Answer
The two curves intersect. The point of intersection is $\left( {2,4,8} \right)$.
Work Step by Step
1. Check if ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ collide.
Suppose there exists some ${t_0}$ such that ${{\bf{r}}_1}\left( {{t_0}} \right) = {{\bf{r}}_2}\left( {{t_0}} \right)$. So,
${{\bf{r}}_1}\left( {{t_0}} \right) = \left( {{t_0},{t_0}^2,{t_0}^3} \right)$
${{\bf{r}}_2}\left( {{t_0}} \right) = \left( {4{t_0} + 6,4{t_0}^2,7 - {t_0}} \right)$
$\left( {{t_0},{t_0}^2,{t_0}^3} \right) = \left( {4{t_0} + 6,4{t_0}^2,7 - {t_0}} \right)$
We obtain
$x = {t_0} = 4{t_0} + 6$, ${\ \ }$ $y = {t_0}^2 = 4{t_0}^2$,
$z = {t_0}^3 = 7 - {t_0}$
From the first equation we get ${t_0} = - 2$. However, this value does not satisfy the second and the third equation. Thus, we conclude that there does not exist ${t_0}$ such that ${{\bf{r}}_1}\left( {{t_0}} \right) = {{\bf{r}}_2}\left( {{t_0}} \right)$. Therefore, ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ do not collide.
2. Check if ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( s \right)$ intersect.
Now, we check if they intersect. Two curves intersect if there exist parameter values $t$ and $s$ such that ${{\bf{r}}_1}\left( t \right) = {{\bf{r}}_2}\left( s \right)$. So,
${{\bf{r}}_1}\left( t \right) = \left( {t,{t^2},{t^3}} \right)$
${{\bf{r}}_2}\left( s \right) = \left( {4s + 6,4{s^2},7 - s} \right)$
We obtain
$x = t = 4s + 6$, ${\ \ }$ $y = {t^2} = 4{s^2}$,
$z = {t^3} = 7 - s$
From the second equation we obtain $t = \pm 2s$. Write ${t_1} = 2{s_1}$ and ${t_2} = - 2{s_2}$.
Substituting ${t_1} = 2{s_1}$ in the first equation gives $2{s_1} = 4{s_1} + 6$. It follows that ${s_1} = - 3$ and ${t_1} = - 6$.
However, substituting ${s_1} = - 3$ and ${t_1} = - 6$ in the third equation do not give the same $z$-coordinates. Therefore, ${s_1} = - 3$ and ${t_1} = - 6$ have no solution.
Next, we investigate ${t_2} = - 2{s_2}$.
Substituting ${t_2} = - 2{s_2}$ in the first equation gives $ - 2{s_2} = 4{s_2} + 6$. It follows that ${s_2} = - 1$ and ${t_2} = 2$. Substituting them in the third equation give $z=8$. It follows that $x=2$ and $y=4$.
Therefore, the point of intersection is $\left( {2,4,8} \right)$.
