Find a basis for $U\cap V$ 
Let $$a = (0,2,3,-1)^T \quad b=(0,2,7,-2)^T \quad c = (0,-2,1,0)^T \quad u = (1,2,0,1)^T\quad v = (2,2,1,2)^T$$
Let $U= \langle a,b,c \rangle, V = \langle u,v\rangle$
Then a) find a basis for $U$ and $V$, b) find a basis for $U\cap V$.

I have solved a) and found a basis $\{a,b\}$ for $U$ and $\{u,v\}$ for V. But How can I combine this information to find a basis for $U\cap V$?
I have tried:
Let $x \in U\cap V$, then $x = \lambda_1 a+\lambda_2 b = \mu_1u+\mu_2 v$.
Solving $\lambda_1 a + \lambda_2 b - \mu_1 u - \mu_2 v = 0$ results in
$$\lambda_1 = -2r\quad \lambda_2 = r\quad \mu_1 = -2r\quad \mu_2 = r\quad (\forall r \in  \mathbb{Q})$$
This means $a,b,u, v$ are lineair dependent. But now what? I know I can remove one basisvector and this would result in set of independent basis vectors, but doesn't that break the spanning property?
I'm somewhat confused, could someone clarify, no solutions please.

Now you have two ways of expressing your vector $x$ in terms of a single arbitrary value $r$. Since $x$ was arbitrary any element of the intersection has this form. So use one of the two ways to find a fixed vector $w$ such that your arbitrary $x = rw$ for some $r$. As a check to your work so far, you should get the same $w$ from using either $a, b$ or from $u, v$, (up to a constant multiplier). Now can you figure out what the basis of $U \cup V$ is?
