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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?

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up vote 2 down vote accepted

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.

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But you shouldn't add $(3\phi_o,3\phi_0^2,2\phi_0^3)$ since that's not part of the direction vector of the tangent line. – Grumpy Parsnip Aug 9 '11 at 20:37
@Jim: Ach, of course. I'll edit. – Zev Chonoles Aug 9 '11 at 20:39
Zev, how did you get that it has a constant angle from your answer? I get how you got $a$ and $b$ but I am not sure how to show it forms a constant angle? – Lays Aug 18 '13 at 8:23

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