The span of (1, 1, 2) and (2, 2, 1) is the set of all vectors of the form a(1, 1, 2)+ b(2, 2, 1)= (a+ 2b, a+2b, 2a+ b). We can write that as (x, x, y).
The span of (1, 3, 4) and (2, 5, 1) is the set of all vectors of the form p(1, 3, 4)+ q(2, 5, 1)= (p+ 2q, 3p+ 5q, 4p+ q). If we write u= p+ 2q and v= 3p+ 5q then, subtracting v- 3u= (3p+ 5q)- (3p+ 6q)= -q so that q= 3u- v. Then u= p+ 2q= p+ 5u- 2v. p= 2v- 4u. 4p+ q= 8u- 15v+ 3u- v= 11u- 16v. So (p+3q, 3p+5q, 4p+q)= (u, v, 11u- 16v).
Vectors in both sets must have u= v so that 11u- 16v= -5u. That is, vectors in U that are also in W must have (x, x, y)= (u, u, -5u). That is the line y= -5x. Removing (cutting) U from W leaves all (x, x, y) such that y is NOT -5x.
Yes, I thought initially that "W cutting U" meant the intersection of W and U but then romsek asked "does W cutting U mean {x:x∈U∧x∉W}?" (That is, W with U removed) and the OP, yossa, responded "yes, my friend". The first is, as you say, a one dimensional subspace of W while the second is not a subspace.
Yes, I thought initially that "W cutting U" meant the intersection of W and U but then romsek asked "does W cutting U mean {x:x∈U∧x∉W}?" (That is, U with its intersection with W removed) and the OP, yossa, responded "yes, my friend". The first is, as you say, a one dimensional subspace of W while the second is not a subspace.