Vector analysis : following given trajectory, will particles collide?

Question

Let two particles move by a trajectory respectively given by $\vec{r_1}(t)=t\vec{i}+t^2\vec{j}+t^3\vec{k}$ and $\vec{r_2}(t)=(1+2t)\vec{i}+(1+6t)\vec{j}+(1+14t)\vec{k}$.

In my vector analysis course, I need to determine if particles will collide. I guess that I need to solve the following system of equations : \begin{cases} t=1+2t \\ t^2=1+6t \\ t^3=1+14t \end{cases} Which has no solution, right? Is that so simple?

Thanks !

Answer 1

Your conclusion is true if the particles have the same $t$. But, if you are just interested in whether or not the two trajectories intersect, you want different parameters.

In the latter case, you want to see if there is a solution to $r_1(s) = r_2(t)$, or, looking at the components,

$\begin{array}\\ s = 1+2t\\ s^2 = 1+6t\\ s^3 = 1+14t\\ \end{array} $

For this, $t = (s-1)/2$, so $s^2 =1+6((s-1)/2) =1+3(s-1) =3s-2 $, or $s^2-3s+2 = 0 $. This has solutions $s = 2$ and $s = 1$.

From the third coordinate, $s^3 =1+14((s-1)/2) =1 + 7(s-1) = 7s-6 $. Of the two possible values for $s$, only $s=1$ works. For this, $t = 0$.

Therefore, the two trajectories intersect for $r_1(1)$ and $r_2(0)$.

Looking at the three equations, this becomes obvious, but that's math biz.

Answer 2

Yes, it is really that simple.

Answer 3

EDIT1:

The time to collision is same, is quotient of relative distance and relative velocity.

Time =

$$ \dfrac{r_2 - r_1}{ \dot{r_2} -\dot{r_1} }= \dfrac{r_2 }{ \dot{r_2} }= \dfrac{ r_1}{ \dot{r_1} } $$