$\mathbb{R}^3$ 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 $\mathbb{R}^3$.
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 $\mathbb{R}^3$.
Denote $L_1 =\{x\in\mathbb{R}^2 : \exists \lambda  :x=a+\lambda  b \} ,$ $L_2 =\{x\in\mathbb{R}^2 : \exists \mu:  x=c+\mu d \} ,$ then
$$L_1 +L_2 =\mathbb{R}^2$$