Vector analysis : following given trajectory, will particles collide? 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 !
 A: Yes, it is really that simple.
A: 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.
A: EDIT1:
The time to collision is same, is quotient of relative distance and relative velocity.
Time =
$$ \dfrac{r_2 - r_1}{ \dot{r_2} -\dot{r_1} }= \dfrac{r_2 }{ \dot{r_2}  }=  \dfrac{ r_1}{ \dot{r_1} }  $$
