1. **State the problem:** We have triangle OAD with vectors \(\vec{OA} = \mathbf{a}\) and \(\vec{OD} = \mathbf{a} + \mathbf{d}\). Point C lies on line AD such that \(AC : AD = 2 : 3\). Points O, C, B are collinear with \(OC : CB = 3 : 5\). We want to express \(\vec{AB} = \lambda \mathbf{a} + \mu \mathbf{d}\) and find the ratio \(\lambda : \mu\) in simplest form. 2. **Find vector \(\vec{AC}\):** Since \(C\) divides \(AD\) in ratio \(2:3\), $$\vec{AD} = \vec{OD} - \vec{OA} = (\mathbf{a} + \mathbf{d}) - \mathbf{a} = \mathbf{d}$$ So, $$\vec{AC} = \frac{2}{3} \vec{AD} = \frac{2}{3} \mathbf{d}$$ 3. **Find position vector of C:** $$\vec{OC} = \vec{OA} + \vec{AC} = \mathbf{a} + \frac{2}{3} \mathbf{d}$$ 4. **Express vector \(\vec{OB}\):** Since O, C, B are collinear and \(OC : CB = 3 : 5\), point B divides segment OC extended beyond C in ratio 3:5. Let \(\vec{OB} = \vec{OC} + t(\vec{CB})\). Since \(OC : CB = 3 : 5\), the total length OB is \(OC + CB = 3 + 5 = 8\) parts. Using section formula for external division, $$\vec{OB} = \vec{O} + \frac{8}{3} (\vec{OC} - \vec{O}) = \frac{8}{3} \vec{OC}$$ Since \(\vec{O} = \mathbf{0}\), $$\vec{OB} = \frac{8}{3} \left( \mathbf{a} + \frac{2}{3} \mathbf{d} \right) = \frac{8}{3} \mathbf{a} + \frac{16}{9} \mathbf{d}$$ 5. **Find vector \(\vec{AB}\):** $$\vec{AB} = \vec{OB} - \vec{OA} = \left( \frac{8}{3} \mathbf{a} + \frac{16}{9} \mathbf{d} \right) - \mathbf{a} = \left( \frac{8}{3} - 1 \right) \mathbf{a} + \frac{16}{9} \mathbf{d} = \frac{5}{3} \mathbf{a} + \frac{16}{9} \mathbf{d}$$ 6. **Find ratio \(\lambda : \mu\):** $$\lambda : \mu = \frac{5}{3} : \frac{16}{9} = \frac{5}{3} \times \frac{9}{16} : 1 = \frac{15}{16} : 1$$ Multiplying both sides by 16 to clear denominator: $$15 : 16$$ **Final answer:** \(\lambda : \mu = 15 : 16\)