1. **Problem 1:** Given triangle OAB with vectors $\vec{OA} = 5\vec{a}$ and $\vec{OB} = 2\vec{b}$, point T lies on AB such that $AT : TB = 5 : 1$. Show that $\vec{OT}$ is parallel to $\vec{a} + 2\vec{b}$. 2. **Step 1:** Express $\vec{AB}$ in terms of $\vec{a}$ and $\vec{b}$. $$\vec{AB} = \vec{OB} - \vec{OA} = 2\vec{b} - 5\vec{a}$$ 3. **Step 2:** Since T divides AB in ratio 5:1 closer to B, the position vector of T relative to A is: $$\vec{AT} = \frac{5}{5+1} \vec{AB} = \frac{5}{6}(2\vec{b} - 5\vec{a})$$ 4. **Step 3:** Find $\vec{OT}$ by adding $\vec{OA}$ and $\vec{AT}$: $$\vec{OT} = \vec{OA} + \vec{AT} = 5\vec{a} + \frac{5}{6}(2\vec{b} - 5\vec{a}) = 5\vec{a} + \frac{10}{6}\vec{b} - \frac{25}{6}\vec{a}$$ 5. **Step 4:** Simplify $\vec{OT}$: $$\vec{OT} = \left(5 - \frac{25}{6}\right)\vec{a} + \frac{10}{6}\vec{b} = \frac{30}{6} - \frac{25}{6}\vec{a} + \frac{10}{6}\vec{b} = \frac{5}{6}\vec{a} + \frac{10}{6}\vec{b} = \frac{5}{6}(\vec{a} + 2\vec{b})$$ 6. **Step 5:** Since $\vec{OT} = \frac{5}{6}(\vec{a} + 2\vec{b})$, $\vec{OT}$ is a scalar multiple of $\vec{a} + 2\vec{b}$, so $\vec{OT}$ is parallel to $\vec{a} + 2\vec{b}$. **Final answer:** $\vec{OT}$ is parallel to $\vec{a} + 2\vec{b}$.