1. **Problem statement:** We have trapezium OACB with vectors $\overrightarrow{OA} = 2\mathbf{a}$, $\overrightarrow{AB} = 5\mathbf{b}$, and $\overrightarrow{AC} = 3\mathbf{b}$. We want to find and simplify the vector $\overrightarrow{OP}$ where point P is the intersection of the diagonals. 2. **Key property:** In a trapezium, the diagonals intersect at a point P that divides them proportionally. We use the fact that $\overrightarrow{OP}$ lies on both diagonals $\overrightarrow{OC}$ and $\overrightarrow{AB}$. 3. **Express vectors:** - $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = 2\mathbf{a} + 3\mathbf{b}$ - $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = 2\mathbf{a} + 5\mathbf{b}$ 4. **Parameterize diagonals:** - Let $\overrightarrow{OP} = \lambda \overrightarrow{OC} = \lambda (2\mathbf{a} + 3\mathbf{b})$ - Also, $\overrightarrow{OP} = \overrightarrow{OA} + \mu \overrightarrow{AB} = 2\mathbf{a} + \mu (5\mathbf{b})$ 5. **Equate the two expressions for $\overrightarrow{OP}$:** $$\lambda (2\mathbf{a} + 3\mathbf{b}) = 2\mathbf{a} + 5\mu \mathbf{b}$$ 6. **Equate components:** - For $\mathbf{a}$: $2\lambda = 2 \Rightarrow \lambda = 1$ - For $\mathbf{b}$: $3\lambda = 5\mu \Rightarrow 3(1) = 5\mu \Rightarrow \mu = \frac{3}{5}$ 7. **Find $\overrightarrow{OP}$:** $$\overrightarrow{OP} = \lambda (2\mathbf{a} + 3\mathbf{b}) = 1 \times (2\mathbf{a} + 3\mathbf{b}) = 2\mathbf{a} + 3\mathbf{b}$$ **Final answer:** $$\boxed{\overrightarrow{OP} = 2\mathbf{a} + 3\mathbf{b}}$$