$\mathbb{R}^3$ spanned by skew lines Let $a,b,c,d$ be vectors in $\mathbb{R}^3$ such that $a+\lambda b$ ($\lambda\in \mathbb{R}$) , and $c+\mu d$ ($\mu\in \mathbb{R}$) are skew lines, i.e $b$ and $d$ are not proportional and  $a+\lambda b\neq c+\mu d$ for any $\lambda, \mu\in \mathbb{R}$, in other words both the lines are niether parallel nor intersect each other in $\mathbb{R}^3$.
Can any one tell how these skew lines span the whole of $\mathbb{R}^3$.
 A: Short answer:
Two different lines will at the very least span a 2-dimensional space. This however, can only be the case if these lines lie in the same plane. So the lines must span a 3-dimensional space.
Longer answer:
Let $\langle.,.\rangle:\mathbb{R^3}\times\mathbb{R^3}\to\mathbb{R}$ denote the inner product on $\mathbb{R^3}$. Define $u:=a+\lambda b$, where $\lambda=-\frac{\langle a,a\rangle}{\langle a,b\rangle}$. Then $\langle a,u\rangle=\langle a,a\rangle-\frac{\langle a,a\rangle}{\langle a,b\rangle}\langle a,b\rangle=0$ (by properties of the inner product). This means that $a$ and $u$ are orthogonal i.e. they are linearly independent. Now if both $c$ and $d$ lie in the span of $a$ and $u$, then we have two points on some plane, which means the line between the points $c$ and $d$ must also lie on this plane. But then the line $c+\mu d$ must lie in the same plane as the span of $a$ and $u$ (which contains the line $a+\lambda b$), which means the lines would intersect. As a result, c and/or $d$ isn't in the span of $a,u$. Without loss of generality, assume $d$. Then $a,u,d$ are 3 linearly independent vectors, which obviously must span $\mathbb{R}^3$. $a$ and $u$ come from the line $a+\lambda b$ and $d$ comes from the line $c+\mu d$, which would also imply that these lines span $\mathbb{R}^3$.
A: More rather than a very explicit answer may this be taken as an indication of a very general property. Consider an arbitrary field $\mathbf{K}$, a left $\mathbf{K}$-vector space $\mathbf{V}$ and an affine space $\mathscr{A}$ having $\mathbf{V}$ as its space of translations (also referred to as director space in some languages, English not being among them to my knowledge). For any affine subspace $\mathscr{R} \subseteq \mathscr{A}$ let us write $\mathrm{Dir}\mathscr{R}$ for its director subspace.
Furthermore, consider two affine subspaces $\mathscr{M}, \mathscr{N} \subseteq \mathscr{A}$ that are disjoint. It is then the case that for any pair of points $(P, Q) \in \mathscr{M} \times \mathscr{N}$ we have the description:
$$\mathrm{Dir}(\mathscr{M} \vee \mathscr{N})=\mathrm{Dir}\mathscr{M}+\mathrm{Dir}\mathrm{\mathscr{N}}+\mathbf{K}(\overrightarrow{PQ}),$$
in other words the director subspace of the affine subspace generated by the union of $\mathscr{M}$ and $\mathscr{N}$ is generated by their respective director subspaces together with the vector line generated by the vector given by any two points, each in one of the affine subspaces in question. It follows that:
$$\mathrm{dim}_K(\mathscr{M} \vee \mathscr{N})=1+\mathrm{dim}_K(\mathrm{Dir}\mathscr{M}+\mathrm{Dir}\mathrm{\mathscr{N}}).$$
In your particular case, the scalar field can be taken to be an arbitrary $\mathbf{K}$, $\mathbf{V}=\mathscr{A}=\mathbf{K}^3$ (where in each instance we are considering the respective canonical structure, of vector respectively affine space on $\mathbf{K}^3$) and $\mathscr{M}$, $\mathscr{N}$ are two disjoint and nonparallel lines. The fact they are disjoint means the general relation(s) above apply in this instance, and the hypothesis of non-parallelism means that the two director vector lines for each of the affine lines in question are distinct. Their sum is therefore a vector plane (a subspace of dimension $2$), hence by virtue of the dimensional relation above the affine subspace generated by the two affine lines is of dimension $3$. Since the ambient space itself is of dimension $3$, this means that the two lines in question span the whole space.
This is immediately generalised to
an ambient space of dimension $2n+1$ ($n \in \mathbb{N}$, of course) with two disjoint subspaces of dimension $n$ whose director subspaces intersect trivially: these subspaces will span the whole ambient space.
