space curve question

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

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

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

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

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

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

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 $t_2 = 0\text{ or }0.5$ .