1. **Problem statement:** Given two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $|\vec{a}|=\sqrt{7}$ and $|\vec{b}|=6$, and their dot product $\vec{a} \cdot \vec{b} = \frac{13}{2}$, find the angle $\theta$ between them if it exists. 2. **Formula used:** The dot product of two vectors is related to the angle between them by the formula: $$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$. 3. **Calculate $\cos \theta$:** Substitute the known values: $$\frac{13}{2} = \sqrt{7} \times 6 \times \cos \theta$$ Simplify the right side: $$\frac{13}{2} = 6\sqrt{7} \cos \theta$$ 4. **Solve for $\cos \theta$:** $$\cos \theta = \frac{\frac{13}{2}}{6\sqrt{7}} = \frac{13}{2} \times \frac{1}{6\sqrt{7}} = \frac{13}{12\sqrt{7}}$$ 5. **Check if $\cos \theta$ is valid:** Since $\cos \theta$ must be between $-1$ and $1$, check the value: Calculate approximate value: $$\sqrt{7} \approx 2.6458$$ $$\cos \theta \approx \frac{13}{12 \times 2.6458} = \frac{13}{31.7496} \approx 0.409$$ This is within the valid range. 6. **Find the angle $\theta$:** $$\theta = \cos^{-1}(0.409)$$ Using a calculator: $$\theta \approx 66.8^\circ$$ **Final answer:** The angle between the vectors $\vec{a}$ and $\vec{b}$ is approximately $66.8^\circ$.