Consider the following curves:

\(r_1(t)=\)

\(r_2(s)=<7-s,s-3,s^2>\)

Find the point of intersection of the curves and angle of intersection \(\theta\).

The parametric equations corresponding to the curve \(r_1(t)\) are as follows:

x=t

\(y=4-t\)

\(z=35+t^2\)

The parametric equations corresponding to the curve \(r_2(s)\) are as follows:

\(x=7-s\)

\(y=s-3\)

\(z=s^2\)

At the point of intersection of two curves, the corresponding coordinates of parametric lines are same.

\(t=7-s\Rightarrow s+t=7\)

\(4-t=s-3\Rightarrow s+t=7\)

\(35+t^2=s^2\Rightarrow s^2-t^2=35\)

That is,

\(s^2-t^2=35\)

\((s-t)(s+t)=35\)

\((s-t)7=35\)

\(s-t=5\)

Solve the equations \(s-t=5\) and \(s+t=7\), to find the values of s and t.

\(s-t+s+t=5+7\Rightarrow2s=12\Rightarrow s=6\)

The corresponding value of t is as follows:

\(t=7-s\Rightarrow t=7-6\Rightarrow t=1\)

Thus, the values are t=1 and s=6

Substitute t=1 in \(r_1(t)\), to find the point of intersection.

\(r_1(t)=<1,4-1,35+1^2>=<1,3,36>\)

\(r_1(t)=<1,3,36>\)

The angle between the normal vectors is given as \(\cos\theta=\frac{r_1'(t)\cdot r_2'(s)}{|r_1'(t)||r_2'(s)|}\)

The direction vector of the line \(r_1(t)\) is \(\)

The direction vector of the line \(r_2(s)\) is \(<-s,s,s^2>\)

At t=1,

\(r_1'(t)=<1,-1,2t>\)

\(r_1'(1)=<1,-1,2>\)

At s=6,

\(r_2'(s)=<-1,1,2s>\)

\(r_2'(6)=<-1,1,12>\)

\(\cos\theta=\frac{r_1'(t)\cdot r_2'(s)}{|r_1'(t)||r_2'(s)|}\)

\(=\frac{r_1'(1)\cdot r_2'(6)}{|r_1'(1)||r_2'(6)|}\)

\(=\frac{<1,-1,2>\cdot<-1,1,12>}{\sqrt{1+1+4}\cdot\sqrt{1+1+144}}\)

\(=\frac{-1-1+24}{\sqrt{6}\cdot\sqrt{146}}\)

\(=\frac{22}{\sqrt{6}\cdot\sqrt{146}}\)

\(=\frac{11}{\sqrt{219}}\)

Thus, \(\theta=\cos^{-1}\left(\frac{11}{\sqrt{219}}\right)\)

\(r_1(t)=\)

\(r_2(s)=<7-s,s-3,s^2>\)

Find the point of intersection of the curves and angle of intersection \(\theta\).

The parametric equations corresponding to the curve \(r_1(t)\) are as follows:

x=t

\(y=4-t\)

\(z=35+t^2\)

The parametric equations corresponding to the curve \(r_2(s)\) are as follows:

\(x=7-s\)

\(y=s-3\)

\(z=s^2\)

At the point of intersection of two curves, the corresponding coordinates of parametric lines are same.

\(t=7-s\Rightarrow s+t=7\)

\(4-t=s-3\Rightarrow s+t=7\)

\(35+t^2=s^2\Rightarrow s^2-t^2=35\)

That is,

\(s^2-t^2=35\)

\((s-t)(s+t)=35\)

\((s-t)7=35\)

\(s-t=5\)

Solve the equations \(s-t=5\) and \(s+t=7\), to find the values of s and t.

\(s-t+s+t=5+7\Rightarrow2s=12\Rightarrow s=6\)

The corresponding value of t is as follows:

\(t=7-s\Rightarrow t=7-6\Rightarrow t=1\)

Thus, the values are t=1 and s=6

Substitute t=1 in \(r_1(t)\), to find the point of intersection.

\(r_1(t)=<1,4-1,35+1^2>=<1,3,36>\)

\(r_1(t)=<1,3,36>\)

The angle between the normal vectors is given as \(\cos\theta=\frac{r_1'(t)\cdot r_2'(s)}{|r_1'(t)||r_2'(s)|}\)

The direction vector of the line \(r_1(t)\) is \(\)

The direction vector of the line \(r_2(s)\) is \(<-s,s,s^2>\)

At t=1,

\(r_1'(t)=<1,-1,2t>\)

\(r_1'(1)=<1,-1,2>\)

At s=6,

\(r_2'(s)=<-1,1,2s>\)

\(r_2'(6)=<-1,1,12>\)

\(\cos\theta=\frac{r_1'(t)\cdot r_2'(s)}{|r_1'(t)||r_2'(s)|}\)

\(=\frac{r_1'(1)\cdot r_2'(6)}{|r_1'(1)||r_2'(6)|}\)

\(=\frac{<1,-1,2>\cdot<-1,1,12>}{\sqrt{1+1+4}\cdot\sqrt{1+1+144}}\)

\(=\frac{-1-1+24}{\sqrt{6}\cdot\sqrt{146}}\)

\(=\frac{22}{\sqrt{6}\cdot\sqrt{146}}\)

\(=\frac{11}{\sqrt{219}}\)

Thus, \(\theta=\cos^{-1}\left(\frac{11}{\sqrt{219}}\right)\)