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

allhvasstH
2021-10-29
Answered

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

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Layton

Answered 2021-10-30
Author has **89** answers

Consider the following curves:

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

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

Find the point of intersection of the curves and angle of intersection$\theta$ .

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

x=t

$y=4-t$

$z=35+{t}^{2}$

The parametric equations corresponding to the curve${r}_{2}\left(s\right)$ 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.

$s-t+s+t=5+7\Rightarrow 2s=12\Rightarrow s=6$

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${r}_{1}\left(t\right)$ , to find the point of intersection.

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

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

The angle between the normal vectors is given as$\mathrm{cos}\theta =\frac{{r}_{1}^{\prime}\left(t\right)\cdot {r}_{2}^{\prime}\left(s\right)}{\left|{r}_{1}^{\prime}\left(t\right)\right|\left|{r}_{2}^{\prime}\left(s\right)\right|}$

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

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

At t=1,

${r}_{1}^{\prime}\left(t\right)=<1,-1,2t>$

${r}_{1}^{\prime}\left(1\right)=<1,-1,2>$

At s=6,

Find the point of intersection of the curves and angle of intersection

The parametric equations corresponding to the curve

x=t

The parametric equations corresponding to the curve

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

That is,

Solve the equations

The corresponding value of t is as follows:

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

Substitute t=1 in

The angle between the normal vectors is given as

The direction vector of the line

The direction vector of the line

At t=1,

At s=6,

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