# At what point do the curves r_1(t)=t,4-t,35+t^2 and r_2(s)=7-s,s-3,s^2Z

At what point do the curves $$\displaystyle{r}_{{1}}{\left({t}\right)}={t},{4}-{t},{35}+{t}^{{2}}$$ and $$\displaystyle{r}_{{2}}{\left({s}\right)}={7}-{s},{s}-{3},{s}^{{2}}$$ intersect? (x,y,z)= Find angle of intersection, $$\displaystyle\theta$$, correct to the nearest degree.

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