Finding a basis for the intersection of two vector subspaces. 
Suppose:
$V_1$ is the subspace of $\mathbb{R}^3$ given by $V_1 =
 \{(2t-s,t,t+s)|t,s\in\mathbb{R}\}$
and
$V_2$ is the subspace of $\mathbb{R}^3$ given by $V_2 =
 \{(s,t,s)|t,s\in\mathbb{R}\}$.

How could a basis and dimension for $V_1 \cap V_2$ be determined?
I thought of maybe setting the respective components of the two spaces equal to each other such that $2t-s = s$, $t=t$, and $t+s=s$.
Solving each of these equations it can be found that $2t=2s$, $t=t$, and $t=0 \implies s=0 \implies V_1 \cap V_2 = \{(0,0,0)\} \implies \text{dim}(V_1 \cap V_2)=0$.
This seems like it isn't right though, but I'm really uncertain.

Another possibility:
Let us first consider the set of points in $V_1\cap V_2$. Suppose $v\in V_1\cap V_2$. Then $v$ must be in the form of $(a,b,a)$ for some $a,b\in \mathbb{R}$. In addition. $(a,b,a)=(2t-s,t,t+s)$ for some $t,s\in \mathbb{R}$. Then we obtain the relation $2a=3b$. This shows that $V_1\cap V_2\subseteq \{ (a,2a/3,a) | a\in \mathbb{R} \}$.
Now we will show that $\{ (a,2a/3,a) | a\in \mathbb{R} \}\subseteq V_1\cap V_2$. Note that for $(t,s)=(2/3,1/3)$, we have that $(2t-s,t,t+s)=(1,2/3,1)$. Thus $(1,2/3,1)\in V_1\cap V_2$. Since $V_1\cap V_2$ is a subspace, we have that $\{ (a,2a/3,a) | a\in \mathbb{R} \}\subseteq V_1\cap V_2$
Hence $V_1\cap V_2= \{ (a,2a/3,a) | a\in \mathbb{R} \}$. This gives us the basis and dimension for $V_1\cap V_2$.
