1. **Problem:** Find the angle between vectors $\vec{a} = (3,0,4)$ and $\vec{b} = (5,1,-1)$. 2. **Formula:** The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the cosine formula: $$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$$ where $\vec{a} \cdot \vec{b}$ is the dot product and $|\vec{a}|$, $|\vec{b}|$ are the magnitudes (lengths) of the vectors. 3. **Calculate the dot product:** $$\vec{a} \cdot \vec{b} = 3 \times 5 + 0 \times 1 + 4 \times (-1) = 15 + 0 - 4 = 11$$ 4. **Calculate the magnitudes:** $$|\vec{a}| = \sqrt{3^2 + 0^2 + 4^2} = \sqrt{9 + 0 + 16} = \sqrt{25} = 5$$ $$|\vec{b}| = \sqrt{5^2 + 1^2 + (-1)^2} = \sqrt{25 + 1 + 1} = \sqrt{27} = 3\sqrt{3}$$ 5. **Calculate cosine of the angle:** $$\cos \theta = \frac{11}{5 \times 3\sqrt{3}} = \frac{11}{15\sqrt{3}}$$ 6. **Calculate the angle $\theta$:** $$\theta = \arccos\left(\frac{11}{15\sqrt{3}}\right)$$ Using a calculator, approximate: $$\theta \approx \arccos(0.423) \approx 65.0^\circ$$ **Final answer:** The angle between the vectors is approximately $65.0^\circ$.