Is there a plane that passes through a pair of lines that have no points in common? I'm reading a book on geometry in Spanish by Ana Berenice Guerrero (see here, p. 18,19). So, there's this theorem that says that given a pair of lines with no points in common there's only one plane that have both of them. 
I have read the proof a lot of times and I feel like there's something wrong. Then my frustration comes from the fact that I can imagine a pair of lines in the space that are not parallel and with no points in common. I cannot visualize the plane that contain both lines. Maybe someone can help me understand this. 

 A: As far as I can see, the proof is incorrect. I have only a rudimentary understanding of Spanish, but it seems the proof assumes that both the planes $\alpha$ and $\beta$ contain all the points of both the lines $a$ and $c$. This is only an assumption, and is not valid. If this assumption were part of the statement of the theorem, it would be okay, for then the theorem would be "If at all there is a plane containing two lines with no common points, then this plane is unique".
A: Interesting logic. I'm not sure what they were thinking. Here is a translation of the false theorem:

Theorem 1.3.6. For two non-intersecting lines, there is a unique intersecting plane
Let $a,c$ be non-intersecting lines. Then there exists a unique plane containing $a,c.$
  
  
*
  
*Take unique points $A,B$ on $a$, and $C,D$ on $c$. Now $ABD$ are not all on the same line. Likewise $CDB$ are not all on the same line. 
  
*By 'axiom 1.2.5,' $ABD$ determine a unique plane $\alpha$ and $CDB$ determine a unique plane $\beta$.
  
*Assume WLOG that $a$ is contained in $\alpha$ and $c$ is contained in $\beta$.
  
*Now both $\alpha$ and $\beta$ contain both $ABD$ and $CDB$. But since the planes defined by $ABD$ and $CDB$ are unique, $\alpha$ and $\beta$ are the same.
  
*Therefore there is a unique plane containing $a,c$.
  

The error occurs on the 4th step. There is no reason that $ABD$ should also contain $C$, or that $CDB$ should contain $A$. For that we would need the additional assumption that the lines are parallel. (Or, if we are more concerned about the uniqueness of the plane, we would instead need to assume something like 'there exists some plane with all 4 points in it')
Unless its not a focus of the course, I would be extremely wary of a textbook with this level of errors.
A: Well. It is only possible to construct a View where the two lines appear to be parallel, and there exists only one such view. So there is some truth in it .. However if you were to have a PLANE.. it will have only 2 points as the two lines are skewed ... so in conclusion, view = possible, Plane = impossible. 
