Question: In triangle $OAB$ $\vec{OA} = 8a$ $\vec{OB} = 4b$ $C$ is the midpoint of $AB$. $D$ is the point on $OC$ such that $OD : DC = 1 : 3$. Find, in terms of $a$ and $b$, the vector $\vec{AD}$.

1. **State the problem:** We need to find the vector $\vec{AD}$ in terms of $a$ and $b$ given $\vec{OA} = 8a$, $\vec{OB} = 4b$, $C$ is midpoint of $AB$, and $D$ lies on $OC$ with ratio $OD : DC = 1 : 3$. 2. **Find $\vec{AB}$:** $$\vec{AB} = \vec{OB} - \vec{OA} = 4b - 8a$$ 3. **Find $\vec{OC}$:** Since $C$ is midpoint of $AB$, $$\vec{OC} = \vec{OA} + \frac{1}{2} \vec{AB} = 8a + \frac{1}{2}(4b - 8a) = 8a + 2b - 4a = 4a + 2b$$ 4. **Find $\vec{OD}$:** Given $D$ divides $OC$ in ratio $1:3$, $OD = \frac{1}{1+3} \vec{OC} = \frac{1}{4} \vec{OC}$, $$\vec{OD} = \frac{1}{4}(4a + 2b) = a + \frac{1}{2}b$$ 5. **Find $\vec{AD}$:** $$\vec{AD} = \vec{OD} - \vec{OA} = \left(a + \frac{1}{2}b\right) - 8a = -7a + \frac{1}{2}b$$ **Final answer:** $$\boxed{\vec{AD} = -7a + \frac{1}{2}b}$$