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I would like to ask how to solve following problem:

We have 2 "super heroes". Their path are given by parameterized function where t means time: $r_{1}(t)=$$\begin{bmatrix}t\\t^2\\t^3\end{bmatrix}$ and $r_{2}(t)=$$\begin{bmatrix} 1+2t \\1+6t\\1+14t\end{bmatrix}$

The question is: Where do their path intersect (if they)? Are they ever going to meet if they start at the same time?

I don't know how to approach the problem. For first question I tried to set system of equations:

$t=1+2s$

$t^2=1+6s$

$t^3=1+14s$

But I wasn't able to solve it and I think similar approach is needed for next question

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  • $\begingroup$ Their paths definitely intersect at (1,1,1) though not at the same time. As the one curve is a line it seems unlikely that the two curves would encounter one another again. $\endgroup$ – Doug M Jun 6 '18 at 0:09
  • $\begingroup$ $r_1(1) = r_2(0) = (1,1,1)$ $\endgroup$ – Doug M Jun 6 '18 at 0:16
  • $\begingroup$ Are the $t$’s in the two parameterizations meant to be the same or not? $\endgroup$ – amd Jun 6 '18 at 1:52
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We substitute the second equation with the first one to get $(1+2s)^2=1+6s$, and hence $s(2s-1)=0$. It follows from the two first equations that either $s=0, t=1$ or $s=1/2, t=2$.

Both solutions are coherent with the third equation, so that the points (1,1,1) and (2,4,8) really are the only two intersection points of both paths.

Now, if they start at the same time, we see that they will never meet eachother, because for each intersection point, we do not have $t=s$.

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  • $\begingroup$ Thank you for explanation! $\endgroup$ – Mirjan Pecenko Jun 6 '18 at 0:17
  • $\begingroup$ You're welcome :) $\endgroup$ – Suzet Jun 6 '18 at 0:17

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