1. **Problem Statement:** Find the angle between the two lines given by the parametric equations: $$x=9-6t,\quad y=-5t-10,\quad z=0$$ and $$x=-5t-7,\quad y=2t,\quad z=-4t-5$$ 2. **Formula:** The angle $\theta$ between two lines with direction vectors $\mathbf{a}$ and $\mathbf{b}$ is given by: $$\cos \theta = \frac{|\mathbf{a} \cdot \mathbf{b}|}{\|\mathbf{a}\| \|\mathbf{b}\|}$$ where $\mathbf{a} \cdot \mathbf{b}$ is the dot product and $\|\mathbf{a}\|$, $\|\mathbf{b}\|$ are the magnitudes of the vectors. 3. **Find direction vectors:** From the first line, direction vector $\mathbf{a} = \langle -6, -5, 0 \rangle$ (coefficients of $t$). From the second line, direction vector $\mathbf{b} = \langle -5, 2, -4 \rangle$. 4. **Calculate dot product:** $$\mathbf{a} \cdot \mathbf{b} = (-6)(-5) + (-5)(2) + (0)(-4) = 30 - 10 + 0 = 20$$ 5. **Calculate magnitudes:** $$\|\mathbf{a}\| = \sqrt{(-6)^2 + (-5)^2 + 0^2} = \sqrt{36 + 25} = \sqrt{61}$$ $$\|\mathbf{b}\| = \sqrt{(-5)^2 + 2^2 + (-4)^2} = \sqrt{25 + 4 + 16} = \sqrt{45}$$ 6. **Calculate cosine of angle:** $$\cos \theta = \frac{|20|}{\sqrt{61} \times \sqrt{45}} = \frac{20}{\sqrt{2745}}$$ 7. **Calculate angle $\theta$:** $$\theta = \cos^{-1} \left( \frac{20}{\sqrt{2745}} \right)$$ Calculate numeric value: $$\sqrt{2745} \approx 52.38$$ $$\cos \theta \approx \frac{20}{52.38} \approx 0.3819$$ $$\theta \approx \cos^{-1}(0.3819) \approx 67.5^\circ$$ **Final answer:** The angle between the two lines is approximately **68 degrees** to the nearest degree.