1. **Problem statement:** Given triangle OAB with points P on AB and C on OB such that $AP : PB = 2 : 3$ and $OC : CB = 1 : 2$, find the values of $r$ and $s$ in the vector expression $\mathbf{OP} = r\mathbf{OA} + s\mathbf{OB}$. 2. **Understanding the problem:** We want to express the position vector of point P in terms of vectors $\mathbf{OA}$ and $\mathbf{OB}$. Since P lies on segment AB, we can write $\mathbf{OP}$ as a linear combination of $\mathbf{OA}$ and $\mathbf{OB}$. 3. **Expressing vectors on segments:** - Since $P$ lies on $AB$, and $AP : PB = 2 : 3$, the ratio divides $AB$ into 5 parts. So, $$\mathbf{AP} = \frac{2}{5} \mathbf{AB}$$ - Vector $\mathbf{AB} = \mathbf{OB} - \mathbf{OA}$. 4. **Write $\mathbf{OP}$ in terms of $\mathbf{OA}$ and $\mathbf{OB}$:** $$\mathbf{OP} = \mathbf{OA} + \mathbf{AP} = \mathbf{OA} + \frac{2}{5}(\mathbf{OB} - \mathbf{OA})$$ 5. **Simplify the expression:** $$\mathbf{OP} = \mathbf{OA} + \frac{2}{5}\mathbf{OB} - \frac{2}{5}\mathbf{OA} = \left(1 - \frac{2}{5}\right)\mathbf{OA} + \frac{2}{5}\mathbf{OB} = \frac{3}{5}\mathbf{OA} + \frac{2}{5}\mathbf{OB}$$ 6. **Therefore, comparing with $\mathbf{OP} = r\mathbf{OA} + s\mathbf{OB}$, we get:** $$r = \frac{3}{5}, \quad s = \frac{2}{5}$$ 7. **Note:** The ratio $OC : CB = 1 : 2$ is given but not needed to find $r$ and $s$ for $\mathbf{OP}$ in this problem. **Final answer:** $$r = \frac{3}{5}, \quad s = \frac{2}{5}$$