if $V=\mathbb{R^3}$ and $S=\{(x,y,0)\in\mathbb{R}^3\}$ why the coset v +S represents parallels planes to xy plane? if $V=\mathbb{R^3}$ and $S=\{(x,y,0)\in\mathbb{R}^3\}$ why the coset v +S represents parallels planes to xy plane?
I know a plane $h$ is parallel to other plane  $z$ if $h=\lambda z$ how can I prove it?
I don't have any idea in how to justify it any hint?
 A: The $(x,y)$ plane $S\,$ has equation $z=0$, and and any plane parallel to $S$ has equation $z=k,\enspace k\in\mathbf R$.
Now if $v=(a,b,c)$, the coset $v+S$ comprises all points with coordinates $(x+a,y+b, c)$, so it has equation $z=c$.
Another explanation:
any two points in the coset have difference in $S$, which corresponds to the definition of an affine subspace with direction $S$.
A: Your definition of $S$ is understood as:
$$
S=\{(x,y,0)\in\mathbb{R}^3\mid x,y\in\mathbb{R}\}.
$$
This is a subspace of $\mathbb{R}^3$. Given any vector $v=(v_1,v_2,v_3)$, the coset
$$
v+S=\{v+s\mid s\in S\}.
$$
So basically you are shifting every vector in $S$ by the same vector $v$, which is the same as shifting the $x$-$y$ plane by the vector $v$. So what you get is a plane that is parallel to the $x$-$y$ plane.
To see a simple example, consider $v=(0,0,1)$. Then $v+S$ is the plane $z=1$, which is parallel to the plane $z=0$ (i.e. S).
A: Let's say we can take it as given that $v+S$ is a plane, and we define parallel planes to be ones that are either equal or disjoint. If $v\in S$ then it's easy to show $v+S=S.$ If $v\not\in S$ suppose there is a vector $w$ such that $w\in S$ and $w\in v+S$. Then for $s\in S,$ $w=v+s$. Then because $S$ is a subspace, $v=w-s\in S,$ a contradiction.
