To determine if the lines $r(t)$ and $R(s)$ intersect, we solve the system of equations derived from equating the two parametric representations:

$r(t) = \langle 2t - 1, 2t + 3, 4t + 1 \rangle, \quad R(s) = \langle s + 1, 3s + 5, -5s + 5 \rangle$

The lines intersect if there exist $t$ and $s$ such that:

$2t - 1 = s + 1, \quad 2t + 3 = 3s + 5, \quad 4t + 1 = -5s + 5$

Step 1: Solve the system of equations

From the first equation:

2t - 1 = s + 1 \implies s = 2t - 2 \tag{1}

Substitute $s = 2t - 2$ into the second equation:

$2t + 3 = 3(2t - 2) + 5$

Simplify:

$2t + 3 = 6t - 6 + 5 \implies 2t + 3 = 6t - 1$

Rearranging terms:

$3 + 1 = 6t - 2t \implies 4t = 4 \implies t = 1$

Step 2: Solve for $s$

Substitute $t = 1$ into $s = 2t - 2$:

$s = 2(1) - 2 = 0$

Step 3: Verify with the third equation

Substitute $t = 1$ and $s = 0$ into the third equation:

$4t + 1 = -5s + 5$

$4(1) + 1 = -5(0) + 5 \implies 4 + 1 = 5$

The third equation is satisfied, confirming that the lines intersect.

Intersection Point

Substitute $t = 1$ into $r(t)$ or $s = 0$ into $R(s)$ to find the intersection point:

$r(1) = \langle 2(1) - 1, 2(1) + 3, 4(1) + 1 \rangle = \langle 1, 5, 5 \rangle$

Thus, the lines intersect at $(1, 5, 5)$.

Part b: Collision

To check if the particles collide, $t$ and $s$ must satisfy the above system with $t \geq 0$ and $s \geq 0$. Here, $t = 1$ and $s = 0$, so the particles collide at $t = 1$ second, and the collision point is $(1, 5, 5)$.

Final Answer:

a. The lines intersect at $(1, 5, 5)$.
b. The particles collide at $(1, 5, 5)$.

Would you like further details or have any questions?

---

1. How can parametric equations represent lines in 3D?
2. How is a system of equations solved for intersection?
3. What if the lines were parallel but distinct—how would we know?
4. How does the third equation confirm intersection validity?
5. What happens if $t$ or $s$ is negative in collision problems?

Tip: To check if two parametric lines intersect, always validate with all three equations for consistency.