Quadric obtained by rotation of a line I have a line of equations:
\begin{align}r:
\begin{cases}
x = 2+2t \\
y = 2 - t \\
z=t
\end{cases}
\end{align}
and another line of equations:
\begin{align}s:
\begin{cases}
x = 3 \\
y = 1 \\
z=k
\end{cases}
\end{align}
I have to determine what quadric I get when i rotate line $r$ about $s$.
Every single point of the line that rotates creates a circumference of equation 
$x^2+y^2=r^2$
hence all those infinte points creates a quadric.
What is the best and fastest way to get to the result?
 A: The squared radius of the circle described on a plane $z=$ const. by a point on line $r$ is $r^2=(2+2z-3)^2+(2-z-1)^2$ (center of circle is $(3,1,z)$). The equation of the quadric is then
$$
(x-3)^2+(y-1)^2=(2z-1)^2+(1-z)^2.
$$
A: When you rotate a line around another you get a (circular) cone or a hyperboloid (including their degenerated versions) .
The axis line $s$ is parallel to the $z$ axis, it passes through the point $(3,1,0)$, and is normal to the planes at $z=const$, that it crosses at $A=(3,1,z)$.
The point at which the line $r$ instead crosses the planes at $z=const$ is $B=(2+2z,2-z,z)$.
Now the distance of a generic point $P=(x,y,z)$ from $A$ shall remain constant and equal to $|PQ|$  during the rotation, hence:
$$(x-3)^2+(y-1)^2+(z-z)^2=(2+2z-3)^2+(2-z-1)^2+(z-z)^2$$ 
Simplify that and you get the equation of the quadric.
A: In general, when you rotate in $\mathbb R^3$ one line around another, you can get one of the following surfaces:


*

*A hyperboloid of one sheet, in case the 2 lines do not intersect, and are not parallel, nor perpendicular. 

*A cylinder, in case the to lines are parallel (and not coincide).

*A cone (2 sided), in case the 2 lines do intersect, and do not coincide and are not perpendicular.

*A plane, in case the 2 lines intersect and are perpendicular.

*A plane with a "hole", in case the 2 lines are perpendicular, and do not intersect.

*Finally, a single line in case when the 2 lines coincide.


I think everything in the above classification is trivial, except the first one:
Why when we rotate one line around the other in $\mathbb R^3$, when the to lines are not parallel nor perpendicular and do no intersect, why do we get a one sheet hyperboloid ?
The answer is this:
Lets say I rotate around the $z$ axis. (It does not matter, right?).
So I am rotating some line (denote it by $l$) around the $z$ axis, and lets say the angle between $l$ and the $XY$ plane is $\alpha$. According to our assumption, $\alpha \ne 0, \pi/2$. 
Now lets say $P$ is the point on l which has minimal distance to the $z$ axis.
I can translate all the $3D$ picture, and assume without loss of generality that $P$ lies in the $XY$ plane.
So $P$ is the intersection of $l$ with the $XY$ plane.
Now it is clear that $l$ is perpendicular to $OP$. (If it is not clear consider the following observation: let $\phi(x,y,z)$ be the distance from (x,y,z) to the z axis. According to our assumption, when we restrict $\phi$ to $l$, $\phi$ has a local minimum at P. Thus, the directional derivative of $\phi$ on $l$ must vanish at $P$, i.e.the gradient of $\phi$ at $P$ is perpendicular to $l$).
So we conclude that $l$ is perpendicular to $OP$.
Now, consider some point Q (any point) on $l$:
Lets denote $$d = distance(P,Q)$$
Now we have: $$z[Q] = d\cdot sin(\alpha)$$
And if we denote by $Q'$ the orthogonal projection of $Q$ on the $XY$ plane, 
then since $l \bot OP$, we get: $PQ' \bot OP$.
But observe that $PQ' = d \cdot cos(\alpha)$
And since $PQ' \bot OP$, we conclude that: $$OQ'^2 = OP^2 + PQ'^2 = OP^2 + d^2 \cdot cos^2(\alpha) = OP^2 + z[Q]^2 / tan^2(\alpha)$$
Which is equivalent to:
$$x[Q]^2 + y[Q]^2 = OP^2 + z[Q]^2 / tan^2(\alpha)$$
or if we write $c$ instead of $OP$, since it is constant and does not depend on $Q$, and actually: $c = OP = min(\{distance(T,Oz) : T \in l\})$
We get the following:
$$x^2 + y^2 - z^2 / tan^2(\alpha) = c^2 $$
Which is equivalent in cylindrical coordinates to the following:
$$\rho^2 - z^2 / tan^2(\alpha) = c^2 $$
And now it is clear that the surface can be obtained by rotating a hyperbola around the z axis, and it is also clear that the obtained hyperboloid will be one sheet.
So now lets consider your case:
You are rotating around line $s$ which is parallel to the $z$ axis, and pass through the point $(3,1,0)$.
The line which is rotated is $r$ which has the above parametrization:
\begin{align}r:
\begin{cases}
x = 2+2t \\
y = 2 - t \\
z=t
\end{cases}
\end{align}
So firstly, lets translate the picture so instead $s$ we will have the $z$ axis.
For this I will add the vector $(-3,-1,0)$ to any point in the picture.
The new line $s'$ is now the $z$ axis, and the new line $r'$ has now the parametrization:
\begin{align}r':
\begin{cases}
x = -1+2t \\
y = 1 - t \\
z=t
\end{cases}
\end{align}
Now the direction vector of the line is $(2,-1,1)$
And the line intersects the $XY$ plane at the point $(-1,1,0)$
Clearly they are not perpendicular, so we should provide a translation, parallel to the z axis, to get the desired $P$ as above.
So lets find where the distance is minimal. This is equivalent to find where the squared distance is minimal, i.e. we want to minimize $$(-1 +2t)^2 + (1-t)^2$$
And if I didn't make a mistake, the minimum is at $$t=3/5$$
So lets translate our picture by the vector: $(0,0,-3/5)$
And we get the following new line:
\begin{align}r'':
\begin{cases}
x = -1+2t \\
y = 1 - t \\
z=t - 3/5
\end{cases}
\end{align}
Hence the new $P$ is: $(1/5, 2/5, 0)$
And now actually $\overrightarrow{OP} = (1/5, 2/5, 0) \bot (2,-1,1)$ = direction vector of r. (This was a sanity check).
Now lets compute $tan^2(\alpha)$:
$$sin^2(\alpha) =cos^2(\pi/2 -\alpha)=\frac{<(2,-1,1),(0,0,1)>^2}{\|(2,-1,1)\|^2 \cdot \|(0,0,1)\|^2} = \frac{1}{6} $$
Hence $$tan^2(\alpha) = \frac{1}{5}$$
Now we have $$c^2 = OP^2 = 1/5$$
So our translated surface looks like:
$$x^2 + y^2 - 5 z^2 = \frac{1}{5}$$
But we want the original hyperboloid! before translation!
So lets translate it back:
$$(x-3)^2 + (y-1)^2 - 5 (z-\frac{3}{5})^2 = \frac{1}{5}$$
And this is the desired hyperboloid!
A: We have line of equation
\begin{align}r:
\begin{cases}
x=f(t)=2+2t\\
y=g(t)=2-t\\
z=h(t)=t
\end{cases}
\end{align}
and the axis of the rotation
\begin{align}s:
\begin{cases}
x=x_0+at=3\\
y=y_0+bt=1\\
z=z_0+ct=k
\end{cases}
\end{align}
Let's choose a generic point on line $s$ and let it be $C(3,1,0)$.
We have to find the plane through $C$ and perpendiular to $r$.
The plane has generic equations
$a(x-f(t))+b(y-g(t))+c(z-h(t))=0$
So our plane has the same parameters as $s$ $(0,0,1)$ so our plane is
$z-t=0$, in particular $t=z$.
Now we need to find the sphere with centre in $P(3,1,0)$ and radius $\overline{CP}$ knowing that $P$ is a point on line $r$ and has equations $P(f(t), g(t), h(t))$.
The generic sphere has equation:
$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=(f(t)-x_0)^2+(g(t)-y_0)^2+(h(t)-z_0)^2$
Then our equations becomes
$(x-3)^2+(y-1)^2 +(z-0)^2 = (2+2t-3)^2+(2-t-1)^2+(t-0)^2$
$x^2-6x+9+y^2-2y+1+z^2 = (2t-1)^2+(1-t)^2+t^2$
$x^2-6x+9+y^2-2y+1+z^2 = 4t^2-4t+1+1+t^2-2t+t^2$
$x^2-6x+9+y^2-2y+1+z^2 = 4z^2-4z+1+1+z^2-2z+z^2$
And the equation that we get is
$x^2+y^2-5z^2-6x-2y+6z+8=0$
It is an hyperboloid as we can verifiy with the matrices
\begin{align}I_3=
\begin{vmatrix}
1 & 0 & 0 & -6 \\
0 & 1 & 0 & -2 \\
0 & 0 & -5 & 6 \\
-6 & -2 & 6 & 8
\end{vmatrix} = 124
\end{align}
\begin{align}I_2=
\begin{vmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -5
\end{vmatrix} = -5
\end{align}
We have $det(I_3)>0$ and $det(I_2)\ne0$, $I_2$'s Eigenvalues are $\lambda_1,2 = -1$ and $\lambda_3 = 5$ which indicates that this equation represents an hyperboloid of one sheets

