Given the following curve, a helix, $$\alpha(t)=(\cos(t),\sin(t),t)$$ I consider, for each $t$, the line that pass through $\alpha(t)$ and it's orthogonal to $z$ axis. Then, I have to parametrize the surface that is the union of all this lines.
I thought about finding $\alpha'(t)$ and take $N(t)$ such that $$N(t)\cdot\alpha'(t)=0$$ and $$N(t)=(x(t),y(t),0)$$
I don't know if what I'm doing is correct and if so, what I'm supposed to do next.
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Hint: Fix an arbitrary $t$ (so you got a point $p(t)$ and parameterize the perpendicular segment from the $z$ axis to $p(t)$. Use a different parameter, like $s$, making sure that $0\leq s\leq 1$. What do you get?
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You need to know very little about the surface (and hence do very little work) to answer the question: first, notice that for any point $p = (x,y,z)$ (not lying on the $z$ axis), the line through $p$ and its orthogonal projection onto the $z$-axis (we need $p$ to not lie on the $z$-axis purely so that this line is well-defined) is parameterised by $(rx,ry,z)$ (with $r$ varying through the reals). Now, you have a parameterisation of your curve given to you, so all you need to do is substitute that parameterisation into this one. Thus, this surface is parameterised by $$\beta(r,t) = (r\cos(t),r\sin(t),t).$$
I haven't done this stuff in about a decade, but it's not that hard. My answer is similar to user3482749's.
Have a quick glance at the top answer to this question to get started. In particular, an example of an equation of a straight line in 3D is:
$$( x,y,z) = ( 5,5,5) +t( -4,-3,-2),$$
where it should be pretty obvious what is going on: $( 5,5,5)$ is just a point (position vector) on the line - you could have used any other point on the line - and $( -4,-3,-2)$ is the "gradient" of the line (the direction vector).
For your question, for each $t$, you want to draw a straight line from $(0,0,t)$ to $ (\cos(t),\sin(t),t)$. So for each $t$, you can use $(0,0,t)$ as a point on your line and your line will have gradient $(\cos(t),\sin(t),t) - (0,0,t) = (\cos(t),\sin(t),0)$. Therefore for each $t$, the equation of the straight line orthogonal to the z-axis is:
$$( x,y,z) = ( 0,0,t) +s( \cos(t),\sin(t),0),s \in \mathbb{R}$$
Therefore the parameterised equation of your surface is this but for $every$ real value of $t$:
$$( x,y,z) = ( 0,0,t) +s( \cos(t),\sin(t),0),s \in \mathbb{R}, t \in \mathbb{R}$$
which is similar to user3482749's answer...