Find the solid of revolution, given some clues. Recently, I have encountered 2 problems related to solids of revolution. Given a normal 3-dimensional Euclidean space, with a given regular coordinate system.
1) Find the equation of the cone of revolution which adopts three coordinate axes as its line and its axe of revolution being in the first octant.
2) Find the equation of the cylinder of revolution which has the radius of $1$ and its axe of revolution being the bisector of $\widehat{yOz}$.


*

*For the first problem, it's not hard to guess that the axe of revolution is $x = y = z$. But in general cases, when given three intersect lines, is it possible to get the axe of revolution by a guaranteed method? 

*And for the second problem, I solve it like this: 


*

*Get the equation of the axe of revolution

*Get the general form of a plane which is perpendicular to the axe, then get the intersection of the plane and the axe

*Write the intersection of that plane and the sphere of radius $1$ which adopts the found intersecton as its center.

*Let the plane run through the space



Is my way correct?
Thank you for reading. I appreciate any help.
 A: Of course there are many different possible approaches to solve the questions you pose.
a) given three intersecting lines, find the axis of the circular cone having those lines as generatrices
For this question, one of the straightest method might possibly be the following.
Denote by $\bf a$, $\bf b$, $\bf c$ the unit vectors corresponding to the generatrices, and by $\bf v$ a generic
vector parallel to the axis.
Imagine these as  directed segments applied at the  line intersection point ( the vertex of the cone).
Then it is clear that a necessary and sufficient condition for $\bf v$ to be on the axis is that it makes 
the same angle with the three generatrices, i.e.
$$
\left\{ \matrix{
  a_{\,x} v_{\,x}  + a_{\,y} v_{\,y}  + a_{\,z} v_{\,z}  = h = \left| {\bf v} \right|\cos \vartheta  \hfill \cr 
  b_{\,x} v_{\,x}  + b_{\,y} v_{\,y}  + b_{\,z} v_{\,z}  = h = \left| {\bf v} \right|\cos \vartheta  \hfill \cr 
  c_{\,x} v_{\,x}  + c_{\,y} v_{\,y}  + c_{\,z} v_{\,z}  = h = \left| {\bf v} \right|\cos \vartheta  \hfill \cr}  \right.
$$
and, fixing a value for $h$,  you can solve this linear system for $\bf v$, and then find $\cos \vartheta$.
Then the equation of the cone will just be
$$
\left( {x - x_{\,0} } \right)v_{\,x}  + \left( {y - y_{\,0} } \right)v_{\,y}  + \left( {z - z_{\,0} } \right)v_{\,z}  = \left| {\bf x}-{\bf x_0} \right| \left| {\bf v} \right|\cos \vartheta 
$$
with ${\bf x}-{\bf x_0}=
\left(\left( {x - x_{\,0} } \right),\left( {y - y_{\,0} } \right), \left( {z - z_{\,0} } \right)\right)$
b) equation of a cyrcular cylinder, given its axis and radius
In this case the straightest way is to impose that the generic point $ {\bf p} = (x,y,z)$ be at constant distance $r$ from the
axis, defined by the unit vector $\bf a$ and one point through which the axis is passing (${\bf p_{a}} =(x_a,y_a,z_a)$), i.e.:
$$
\left| {\;\left( {{\bf p} - {\bf p}_{\,{\bf a}} } \right) \times {\bf a}\;} \right| = r
$$
Otherwise, if you want to express it in a parametric form, take two unit vectors ($\bf n$ and $\bf m$) normal to $\bf a$ and between themselves,
express the point on the directing circle as 
${\bf n}\,r\,\cos \vartheta  + {\bf m}\,r\,\sin \vartheta $
and add a translation proportional to $\bf a$, thus
$$
{\bf p} - {\bf p}_{\,{\bf a}}  = {\bf n}\,r\,\cos \vartheta  + {\bf m}\,r\,\sin \vartheta  + t\,{\bf a}
$$
