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.

Here is one way:
The idea is to write $V_k$ as $\ker u_k^T$ for some $u_k$,
then
$v \in V_1 \cap V_2$ iff  $v \in \ker \begin{bmatrix} u_1^T \\ u_2^T \end{bmatrix}$. If $u_k$ are linearly independent, then if $u_3\neq 0$  is
orthogonal to $u_1,u_2$, we see that $V_1 \cap V_2 = \operatorname{sp} \{ u_3 \}$.
In the following we use the fact that $(\ker C)^\bot = {\cal R} C^T$.
It should be straightforward to show that $\dim V_1 = \dim V_2 = 2$, 
Hence $\dim V_1^\bot = \dim V_2^\bot = 1$.
Suppose $(x,y,z)^T \in V_1^\bot$, then we must 
have $t(2x+y+z) + s(z-x) = 0$ for all $s,t$. Hence $z=x$ and $y=-3x$. Letting
$x=1$ gives $u_1=(1,-3,1)^T$ and hence $V_1^\bot = \operatorname{sp} \{u_1\}$, and so $V_1 = \ker u_1^T$.
Suppose $(x,y,z)^T \in V_2^\bot$, then we must have $s(x+z)+ty=0$ for all $s,t$. Hence $y=0, z=-x$. Letting $x=1$ gives $u_2=(1, 0 ,-1)^T$
and hence $V_1^\bot = \operatorname{sp} \{u_2\}$, and so $V_2 = \ker u_2^T$.
Hence $v \in V_1 \cap V_2$ iff $v \in \ker u_1^T$ and $v \in \ker u_2^T$ iff
$u \in \ker \begin{bmatrix} u_1^T \\ u_2^T \end{bmatrix}$.
Solving 
$\begin{bmatrix} u_1^T \\ u_2^T \end{bmatrix} (x,y,z)^T = 0$ gives
$z=x, y = {1 \over 3} 2x$. Choosing $x=3$ gives $u_3 = (3,2,3)^T$.
Hence $v \in V_1 \cap V_2$ iff $v \in \operatorname{sp} \{ u_3 \}$.
