Say that the vectors $\vec{v_1}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} , \vec{v_2}=\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}$. The two vectors are linearly independent.
I want to try to find all vectors $\vec{y}$ that will make the $span \left \{ \vec{v_1}, \vec{v_2}, \vec{y} \right \} = \mathbb{R^3}$. There will be many such vectors $\vec{y}$. At first, I thought I could do a projection onto the column space of $\vec{v_1}$ and $\vec{v_2}$ and get a perpendicular vector. But even before I could do such a projection, I need a vector that is not in the column space of $\vec{v_1}$ and $\vec{v_2}$ to project from. And if I could find one such vector, then I wouldn't even need to do a projection because that will just fit in well as a vector for $\vec{y}$ to span the whole of $\mathbb{R^3}$.
Subsequently, base on the above thought, I begin to think that $\vec{y}$ not necessarily has to be perpendicular to $\vec{v_1}$ and $\vec{v_2}$. As long as $\vec{y}$ is not in the column space of $\vec{v_1}$ and $\vec{v_2}$, it will work. So as long as $\vec{y} \neq c_1\vec{v_1} + c_2\vec{v_2}$, I got all all the vectors that can make the span equals to the whole of $\mathbb{R^3}$. But expressing it this way isn't very right. How can I express this in the more usual explicit form from this idea?