TITLE: basic question about angles of lines
QUESTION [5 upvotes]: Show that the tangent lines to the regular parametrized curve $$
\alpha \left( t \right) = \left( {3t,3t^2 ,2t^3 } \right)
$$
   make a constant angle with the line $y=0$ , $z=x$

First of all, the derivate of that curve is $$
\left( {3,6t,6t^2 } \right)
$$
So in an arbitrary point of the curve at $ t=$$
\varphi 
$$
_0 $ the tangent line is
$$
\eqalign{
  & \left( {3\varphi _0 ,3\varphi _0 ^2 ,2\varphi _0 ^3 } \right) + t\left( {3,6\varphi _0 ,6\varphi _0 ^2 } \right)  \cr 
  &  = \left( {3\varphi _0  + 3\,t\,\,,\,\,\,3\varphi _0 ^2  + 6\,t\,\varphi _0 ,\,\,\,2\varphi _0 ^3  + 2\,t\,\varphi _0 ^2 } \right) \cr} 
$$
The other line is $ (u,0,u) $   but the dot product betweem this two lines is not constant, what is bad?

REPLY [2 votes]: The vectors we want to take the dot product of are
$$\mathbf{a}=(u,0,u),$$
$$\mathbf{b}=(3,6t,6t^2).$$
(As Jim points out below, we want to take the dot product of the direction vectors of the lines; what I wrote before was the dot product of $(u,0,u)$ and the position vector of a point on the tangent line). Recall that 
$$\mathbf{a}\cdot\mathbf{b}=||\mathbf{a}||\,\,||\mathbf{b}||\cos(\theta).$$
The fact that the dot product you're getting isn't constant is due to factor of $||\mathbf{a}||\,\,||\mathbf{b}||$ that you are forgetting to compensate for (or at least, I assume this is the problem that's occuring). Once you divide your answer by $||(u,0,u)||=u\sqrt{2}$ and $\left|\left|\left( {3\,\,,\,6\,t\,,6\,t ^2 } \right)\right|\right|=\sqrt{9+36t^2+36t^4}$, it should be constant, i.e. not have any $t$'s or $u$'s.