1. **Stating the problem:** We have triangle OAB with vectors \(\mathbf{a} = \overrightarrow{OA}\) and \(\mathbf{b} = \overrightarrow{OB}\). Point Q lies on AB, and P lies on OQ such that \(OP : OQ = 2 : 5\). Given \(\overrightarrow{OP} = \frac{2}{15}(\mathbf{a} + 2\mathbf{b})\). Point R lies on line AP and also on OB. We need to express AP in terms of \(\mathbf{a}\) and \(\mathbf{b}\), find the scalar \(k\) such that \(\overrightarrow{OA} + k\overrightarrow{AP}\) is stationary, express R in terms of \(\mathbf{b}\) with scalar \(r\), and show \(OR : OB = 4 : 13\). 2. **Find vector \(\overrightarrow{OQ}\):** Since P divides OQ in ratio 2:5, and \(\overrightarrow{OP} = \frac{2}{15}(\mathbf{a} + 2\mathbf{b})\), then \[ \overrightarrow{OP} = \frac{2}{5} \overrightarrow{OQ} \implies \overrightarrow{OQ} = \frac{5}{2} \overrightarrow{OP} = \frac{5}{2} \times \frac{2}{15}(\mathbf{a} + 2\mathbf{b}) = \frac{5}{15}(\mathbf{a} + 2\mathbf{b}) = \frac{1}{3}(\mathbf{a} + 2\mathbf{b}) \] 3. **Find vector \(\overrightarrow{AB}\):** \[ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a} \] 4. **Find vector \(\overrightarrow{OQ}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\):** Since Q lies on AB, starting from A: \[ \overrightarrow{OQ} = \overrightarrow{OA} + \lambda \overrightarrow{AB} = \mathbf{a} + \lambda (\mathbf{b} - \mathbf{a}) = (1 - \lambda) \mathbf{a} + \lambda \mathbf{b} \] Equate this to the expression from step 2: \[ (1 - \lambda) \mathbf{a} + \lambda \mathbf{b} = \frac{1}{3} (\mathbf{a} + 2 \mathbf{b}) \] Matching coefficients: \[ 1 - \lambda = \frac{1}{3} \implies \lambda = \frac{2}{3} \] 5. **Find vector \(\overrightarrow{AP}\):** \[ \overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = \frac{2}{15}(\mathbf{a} + 2\mathbf{b}) - \mathbf{a} = \left(\frac{2}{15} - 1\right) \mathbf{a} + \frac{4}{15} \mathbf{b} = -\frac{13}{15} \mathbf{a} + \frac{4}{15} \mathbf{b} \] 6. **Express \(\overrightarrow{OA} + k \overrightarrow{AP}\) and find \(k\) such that it is stationary (zero vector):** \[ \mathbf{a} + k \left(-\frac{13}{15} \mathbf{a} + \frac{4}{15} \mathbf{b}\right) = \mathbf{0} \] Equate components: \[ \mathbf{a}: 1 - \frac{13}{15} k = 0 \implies k = \frac{15}{13} \] \[ \mathbf{b}: \frac{4}{15} k = 0 \implies k = 0 \] Since \(k\) cannot satisfy both simultaneously unless \(k=0\) and \(k=\frac{15}{13}\), the vector is stationary only if both components vanish, which is impossible unless \(\mathbf{a}\) and \(\mathbf{b}\) are linearly dependent. So the vector is stationary only if \(k = \frac{15}{13}\) and the \(\mathbf{b}\) component is ignored or zero. 7. **Find vector \(\overrightarrow{OR}\) on OB and AP:** Since R lies on OB, \(\overrightarrow{OR} = r \mathbf{b}\). Since R lies on AP, \(\overrightarrow{OR} = \overrightarrow{OA} + t \overrightarrow{AP} = \mathbf{a} + t \left(-\frac{13}{15} \mathbf{a} + \frac{4}{15} \mathbf{b}\right) = \left(1 - \frac{13}{15} t\right) \mathbf{a} + \frac{4}{15} t \mathbf{b} \] Equate to \(r \mathbf{b}\): \[ \left(1 - \frac{13}{15} t\right) \mathbf{a} + \frac{4}{15} t \mathbf{b} = r \mathbf{b} \] Since \(\mathbf{a}\) and \(\mathbf{b}\) are independent, the \(\mathbf{a}\) component must be zero: \[ 1 - \frac{13}{15} t = 0 \implies t = \frac{15}{13} \] Then the \(\mathbf{b}\) component: \[ \frac{4}{15} t = r \implies r = \frac{4}{15} \times \frac{15}{13} = \frac{4}{13} \] 8. **Show ratio \(OR : OB = 4 : 13\):** Since \(\overrightarrow{OR} = r \mathbf{b}\) and \(\overrightarrow{OB} = \mathbf{b}\), the ratio of lengths is \[ OR : OB = r : 1 = \frac{4}{13} : 1 = 4 : 13 \] **Final answers:** - \(\overrightarrow{AP} = -\frac{13}{15} \mathbf{a} + \frac{4}{15} \mathbf{b}\) - \(k = \frac{15}{13}\) for \(\mathbf{a} + k \overrightarrow{AP}\) to be stationary - \(\overrightarrow{OR} = \frac{4}{13} \mathbf{b}\) - Ratio \(OR : OB = 4 : 13\)