intersection of to curves

[Guest]

At what point do the curves r1(t)=<t,1-t,3+t^2> and r2(s)=<3-s,s-2,s^2>.
Find their angle of intersection correct to the nearest degree.

I'm not sure what to do. I thought i take each x,y, and z value, set them equal to each other, and find their intersection point. In that case i would have:

t = 3-s
the intersection point would be at t=1.5

1-t = s-2
the intersection point would be at t=1.5

3+t^2 = s^2
but these do not intersect

I'm guessing i'm not suppost to find the intersection point this way.

 

[stapel]

At what point do the curves r1(t)=<t,1-t,3+t^2> and r2(s)=<3-s,s-2,s^2>.

"At what point do the curves r1 and r2".... what? "Intersect"? Or something else?

Thank you.

Eliz.

 

[Guest]

Sorry, accidently left a word out.

At what point do the curves intersect.

edit*
nvm, i figured it out.

 

[soroban]

Hello, almostacasper!

(1) At what point do these curves intersect?
. . \(\displaystyle f(t)\:=\:\langle t,\,1-t\,,3+t^2\rangle\) and \(\displaystyle g(s)\:=\:\langle3-s,\,s-2,\.s^2\rangle\)
(2) Find their angle of intersection correct to the nearest degree.

Set the \(\displaystyle x\)'s equal, the \(\displaystyle y\)'s equal, and the \(\displaystyle z\)'s equal.

. . \(\displaystyle \begin{array}t & = & 3\,-\,s & \;\Rightarrow\; & t + s & = & 3 \\
1\,-\,t & = & s\,-\,2 & \;\Rightarrow\; & t\,+\,s & = & 3 \\
3\,+\,t^2 & = & s^2 & \;\Rightarrow\; & t^2\,-\,s^2 & = & -3 \end{array}\)

Solve the second equation for \(\displaystyle s:\;\;s \:=\:3\,-\,t\)

Substitute into the third equation: \(\displaystyle \:t^2\,-\,(3\,-\,t)^2 \:=\:3\)

Simplify: \(\displaystyle \,t^2\,-\,9\,+\,6t\,-\,t^2\:=\:-3\;\;\Rightarrow\;\;6t\:=\:6\;\;\Rightarrow\;\;t\,=\,1\)

Substitute into the first equation: \(\displaystyle \:1\,+\,s\:=\:3\;\;\Rightarrow\;\;s\,=\,2\)

(1) The point of intersection is: \(\displaystyle \:f(1)\:=\:g(2)\:=\:\L\fbox{\langle1,\,0,\,4\rangle}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

\(\displaystyle f'(t)\:=\:\langle 1,\,-1,\,2t\rangle\)
. . At \(\displaystyle t\,=\,1:\;f'(1)\:=\:\langle1,\,-1,\,2\rangle\)

\(\displaystyle g'(t)\:=\:\langle-1,\,1,\,2s\rangle\)
. . At \(\displaystyle s\,=\,2:\;g'(2)\:=\:\langle -1,\,1,\,4\rangle\)

The angle between two vectors, \(\displaystyle \vec{u}\) and \(\displaystyle \vec{v}\), is given by: \(\displaystyle \L\:\cos\theta \:=\:\frac{|\vec{u}\cdot\vec{v}|}{|\vec{u}||\vec{v}|}\)


We have: \(\displaystyle \:\:\cos\theta \;=\;\frac{\langle1,\,-1,\,2\rangle\cdot\langle-1,\,1,\,4\rangle}{\sqrt{1^2+1^2+2^2}\cdot\sqrt{1^2+1^2+4^2}} \;=\;\frac{-1\,-\,1\,+\,8}{\sqrt{6}\cdot\sqrt{18}} \;=\;\frac{6}{6\sqrt{3}}\;=\;\frac{1}{\sqrt{3}}\)

(b) Therefore: \(\displaystyle \theta \:=\:\cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \:\approx\:\L\fbox{54.74^o}\)