For which $a,b$ it is true that $\mathbb{R}^3=V_a\oplus V_b$?

Question

$a,b\in\mathbb{R}$. In $\mathbb{R}^3$ we have $V_a=span(\begin{bmatrix} 1 \\ 1+a \\ -2\end{bmatrix},\begin{bmatrix} 2 \\ 6 \\ -2-a\end{bmatrix})$, $V_b=span(\begin{bmatrix} 0 \\ 3 \\ -1-b\end{bmatrix},\begin{bmatrix} 2 \\ 2+b \\ -2\end{bmatrix})$. For which $a,b$ it is true that $\mathbb{R}^3=V_a\oplus V_b$? Obviously,$\begin{bmatrix} 0 \\ 3 \\ -1-b\end{bmatrix},\begin{bmatrix} 2 \\ 2+b \\ -2\end{bmatrix} $ are linearly indepentent because in first coordinate firt has $0$ and second $2$ and it's easy to prove this for $\begin{bmatrix} 1 \\ 1+a \\ -2\end{bmatrix},\begin{bmatrix} 2 \\ 6 \\ -2-a\end{bmatrix}$. So, $dim(V_a)=dim(V_b)=2$, so $dim(V_a\cap V_b)>0$, so it is not possible that $\mathbb{R}^3=V_a\oplus V_b$. Am I right?

Answer 1

You're not quite right, because of one case (which, naturally enough, is the part that says "it's easy to prove...").

If $a = 2$, then $V_a$ is one dimensional because the second vector is a scalar multiple of the first. A dimension argument is insufficient in that case.