To determine whether the given lines intersect, we need to check if there exists a common point where both lines coincide.

The two lines are given in parametric form:

1. Line 1: $(-3, -1, 1)t + (-1, -4, -3)$

   • Parametric equation: $\mathbf{r}_1(t) = (-3t - 1, -1t - 4, 1t - 3)$

2. Line 2: $(0, -2, 1)s + (-13, -6, 0)$

   • Parametric equation: $\mathbf{r}_2(s) = (0s - 13, -2s - 6, 1s + 0)$

The lines intersect if there exists a point where $\mathbf{r}_1(t) = \mathbf{r}_2(s)$. This leads to the following system of equations:

-3t - 1 &= -13 \\ -t - 4 &= -2s - 6 \\ t - 3 &= s \end{aligned}$$ ### Solving this system of equations: 1. From the first equation: $$-3t - 1 = -13 \implies -3t = -12 \implies t = 4$$ 2. Substitute $$t = 4$$ into the third equation: $$4 - 3 = s \implies s = 1$$ 3. Now, substitute $$t = 4$$ and $$s = 1$$ into the second equation: $$-(4) - 4 = -2(1) - 6 \implies -8 = -8$$ This is true, so the system is consistent. ### Conclusion: Since the system of equations is consistent, the lines intersect. The intersection occurs when $$t = 4$$ and $$s = 1$$. We can now verify the intersection point by substituting $$t = 4$$ into the parametric equation for Line 1 or $$s = 1$$ into the parametric equation for Line 2. Both will give the same point of intersection: $$\mathbf{r}_1(4) = (-3(4) - 1, -4 - 4, 1(4) - 3) = (-13, -8, 1)$$ $$\mathbf{r}_2(1) = (0(1) - 13, -2(1) - 6, 1(1) + 0) = (-13, -8, 1)$$ Thus, the lines intersect at the point $$(-13, -8, 1)$$. Would you like more details or further clarifications? Here are some related questions to consider: 1. What are the conditions for two lines to be parallel? 2. How do you find the angle between two lines in 3D? 3. Can two skew lines in 3D space ever intersect? 4. How do you find the distance between two lines in space? 5. What is the equation of a plane passing through two intersecting lines? **Tip:** When solving systems of parametric equations, always ensure that all components of the vectors are satisfied for the same parameters.