
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 $\displaystyle \blacksquare$

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...