Consider the following curves:

\(\displaystyle{r}_{{1}}{\left({t}\right)}={<}{t},{4}-{t},{35}+{t}^{{2}}{>}\)

\(\displaystyle{r}_{{2}}{\left({s}\right)}={<}{7}-{s},{s}-{3},{s}^{{2}}{>}\)

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

The parametric equations corresponding to the curve \(\displaystyle{r}_{{1}}{\left({t}\right)}\) are as follows:

x=t

\(\displaystyle{y}={4}-{t}\)

\(\displaystyle{z}={35}+{t}^{{2}}\)

The parametric equations corresponding to the curve \(\displaystyle{r}_{{2}}{\left({s}\right)}\) are as follows:

\(\displaystyle{x}={7}-{s}\)

\(\displaystyle{y}={s}-{3}\)

\(\displaystyle{z}={s}^{{2}}\)

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

\(\displaystyle{t}={7}-{s}\Rightarrow{s}+{t}={7}\)

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

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

That is,

\(\displaystyle{s}^{{2}}-{t}^{{2}}={35}\)

\(\displaystyle{\left({s}-{t}\right)}{\left({s}+{t}\right)}={35}\)

\(\displaystyle{\left({s}-{t}\right)}{7}={35}\)

\(\displaystyle{s}-{t}={5}\)

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

\(\displaystyle{s}-{t}+{s}+{t}={5}+{7}\Rightarrow{2}{s}={12}\Rightarrow{s}={6}\)

The corresponding value of t is as follows:

\(\displaystyle{t}={7}-{s}\Rightarrow{t}={7}-{6}\Rightarrow{t}={1}\)

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

Substitute t=1 in \(\displaystyle{r}_{{1}}{\left({t}\right)}\), to find the point of intersection.

\(\displaystyle{r}_{{1}}{\left({t}\right)}={<}{1},{4}-{1},{35}+{1}^{{2}}\ge{<}{1},{3},{36}{>}\)

\(\displaystyle{r}_{{1}}{\left({t}\right)}={<}{1},{3},{36}{>}\)

The angle between the normal vectors is given as \(\displaystyle{\cos{\theta}}={\frac{{{r}_{{1}}'{\left({t}\right)}\cdot{r}_{{2}}'{\left({s}\right)}}}{{{\left|{r}_{{1}}'{\left({t}\right)}\right|}{\left|{r}_{{2}}'{\left({s}\right)}\right|}}}}\)

The direction vector of the line \(\displaystyle{r}_{{1}}{\left({t}\right)}\) is \(\displaystyle{<}{t},-{t},{t}^{{2}}{>}\)

The direction vector of the line \(\displaystyle{r}_{{2}}{\left({s}\right)}\) is \(\displaystyle{<}-{s},{s},{s}^{{2}}{>}\)

At t=1,

\(\displaystyle{r}_{{1}}'{\left({t}\right)}={<}{1},-{1},{2}{t}{>}\)

\(\displaystyle{r}_{{1}}'{\left({1}\right)}={<}{1},-{1},{2}{>}\)

At s=6,

\(\displaystyle{r}_{{2}}'{\left({s}\right)}={<}-{1},{1},{2}{s}{>}\)

\(\displaystyle{r}_{{2}}'{\left({6}\right)}={<}-{1},{1},{12}{>}\)

\(\displaystyle{\cos{\theta}}={\frac{{{r}_{{1}}'{\left({t}\right)}\cdot{r}_{{2}}'{\left({s}\right)}}}{{{\left|{r}_{{1}}'{\left({t}\right)}\right|}{\left|{r}_{{2}}'{\left({s}\right)}\right|}}}}\)

\(\displaystyle={\frac{{{r}_{{1}}'{\left({1}\right)}\cdot{r}_{{2}}'{\left({6}\right)}}}{{{\left|{r}_{{1}}'{\left({1}\right)}\right|}{\left|{r}_{{2}}'{\left({6}\right)}\right|}}}}\)

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

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

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

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

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

\(\displaystyle{r}_{{1}}{\left({t}\right)}={<}{t},{4}-{t},{35}+{t}^{{2}}{>}\)

\(\displaystyle{r}_{{2}}{\left({s}\right)}={<}{7}-{s},{s}-{3},{s}^{{2}}{>}\)

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

The parametric equations corresponding to the curve \(\displaystyle{r}_{{1}}{\left({t}\right)}\) are as follows:

x=t

\(\displaystyle{y}={4}-{t}\)

\(\displaystyle{z}={35}+{t}^{{2}}\)

The parametric equations corresponding to the curve \(\displaystyle{r}_{{2}}{\left({s}\right)}\) are as follows:

\(\displaystyle{x}={7}-{s}\)

\(\displaystyle{y}={s}-{3}\)

\(\displaystyle{z}={s}^{{2}}\)

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

\(\displaystyle{t}={7}-{s}\Rightarrow{s}+{t}={7}\)

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

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

That is,

\(\displaystyle{s}^{{2}}-{t}^{{2}}={35}\)

\(\displaystyle{\left({s}-{t}\right)}{\left({s}+{t}\right)}={35}\)

\(\displaystyle{\left({s}-{t}\right)}{7}={35}\)

\(\displaystyle{s}-{t}={5}\)

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

\(\displaystyle{s}-{t}+{s}+{t}={5}+{7}\Rightarrow{2}{s}={12}\Rightarrow{s}={6}\)

The corresponding value of t is as follows:

\(\displaystyle{t}={7}-{s}\Rightarrow{t}={7}-{6}\Rightarrow{t}={1}\)

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

Substitute t=1 in \(\displaystyle{r}_{{1}}{\left({t}\right)}\), to find the point of intersection.

\(\displaystyle{r}_{{1}}{\left({t}\right)}={<}{1},{4}-{1},{35}+{1}^{{2}}\ge{<}{1},{3},{36}{>}\)

\(\displaystyle{r}_{{1}}{\left({t}\right)}={<}{1},{3},{36}{>}\)

The angle between the normal vectors is given as \(\displaystyle{\cos{\theta}}={\frac{{{r}_{{1}}'{\left({t}\right)}\cdot{r}_{{2}}'{\left({s}\right)}}}{{{\left|{r}_{{1}}'{\left({t}\right)}\right|}{\left|{r}_{{2}}'{\left({s}\right)}\right|}}}}\)

The direction vector of the line \(\displaystyle{r}_{{1}}{\left({t}\right)}\) is \(\displaystyle{<}{t},-{t},{t}^{{2}}{>}\)

The direction vector of the line \(\displaystyle{r}_{{2}}{\left({s}\right)}\) is \(\displaystyle{<}-{s},{s},{s}^{{2}}{>}\)

At t=1,

\(\displaystyle{r}_{{1}}'{\left({t}\right)}={<}{1},-{1},{2}{t}{>}\)

\(\displaystyle{r}_{{1}}'{\left({1}\right)}={<}{1},-{1},{2}{>}\)

At s=6,

\(\displaystyle{r}_{{2}}'{\left({s}\right)}={<}-{1},{1},{2}{s}{>}\)

\(\displaystyle{r}_{{2}}'{\left({6}\right)}={<}-{1},{1},{12}{>}\)

\(\displaystyle{\cos{\theta}}={\frac{{{r}_{{1}}'{\left({t}\right)}\cdot{r}_{{2}}'{\left({s}\right)}}}{{{\left|{r}_{{1}}'{\left({t}\right)}\right|}{\left|{r}_{{2}}'{\left({s}\right)}\right|}}}}\)

\(\displaystyle={\frac{{{r}_{{1}}'{\left({1}\right)}\cdot{r}_{{2}}'{\left({6}\right)}}}{{{\left|{r}_{{1}}'{\left({1}\right)}\right|}{\left|{r}_{{2}}'{\left({6}\right)}\right|}}}}\)

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

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

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

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

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