Linear Algebra, Vector Space: how to find intersection of two subspaces? $${ W = Sp\{{(1,3,4),(2,5,1)\}}\\ U = Sp\{{(1,1,2),(2,2,1)}} \}$$
Find a span $$U \cap W$$
First time using Math latex, pretty hard.
 A: Even though this is years after the question, posting following for completeness. This is very similar to solution by Pavan Karjol.
Let $P_u$ and $P_v$ be two projection matrices onto the two sub-spaces spanned by the columns of $U$ and $V$. These are calculated as $P_u=U(U^T U)^{-1} U^T$ and $P_v =V(V^T V)^{-1} V^T.$
Let $z$ be a vector that lies in intersection of these two sub-spaces. Since  $z$ is on both sub-spaces, projecting it onto each subspace doesn't change the vector.
Therefore, $z = P_uz$ and $z = P_vz$.
Substituting one in the other gives $z = P_uP_vz$ therefore, $0 = (P_uP_v-I)z$ .
Meaning, $z$ is in the null space of $M=(P_uP_v - I)$. Perform the eigenvalue decomposition on $M$ and select the eigenvectors corresponding to zero or nearly zero eigenvalues and that forms an orthogonal basis for the intersection of given sub-spaces.
A: Hint: Find the normal vector to each span. Then a vector is in the span if and only if the dot-product with the normal vector is $0$. The formula for the normal vector of a 2-d span in 3-d is the cross product of your two spanning vectors. So for your two spans, you get two normal vectors, say $u,v$, and then you want to solve the system $[u^T \,; \,  v^T]x = 0$ for $x$, which can be done with row elimination, or simply computing the cross product of $u$ and $v$.
A: The answer is Sp(1, 1, -10).
Call the vectors a = (1,1,2); b = (2,2,1); c = (1, 3, 4); d = (2, 5, 1) then any vector in $U \bigcap W$ must be a linear combination of a and b, and at the same time a linear combination of c and d.
This gives you the following simultaneous equations:
$\alpha$(1,1,2) + $\beta$(2,2,1) = $\gamma$(1,3,4) + $\delta$(2,5,1). Solve them by the method of your choice to get a parametric equation for a line.
A: Let $U$ and $V$ be two sub spaces(in matrix form: columns as basis vectors). Let $z$ be a vector that lies in intersection of these two sub spaces. Then $\exists$ two coeff vectors $x,y$ such that 
\begin{align*}
z = Ux &= Vy \\
Ux &= Vy \\
U^T Ux &= U^T Vy \\
x &= (U^T U)^{-1} U^TVy \quad \text{ and similarly } \quad y = (V^T V)^{-1} V^TUx  \\ 
\text{Thus} \quad x &= (U^T U)^{-1} U^TV(V^T V)^{-1} V^TUx \\
x &= \hat{M}x, 
\end{align*}
where, $$\hat{M} =(U^T U)^{-1} U^TV(V^T V)^{-1} V^TU $$ 
We can see that $x$ is the Eigen vector of $\hat{M}$ corresponding to Eigen value $1$. Thus required basis is the set of independent vectors such that
$$\{Ux:\hat{M}x = x\}$$
In another way, let $\hat{M_1} = \Big(U(U^T U)^{-1} U^T \Big)  \Big(V(V^T V)^{-1} V^T \Big) = P_UP_V$. The required basis is the set of independent vectors such that 
$$\{s:\hat{M_1}s = s\}$$
Geometrically $P_U=U(U^T U)^{-1} U^T$ and $P_V =V(V^T V)^{-1} V^T $ are projection matrices onto the sub spaces $U$ and $V$ respectively. So we can see that the basis elements are those independent vectors, which remain unchanged after two  projections, corresponding to the given two sub spaces.
