1. **Problem:** Calculate the dot product $\vec{u} \cdot \vec{v}$ and classify the angle between $\vec{u}$ and $\vec{v}$ for the vectors: (a) $\vec{u} = (2,8)$ and $\vec{v} = (-3,1)$. 2. **Formula:** The dot product of two vectors $\vec{u} = (u_1,u_2)$ and $\vec{v} = (v_1,v_2)$ is given by: $$\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2$$ The angle $\theta$ between vectors satisfies: $$\vec{u} \cdot \vec{v} = \|\vec{u}\| \|\vec{v}\| \cos \theta$$ where $\|\vec{u}\| = \sqrt{u_1^2 + u_2^2}$ and $\|\vec{v}\| = \sqrt{v_1^2 + v_2^2}$. If $\vec{u} \cdot \vec{v} > 0$, angle is acute. If $\vec{u} \cdot \vec{v} = 0$, angle is right (90 degrees). If $\vec{u} \cdot \vec{v} < 0$, angle is obtuse. 3. **Calculation:** $$\vec{u} \cdot \vec{v} = 2 \times (-3) + 8 \times 1 = -6 + 8 = 2$$ 4. **Norms:** $$\|\vec{u}\| = \sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17} \approx 8.25$$ $$\|\vec{v}\| = \sqrt{(-3)^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \approx 3.16$$ 5. **Angle cosine:** $$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \|\vec{v}\|} = \frac{2}{8.25 \times 3.16} = \frac{2}{26.07} \approx 0.077$$ 6. **Angle:** $$\theta = \arccos(0.077) \approx 85.58^\circ$$ 7. **Classification:** Since dot product is positive and close to zero, angle is acute but very close to right angle. **Final answer:** Dot product $= 2$, angle $\approx 85.58^\circ$, angle is acute.