1. **Stating the problem:** We are given quadrilateral OABC with vectors: $$\vec{OA} = \vec{a}, \quad \vec{AB} = 3\vec{b} + \frac{1}{2}\vec{a}, \quad \vec{OC} = 2\vec{b}$$ Point D lies on OB such that $$\vec{OD} = \frac{3}{5} \vec{OB}$$ Point E lies on CB such that $$\vec{CE} = k \vec{CB}$$ Given that points A, D, and E are collinear, we need to find: (b)(i) The value of $$k$$. (b)(ii) The ratio $$AD : DE$$ in integers. --- 2. **Find vectors OB and CB:** Since $$\vec{OA} = \vec{a}$$ and $$\vec{AB} = 3\vec{b} + \frac{1}{2}\vec{a}$$, $$\vec{OB} = \vec{OA} + \vec{AB} = \vec{a} + 3\vec{b} + \frac{1}{2}\vec{a} = \frac{3}{2}\vec{a} + 3\vec{b}$$ Similarly, $$\vec{OC} = 2\vec{b}$$ So, $$\vec{CB} = \vec{OB} - \vec{OC} = \left(\frac{3}{2}\vec{a} + 3\vec{b}\right) - 2\vec{b} = \frac{3}{2}\vec{a} + \vec{b}$$ --- 3. **Find vector OD and AD:** Given $$\vec{OD} = \frac{3}{5} \vec{OB} = \frac{3}{5} \left(\frac{3}{2}\vec{a} + 3\vec{b}\right) = \frac{9}{10}\vec{a} + \frac{9}{5}\vec{b}$$ Vector $$\vec{AD} = \vec{OD} - \vec{OA} = \left(\frac{9}{10}\vec{a} + \frac{9}{5}\vec{b}\right) - \vec{a} = \left(\frac{9}{10} - 1\right)\vec{a} + \frac{9}{5}\vec{b} = -\frac{1}{10}\vec{a} + \frac{9}{5}\vec{b}$$ --- 4. **Express vector AE in terms of k:** Since $$\vec{CE} = k \vec{CB} = k \left(\frac{3}{2}\vec{a} + \vec{b}\right) = \frac{3k}{2}\vec{a} + k\vec{b}$$ Vector $$\vec{AE} = \vec{AC} + \vec{CE}$$ But $$\vec{AC} = \vec{OC} - \vec{OA} = 2\vec{b} - \vec{a} = -\vec{a} + 2\vec{b}$$ So, $$\vec{AE} = (-\vec{a} + 2\vec{b}) + \left(\frac{3k}{2}\vec{a} + k\vec{b}\right) = \left(-1 + \frac{3k}{2}\right)\vec{a} + (2 + k)\vec{b}$$ --- 5. **Collinearity condition of A, D, E:** Vectors $$\vec{AD}$$ and $$\vec{AE}$$ must be parallel, so there exists a scalar $$\lambda$$ such that: $$\vec{AE} = \lambda \vec{AD}$$ Equate components: $$\left(-1 + \frac{3k}{2}\right)\vec{a} + (2 + k)\vec{b} = \lambda \left(-\frac{1}{10}\vec{a} + \frac{9}{5}\vec{b}\right)$$ This gives two equations: $$-1 + \frac{3k}{2} = -\frac{1}{10} \lambda \quad (1)$$ $$2 + k = \frac{9}{5} \lambda \quad (2)$$ --- 6. **Solve for k and $\lambda$:** From (1): $$\lambda = 10 \left(1 - \frac{3k}{2}\right) = 10 - 15k$$ From (2): $$2 + k = \frac{9}{5} \lambda = \frac{9}{5} (10 - 15k) = 18 - 27k$$ Rearranged: $$2 + k = 18 - 27k$$ $$k + 27k = 18 - 2$$ $$28k = 16$$ $$k = \frac{16}{28} = \frac{4}{7}$$ --- 7. **Find $\lambda$:** $$\lambda = 10 - 15 \times \frac{4}{7} = 10 - \frac{60}{7} = \frac{70}{7} - \frac{60}{7} = \frac{10}{7}$$ --- 8. **Find ratio AD : DE:** Since $$\vec{AE} = \lambda \vec{AD}$$, point E divides segment AD in ratio $$\lambda : (1 - \lambda)$$. Vector $$\vec{DE} = \vec{AE} - \vec{AD} = (\lambda - 1) \vec{AD}$$ Length ratio: $$AD : DE = 1 : |\lambda - 1| = 1 : \left|\frac{10}{7} - 1\right| = 1 : \frac{3}{7} = 7 : 3$$ --- **Final answers:** (b)(i) $$k = \frac{4}{7}$$ (b)(ii) $$AD : DE = 7 : 3$$