Answer
The two curves collide and intersect at $\left( {12,4,2} \right)$.
The two curves intersect at another point, which is $\left( {4,0,6} \right)$.
Work Step by Step
1. Check if ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ collide.
Suppose there is some ${t_0}$ such that ${{\bf{r}}_1}\left( {{t_0}} \right) = {{\bf{r}}_2}\left( {{t_0}} \right)$. So,
${{\bf{r}}_1}\left( {{t_0}} \right) = \left( {{t_0}^2 + 3,{t_0} + 1,6{t_0}^{ - 1}} \right)$
${{\bf{r}}_2}\left( {{t_0}} \right) = \left( {4{t_0},2{t_0} - 2,{t_0}^2 - 7} \right)$
$\left( {{t_0}^2 + 3,{t_0} + 1,6{t_0}^{ - 1}} \right) = \left( {4{t_0},2{t_0} - 2,{t_0}^2 - 7} \right)$
We obtain
$x = {t_0}^2 + 3$, ${\ \ }$ $y = {t_0} + 1 = 2{t_0} - 2$,
$z = 6{t_0}^{ - 1} = {t_0}^2 - 7$
From the second equation we obtain ${t_0} = 3$, So, $y=4$. Substituting in the first equation gives $x=12$. Substituting in the third equation gives $z=2$. Since ${{\bf{r}}_1}\left( 3 \right) = {{\bf{r}}_2}\left( 3 \right)$, therefore, by definition ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ collide at ${t_0} = 3$. From Exercise 32, we know that if ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ collide, then they intersect. Thus, the point of intersection is ${{\bf{r}}_1}\left( 3 \right) = {{\bf{r}}_2}\left( 3 \right) = \left( {12,4,2} \right)$.
2. Check if ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ intersect.
We still need to check if they intersect at some other point. 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,
${{\bf{r}}_1}\left( t \right) = \left( {{t^2} + 3,t + 1,6{t^{ - 1}}} \right)$
${{\bf{r}}_2}\left( s \right) = \left( {4s,2s - 2,{s^2} - 7} \right)$
We obtain
$x = {t^2} + 3 = 4s$, ${\ \ }$ $y = t + 1 = 2s - 2$,
$z = 6{t^{ - 1}} = {s^2} - 7$
From the second equation we obtain $t=2s-3$. Substituting it in the first equation gives
${\left( {2s - 3} \right)^2} + 3 = 4s$
$4{s^2} - 12s + 9 + 3 = 4s$
$4{s^2} - 16s + 12 = 0$
${s^2} - 4s + 3 = 0$
$\left( {s - 1} \right)\left( {s - 3} \right) = 0$
The solutions are ${s_1} = 1$ and ${s_2} = 3$. The corresponding $t$ values are ${t_1} = - 1$ and ${t_2} = 3$.
Substituting these values in the first, the second and the third equation give
$\begin{array}{*{20}{c}}
{Parameters}&x&y&z\\
{{s_1} = 1,{t_1} = - 1}&4&0&6\\
{{s_2} = 3,{t_2} = 3}&{12}&4&2
\end{array}$
So, there are two points of intersection $\left( {4,0,6} \right)$ and $\left( {12,4,2} \right)$. In part (a) we have already obtained the point $\left( {12,4,2} \right)$, which is also the point of collision.
