- #1

Pengwuino

Gold Member

- 4,989

- 15

Ok so I have a T vector and N vector...

[tex]\begin{array}{l}

T(1) = \langle \frac{2}{3},\frac{{ - 1}}{3},\frac{2}{3}\rangle \\

N(1) = \langle \frac{2}{3},\frac{2}{3},0\rangle \\

B(1) = T(1) \times N(1) \\

B(1) = \langle \frac{{ - 1}}{3},\frac{2}{3},\frac{2}{3}\rangle \\

\end{array}[/tex]

I also have the coordinate of the original equation at 1...[tex]r(1) = \langle 2,1,0\rangle [/tex]

This left me with the standard equation of the osculating plane...

[tex]- x + 2y + 2z = 0[/tex]

Now I need to find the coordinates of hte center of this circle where t=1. How am i suppose to do this?

[tex]\begin{array}{l}

T(1) = \langle \frac{2}{3},\frac{{ - 1}}{3},\frac{2}{3}\rangle \\

N(1) = \langle \frac{2}{3},\frac{2}{3},0\rangle \\

B(1) = T(1) \times N(1) \\

B(1) = \langle \frac{{ - 1}}{3},\frac{2}{3},\frac{2}{3}\rangle \\

\end{array}[/tex]

I also have the coordinate of the original equation at 1...[tex]r(1) = \langle 2,1,0\rangle [/tex]

This left me with the standard equation of the osculating plane...

[tex]- x + 2y + 2z = 0[/tex]

Now I need to find the coordinates of hte center of this circle where t=1. How am i suppose to do this?

Last edited: