1. Let's consider a challenging O Level vectors problem involving geometry and vector algebra. 2. Problem: In triangle $ABC$, let $\vec{AB}=\mathbf{a}$ and $\vec{AC}=\mathbf{b}$. Point $D$ lies on line segment $BC$ such that $BD:DC=2:1$. Find the vector $\vec{AD}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. 3. Step 1: Express $\vec{BC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. Since $\vec{BC}=\vec{AC}-\vec{AB}=\mathbf{b}-\mathbf{a}$. 4. Step 2: Because $D$ divides $BC$ in ratio $2:1$, we can write the position vector of $D$ as: $$\vec{BD} = \frac{2}{3} \vec{BC} = \frac{2}{3}(\mathbf{b} - \mathbf{a})$$ 5. Step 3: To find $\vec{AD}$, we add $\vec{AB}$ and $\vec{BD}$: $$\vec{AD} = \vec{AB} + \vec{BD} = \mathbf{a} + \frac{2}{3}(\mathbf{b} - \mathbf{a}) = \mathbf{a} + \frac{2}{3}\mathbf{b} - \frac{2}{3}\mathbf{a}$$ 6. Step 4: Combine like terms to simplify: $$\vec{AD} = \left(1 - \frac{2}{3}\right)\mathbf{a} + \frac{2}{3} \mathbf{b} = \frac{1}{3} \mathbf{a} + \frac{2}{3} \mathbf{b}$$ 7. Final answer: $$\boxed{\vec{AD} = \frac{1}{3} \mathbf{a} + \frac{2}{3} \mathbf{b}}$$ This expresses $\vec{AD}$ neatly in terms of the vectors defining triangle $ABC$.