Vector Functions — Question 4

Question 4 – Solution

Strategy. A path intersection permits different parameters $t$ and $s$. A collision additionally requires the same time.

Step 1: Match coordinates Equality of the two position vectors requires $$t=2-s,\qquad t^2=s,\qquad 1-t=s-1.$$ The first and third equations both simplify to $$t+s=2.$$ Using $s=t^2$ gives $$t+t^2=2\quad\Longrightarrow\quad t^2+t-2=0 \quad\Longrightarrow\quad (t-1)(t+2)=0.$$

Step 2: Check both candidates If $t=1$, then $s=t^2=1$, and all equations hold. If $t=-2$, then $s=4$, but $t+s=2$ also holds. Thus there are two path intersections: $$\boxed{\left\langle 1,1,0\right\rangle\ (t,s)=(1,1)},\qquad \boxed{\left\langle -2,4,3\right\rangle\ (t,s)=(-2,4)}.$$

Step 3: Collision test With a common clock, a collision requires $t=s$. Only $(t,s)=(1,1)$ satisfies that requirement. Therefore the objects collide at $$\boxed{t=1\text{ at }\left\langle 1,1,0\right\rangle}.$$

Verification. Direct substitution into both vector functions reproduces each path-intersection point.