Intersection of two subspaces is the span of the vector b Let the vectors $u_1=\begin{bmatrix}1\\2\\3\end{bmatrix}$, $u_2=\begin{bmatrix}2\\3\\4\end{bmatrix}$, $v_1=\begin{bmatrix}1\\1\\2\end{bmatrix}$, $v_2=\begin{bmatrix}2\\2\\3\end{bmatrix}$ $\in \mathbb{R}^3$. 
Let the subspaces be $U=span(u_1,u_2)$, $V=span(v_1,v_2)$
1) Show that $U\cap V=span(b)$, where $b=\begin{bmatrix}1\\1\\1\end{bmatrix}$.
My approach:
So U is the span of the two vectors $u_1, u_2$, which is the linear combination of the vectors. So they span a plane in $\mathbb{R}^3$. Which mean U is the set of vectors on the following form:
$U=\{u=\begin{bmatrix}\alpha'+2\beta'\\2\alpha'+3\beta'\\3\alpha'+4\beta'\end{bmatrix} | \alpha,\beta\in \mathbb{R} \}$,
The same approach goes for V, so V is the set of following vectors:
$V=\{v=\begin{bmatrix}\alpha+2\beta\\\alpha+2\beta\\2\alpha+3\beta\end{bmatrix} | \alpha,\beta\in \mathbb{R} \}$,
So $U\cap V$ is the set of elements contained in both U and V. In that case I should find the $\alpha',\beta',\alpha,\beta$ where $u=v$.
Doing this creating a total matrix: $(u-v|0)$ I get the following equations:
$\alpha'+\alpha$=0, $\beta'-\beta=0$ and $\alpha+\beta=0$. 
So honestly, now I'm a bit lost. I was hoping putting these equations back in the vectors v and u, that i would get some scalar ($\alpha',\beta',\alpha,\beta$) times the vector b: $k\begin{bmatrix}1\\1\\1\end{bmatrix}$, and then i could conclude that span was the linear combination of that one vector. But that only works for the vector v.
Could somebody help me telling me why my approach is wrong, and what the best way of doing this?
 A: Here's a more algorithmic approach.
We wish to find a basis of the intersection of $U=\operatorname{Span}\{u_1, u_2\}$ and $V=\operatorname{Span}\{v_1, v_2\}$ where
\begin{align*}
u_1 &= \left\langle1,\,2,\,3\right\rangle &
u_2 &= \left\langle2,\,3,\,4\right\rangle &
v_1 &= \left\langle1,\,1,\,2\right\rangle &
v_2 &= \left\langle2,\,2,\,3\right\rangle
\end{align*}
Let's start by inserting our vectors into the columns of a matrix
$$
\left[
\begin{array}{cc|cc}
u_1 & u_1 & v_1 & v_2
\end{array}\right]
=
\left[\begin{array}{rr|rr}
1 & 2 & 1 & 2 \\
2 & 3 & 1 & 2 \\
3 & 4 & 2 & 3
\end{array}\right]
$$
The nullity of this matrix is the dimension of $U\cap V$. 
Row reducing this matrix gives
$$
\operatorname{rref}\left[
\begin{array}{cc|cc}
u_1 & u_1 & v_1 & v_2
\end{array}\right]=\left[\begin{array}{rr|rr}
1 & 0 & 0 & -1 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1
\end{array}\right]
$$
We immediately see that $\dim(U\cap V)=1$. Moreover, this reduced form tells us that 
$$
u_1-u_2-v_1+v_2=0
$$
which is equivalent to
$$
u_1-u_2=v_1-v_2=\left\langle-1,\,-1,\,-1\right\rangle
$$
This gives $U\cap V=\operatorname{Span}\{\left\langle-1,\,-1,\,-1\right\rangle\}$.
A: Never mind i figured it out. I made a mistake reading my equations from the total matrix. Instead of $\alpha' +\alpha=0$ It should be $\alpha' +\beta=0$
