# Normal vector

• Feb 4th 2011, 03:48 PM
Killer
Normal vector
Given the curve $\displaystyle C$ on $\displaystyle \mathbb R^3$ parametrized by $\displaystyle r(t).$ Find the normal vector on $\displaystyle r(1)=(3,1,3)$ if the osculator plane on that point is $\displaystyle 3y-x=0$ and $\displaystyle r(1)\cdot N(1)>0$ with $\displaystyle r'(1)=(6,2,2).$ (I don't know if "osculator" makes sense here.)

I hope you guys can help, I have an exam on march and I'm scared.

I don't remember how to solve this, since I lost my calculus notebook.

Thanks!
• Feb 4th 2011, 10:52 PM
FernandoRevilla
Hint :

$\displaystyle \det [r(t)-r(1),r'(1),r''(1)]=k(3y-x)$

Fernando Revilla
• Feb 5th 2011, 10:30 AM
Killer
Quote:

Originally Posted by FernandoRevilla
$\displaystyle \det [r(t)-r(1),r'(1),r''(1)]=k(3y-x)$

Okay, this may be related to the osculator plane, but I don't get the determinant and their components.

Could you elaborate?

Thanks!
• Feb 7th 2011, 12:12 PM
Killer
Hi Fernando, is it possible that you may help me a bit more?

I'll appreciate that!
• Feb 7th 2011, 12:45 PM
zzzoak
$\displaystyle r(t)=(x(t),y(t),z(t))$

$\displaystyle r(1)=(x(1),y(1),z(1))$

$\displaystyle r'(t)=(x'(t),y'(t),z'(t))$

$\displaystyle r'(1)=(x'(1),y'(1),z'(1)).$