) in terms of the position vectors \( \mathbf{a} \) and \( \mathbf{b} \).

- **Point \( P \):**  
  Since \( AP:PB = 2:1 \), point \( P \) divides \( AB \) internally in the ratio 2:1. Using the section formula:
  \[
  \vec{P} = \frac{1 \cdot \mathbf{a} + 2 \cdot \mathbf{b}}{1+2} = \frac{\mathbf{a} + 2\mathbf{b}}{3}
  \]

- **Point \( Q \):**  
  Since \( OQ:QP = 1:3 \), point \( Q \) divides \( OP \) in the ratio 1:3. Again, using the section formula:
  \[
  \vec{Q} = \frac{3 \cdot \vec{O} + 1 \cdot \vec{P}}{1+3} = \frac{3\vec{O} + \vec{P}}{4} = \frac{3 \cdot 0 + \frac{\mathbf{a} + 2\mathbf{b}}{3}}{4} = \frac{\mathbf{a} + 2\mathbf{b}}{12}
  \]

#### Step 2: Condition for Collinearity (RQA is a Straight Line)
Since \( R \), \( Q \), and \( A \) are collinear, the position vector \( \vec{R} \) must lie on the line \( OB \), and the vector \( \vec{RQ} \) should be parallel to \( \vec{QA} \).

- **Equation for \( R \) on \( OB \):**
  Let \( \vec{R} = \lambda \mathbf{b} \), where \( \lambda \) is a scalar to be determined.

- **Vectors \( \vec{RQ} \) and \( \vec{QA} \):**
  - \( \vec{RQ} = \vec{R} - \vec{Q} = \lambda \mathbf{b} - \frac{\mathbf{a} + 2\mathbf{b}}{12} = \lambda \mathbf{b} - \frac{1}{12} (\mathbf{a} + 2\mathbf{b}) \)
  - \( \vec{QA} = \vec{A} - \vec{Q} = \mathbf{a} - \frac{\mathbf{a} + 2\mathbf{b}}{12} = \mathbf{a} - \frac{1}{12}(\mathbf{a} + 2\mathbf{b}) \)

#### Step 3: Solve for \( \lambda \) (ratio \( OR:RB \))
Using the fact that \( \vec{RQ} \) is parallel to \( \vec{QA} \), we can set up a proportionality condition. Solving this will give the value of \( \lambda \), which represents the ratio \( \frac{OR}{RB} \).

I will now compute the required ratio.The solution gives \( \lambda = 1 \), which means that the point \( R \) divides the line segment \( OB \) such that \( OR:RB = 1:1 \).

Thus, the ratio \( \frac{OR}{RB} \) is \( 1:1 \).

Do you need further clarification or details?

Here are 5 related questions you might find useful:
1. How can we apply section formulas for different types of divisions (internal and external)?
2. How do we verify collinearity using vector methods?
3. What other methods can we use to solve for ratios in geometric problems?
4. How would the solution change if the ratio \( OQ:QP \) was different?
5. What are some practical applications of the section formula in real-life geometry problems?

**Tip:** When solving problems

Calculate the ratio OR:RB based on the given information in triangle OAB, where OA = a and OB = b, and the points P, Q, and R divide segments in specific ratios.

This problem explores how to calculate the ratio OR:RB in triangle OAB using vector algebra, section formulas, and the concept of collinearity. Follow a step-by-step solution for understanding division of line segments and solving vector-based geometric problems. Ideal for students in Grades 10-12.

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