# Easy vector question

• Jun 22nd 2008, 08:55 PM
Mathstud28
Easy vector question
I was rifling through some USSR problem books (with no solutions!) and I found this one, its really easy, but I was dissatisfied with my proof. I think it can be done with less work.

Prove that the principal normal unit vector, denoted $\displaystyle \bold{N(t)}$ does not exist for a linear function $\displaystyle \bold{r(t)}$

My proof is, dont look if you dont want to

$\displaystyle \bold{r(t)}=\langle{at+b,ct+d,et+f\rangle}$

so $\displaystyle \bold{r'(t)}=\langle{a,c,e}\rangle$

and

$\displaystyle ||\bold{r'(t)}||=\sqrt{a^2+c^2+e^2}$

So the tangent unit vector is

$\displaystyle \bold{T(t)}=\frac{\langle{a,c,e}\rangle}{\sqrt{a^2 +c^2+e^2}}$

So now

$\displaystyle \bold{T'(t)}=\langle{0,0,0\rangle}$

and $\displaystyle ||\bold{T'(t)}||=0$

$\displaystyle \therefore\bold{N(t)}=\frac{\bold{T'(t)}}{||\bold{ T'(t)}||}$ does not exist because there will be a division by zero $\displaystyle \blacksquare$
• Jun 22nd 2008, 10:32 PM
Aryth
It looks right and it makes sense. The Tangent Vector Function's components are all constants, and therefore it has to have a magnitude of zero, and in order for the PUV to exist, $\displaystyle ||\bold{T'(t)}||$ can't be zero. That's the way I'd have proven it...