1. **State the problem:** Determine which pairs of lines are skew lines. Skew lines are lines that are not parallel and do not intersect. 2. **Recall definitions and formulas:** - Lines are parallel if their direction vectors are scalar multiples. - Lines intersect if there exists parameters $t$ and $s$ such that their position vectors are equal. - If lines are neither parallel nor intersecting, they are skew. 3. **Analyze each pair:** **Pair 1:** $r_1(t) = \langle 1, 2, 3 \rangle + t \langle 1, 1, 1 \rangle$ $r_2(t) = \langle 2, 3, 4 \rangle + t \langle 1, 1, 1 \rangle$ - Direction vectors: both $\langle 1,1,1 \rangle$ (parallel) - Since direction vectors are the same, lines are parallel. **Pair 2:** $r_1(t) = \langle 1, 2, 3 \rangle + t \langle 1, 0, 0 \rangle$ $r_2(t) = \langle 2, 3, 4 \rangle + t \langle 0, 1, 1 \rangle$ - Direction vectors: $\langle 1,0,0 \rangle$ and $\langle 0,1,1 \rangle$ (not parallel) - Check if they intersect by solving: $$1 + t = 2 + s \cdot 0 = 2$$ $$2 + 0 = 3 + s \cdot 1 = 3 + s$$ $$3 + 0 = 4 + s \cdot 1 = 4 + s$$ - From first: $t = 1$ - From second: $2 = 3 + s \Rightarrow s = -1$ - From third: $3 = 4 + s \Rightarrow s = -1$ - Since $s$ is consistent, lines intersect, so not skew. **Pair 3:** $r_1(t) = \langle 1, 2, 3 \rangle + t \langle 1, 1, 1 \rangle$ $r_2(t) = \langle 1, 2, 3 \rangle + t \langle 0, 0, 0 \rangle$ - Second line is a point (direction vector zero) - So lines intersect at $\langle 1,2,3 \rangle$ **Pair 4:** $r_1(t) = \langle 1, 2, 3 \rangle + t \langle 2, 1, 4 \rangle$ $r_2(t) = \langle 4, 5, 1 \rangle + t \langle 1, 3, 2 \rangle$ - Direction vectors: $\langle 2,1,4 \rangle$ and $\langle 1,3,2 \rangle$ (not parallel) - Check intersection by solving system: $$1 + 2t = 4 + s$$ $$2 + t = 5 + 3s$$ $$3 + 4t = 1 + 2s$$ - From first: $s = -3 + 2t$ - Substitute into second: $$2 + t = 5 + 3(-3 + 2t) = 5 - 9 + 6t = -4 + 6t$$ $$2 + t = -4 + 6t \Rightarrow 6 = 5t \Rightarrow t = \frac{6}{5}$$ - Substitute $t$ into $s$: $$s = -3 + 2 \times \frac{6}{5} = -3 + \frac{12}{5} = -\frac{15}{5} + \frac{12}{5} = -\frac{3}{5}$$ - Check third equation: $$3 + 4 \times \frac{6}{5} = 1 + 2 \times \left(-\frac{3}{5}\right)$$ $$3 + \frac{24}{5} = 1 - \frac{6}{5}$$ $$\frac{15}{5} + \frac{24}{5} = \frac{5}{5} - \frac{6}{5}$$ $$\frac{39}{5} \neq -\frac{1}{5}$$ - No solution, so lines do not intersect and are not parallel, hence skew. 4. **Final answer:** Only pair 4 consists of skew lines. **Answer:** The pair $r_1(t) = \langle 1, 2, 3 \rangle + t \langle 2, 1, 4 \rangle$ and $r_2(t) = \langle 4, 5, 1 \rangle + t \langle 1, 3, 2 \rangle$ are skew lines.