1. **State the problem:** Find the angle between the two lines given by their parametric equations: Line 1: $x = -2t - 9$, $y = 10t - 1$, $z = -2t - 9$ Line 2: $x = 8 - 4t$, $y = -t - 2$, $z = 7t + 1$ 2. **Identify direction vectors:** The direction vector of a line is the vector of coefficients of $t$. For Line 1, direction vector $\vec{v_1} = \langle -2, 10, -2 \rangle$ For Line 2, direction vector $\vec{v_2} = \langle -4, -1, 7 \rangle$ 3. **Formula for angle between two vectors:** $$\cos \theta = \frac{\vec{v_1} \cdot \vec{v_2}}{\|\vec{v_1}\| \|\vec{v_2}\|}$$ where $\vec{v_1} \cdot \vec{v_2}$ is the dot product and $\|\vec{v}\|$ is the magnitude of vector $\vec{v}$. 4. **Calculate dot product:** $$\vec{v_1} \cdot \vec{v_2} = (-2)(-4) + (10)(-1) + (-2)(7) = 8 - 10 - 14 = -16$$ 5. **Calculate magnitudes:** $$\|\vec{v_1}\| = \sqrt{(-2)^2 + 10^2 + (-2)^2} = \sqrt{4 + 100 + 4} = \sqrt{108} = 6\sqrt{3}$$ $$\|\vec{v_2}\| = \sqrt{(-4)^2 + (-1)^2 + 7^2} = \sqrt{16 + 1 + 49} = \sqrt{66}$$ 6. **Calculate cosine of angle:** $$\cos \theta = \frac{-16}{6\sqrt{3} \times \sqrt{66}} = \frac{-16}{6 \sqrt{198}} = \frac{-16}{6 \times 14.07} = \frac{-16}{84.42} \approx -0.1895$$ 7. **Calculate angle:** $$\theta = \cos^{-1}(-0.1895) \approx 101^{\circ}$$ 8. **Final answer:** The angle between the two lines is approximately **101 degrees**.