Can two vectors of 3-Tuples span $\mathbb R^3$? I'm just checking, but when we have $2$ vectors 
$$
V_1=\begin{pmatrix}a\\b\\c\end{pmatrix}\text{ and }V_2=\begin{pmatrix}e\\f\\g\end{pmatrix}
$$
We could theoretically span $\mathbb R^3$ real space with just these two vectors right?
 A: No.
In the normal case ($V_1\ne0$, $V_2\ne 0$ and they are linearly independent) consider $V_3 = V_1\times V_2 = (bg-cf, ce-ag, af-be)$ then you have $V_1\cdot V_3 = abg-acf + bce - abg + acf - bce = 0$ similarily you have $V_2\cdot V_3 = 0$. This makes it impossible that $V_3 = aV_1 + bV_2$.
To see why you consider 
$$0 = aV_1\cdot V_3 = a^2 V_1\cdot V_1 + ab\cdot V_2$$
$$0 = -bV_2\cdot V_3 = -b^2 V_2\cdot V_2 + ab\cdot V_2$$
And sum these equations and get $0 = a^2 V_1\cdot V_1 + b^2 V_2\cdot V_2$, but we know that $V_1\cdot V_1\ge 0$ and equal to zero iff $V_1=0$ so we see that $V_1=0$ or $a=0$ (and $V_2=0$ or $b=0$). So since $V_1\ne 0$ and $V_2\ne 0$ you have $a=b=0$ which means that $V_3=0$ but this contradicts the fact that $V_1$ and $V_2$ are linearly independent.
The cases where either $V_1$, $V_2$ is zero or they are linearly independent is basically the case where you only have one vector. You use the same argument - you only have to find a vector perpendicular to it. 
The last case where both are zero is dead simple. Zero times anything is still zero - just take a non-zero vector.

Another way to see this is that if they spanned $\mathbb R^3$ you could write $e_x$, $e_y$ and $e_z$ as linear combinations of $V_1$ and $V_2$. That is
$$e_x = x_1 V_1 + x_2 V_2 $$
$$e_y = y_1 V_1 + y_2 V_2 $$
Since $e_x$ and $e_y$ are linearly independent then so are $(x_1, x_2)$ and $(y_1, y_2)$ which means that we can solve for $V_1$ and $V_2$ by the above. That is $V_1$ and $V_2$ are linear combinations of $e_x$ and $e_y$. Now if $e_z$ is a linear combination of $V_1$ and $V_2$ it would also be a linear combination of $e_x$ and $e_y$ which clearly isn't the case.
A: If you try to span $\mathbb{R}^{3}$ using these vectors, i.e. for any general $v = (x,y,z) \in \mathbb{R}^{3}$ you want to find $\alpha_{1}, \alpha_{2} \in \mathbb{R}$ such that
$$
\alpha_{1} V_{1} + \alpha_{2} V_{2} = v
$$
this is a system of equations with three equations (one for each component) but just two unknowns ($\alpha_{1}, \alpha_{2}$). So you cannot, for every $v$, find a solution. 
To have a more specific example, try to find $\alpha_{1}, \alpha_{2}$ such that $\alpha_{1} V_{1} + \alpha_{2} V_{2} = v$ for the example @Mundron Schmidt and @skyking posted above, at least in the case that the $V_{1}$, $V_{2}$ are linearly independent (i.e. in this case, not multiples of each other). His method, using the "cross product" provides a way to construct a vector $V_{3}$ that is (again, if $V_{1}, V_{2}$ are linearly independent) always linearly independent of $V_{1},V_{2}$ - in fact it is even orthogonal to them, as he showed - and this cannot be combined linearly from them.
So no, it is not possible to span a 3-dimensional space with 2 vectors.
A: While skyking gave a very elegant explanation for $\mathbb{R}^3$, there is a general fact that $n$ $n+1$ tuples can't span $\mathbb{R}^{n+1}$. One possible proof is this:
1) For $n=1$ this is obviously true.
2) Suppose any element of $\mathbb{R}^{n+1}$ can be represented as a linear combination of your tuples, then you have $e_{n+1} = (0, 0, ..., 0, 1)^T = a_1V_1+a_2V_2+...+a_nV_n$. Let $a_n \neq 0$, then $V_n = b_1V_1 + b_2V_2 + ... + b_{n-1}V_{n-1} + b_ne_{n+1}$. Thus, vectors $e_1, e_2, ..., e_n$ can be expressed using only $V_1, ..., V_{n-1}$ and $e_{n+1}$. So, ignoring the last component, you get that $n-1$ $n$-tuples span $\mathbb{R}^n$.
3) By induction, from 1) and 2) we conclude that the statement holds for all $n$.
