1. ## space curve question

Two particles travel along the space curves

$r_{1}(t) = (t,t^2,t^3) \ \ r_{2}(t) = (1+2t, 1+6t, 1+ 14t)$

Do the particles collide? Do they intersect?

so far I have:

for $r_{1}(t) \ \ x=t \ \ y=t^2 \ \ z=t^3$
therefore $y=x^2 \ \ z=x^3$
for $r_{2}(t) \ \ x= 1 +2t \ \ y = 1+6t \ \ z = 1+14t$
therefore $y=3x-2 \ \ z=7x-6$

I'm not sure if I have to take $y=y=x^2=3x-2$, where $x=1$ and $z=z=x^3=7x-6$ where $x=1$ ?

2. Say they intersect, in that case, particle one intersects in the time $t_1$ and the second in time $t_2$. We now have:

$\left(\begin{array}{c}t_1\\ \\t_1^2\\ \\t_1^3\end{array}\right)=\left(\begin{array}{c}1+ 2t_2\\ \\1+6t_2\\ \\1+14t_2\end{array}\right)\Longleftrightarrow$

$
\left(\begin{array}{c}t_1\\ \\(1+2t_2)^2\\ \\(1+2t_2)^3\end{array}\right)=\left(\begin{array} {c}1+2t_2\\ \\1+6t_2\\ \\1+14t_2\end{array}\right)\Longleftrightarrow$

$\left(\begin{array}{c}t_1\\ \\1+4t_2+4t_2^2\\ \\1+6t_2+12t_2^2+8t_2^3\end{array}\right)=\left(\b egin{array}{c}1+2t_2\\ \\1+6t_2\\ \\1+14t_2\end{array}\right)\Longleftrightarrow$

$
\left(\begin{array}{c}t_1\\ \\t_2\\ \\12t_2^2+8t_2^3\end{array}\right)=\left(\begin{ar ray}{c}1+2t_2\\ \\0\text{ or }0.5\\ \\8t_2\end{array}\right)\Longleftrightarrow$

$
\left(\begin{array}{c}t_1\\ \\t_2\\ \\t_2\end{array}\right)=\left(\begin{array}{c}1+2t _2\\ \\0\text{ or }0.5\\ \\0\text{ or }0.5\text{ or }2\end{array}\right)\Longleftrightarrow\left(\begi n{array}{c}t_1\\ \\t_2\end{array}\right)=\left(\begin{array}{c}1+2t _2\\ \\0\text{ or }0.5\end{array}\right)$

So, they intersect at two places. If they hadn't intersected, the equation(s) wouldn't have been possible to solve. Now if you want to se if they colide, put $t_1=t_2$. Then we have $t_2 = 1+2t_2 \Longleftrightarrow t_2 = -1$, which doesn't fit with $t_2 = 0\text{ or }0.5$.