# Math Help - Finding a unit normal vector

1. ## Finding a unit normal vector

I wanted to make sure i got the right answer
$T=cost*cos(sint) i - cost*sin(sint) j- sint k$
$T'=[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j - costk$
$|T'|=\sqrt{(-sint*cos(sint)-sin(sint)cos^2t)^2 + (sint*sin(sint)-cos^2+cos(sint))^2 - cos^2t}$
I got the answer for |T'| to be $cos^2t$
$\frac{[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j - costk}{cos^2t}$
$\frac{[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j}{cos^2t} - sectk$

I don't know if I should simplify more.

2. Originally Posted by alexgo
I wanted to make sure i got the right answer
$T=cost*cos(sint) i - cost*sin(sint) j- sint k$
$T'=[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j - costk$
$|T'|=\sqrt{(-sint*cos(sint)-sin(sint)cos^2t)^2 + (sint*sin(sint)-cos^2+cos(sint))^2 - cos^2t}$
I got the answer for |T'| to be $cos^2t$
$\frac{[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j - costk}{cos^2t}$
$\frac{[-sint*cos(sint)-sin(sint)cos^2t]i + [sint*sin(sint)-cos^2+cos(sint)]j}{cos^2t} - sectk$
You are making hard on yourself.

$\displaystyle T' = \frac{{R' \times (R'' \times R')}}
{{\left\| {R'} \right\|^3 }}$

3. Originally Posted by Plato
You are making hard on yourself.

$\displaystyle T' = \frac{{R' \times (R'' \times R')}}
{{\left\| {R'} \right\|^3 }}$
ha that is a pain too.

4. Originally Posted by alexgo
ha that is a pain too.
Well it is all a pain if you ask me.