Planes through the origin are subspaces of $\Bbb{R}^3$ I'm reading the book Elementary Linear Algebra by Anton and Rorres, and the following has me a bit confused:
"If $\mathbf{u}$ and $\mathbf{v}$ are vectors in a plane $W$ through the origin of $\Bbb{R}^3$, then it is evident geometrically that $\mathbf{u + v}$ and $k\mathbf{u}$ also lie in the same plane $W$ for any scalar $k$ (Figure 4.2.3). Thus $W$ is closed under addition and scalar multiplication.

It says that it is evident that $k\mathbf{u}$ also lies in the same plane $W$, but I feel like if $k$ is sufficiently large enough, the vector $k\mathbf{u}$ would extend outside of the vector space $W$. Can someone explain this to me a little more clearly please?
 A: Remember that a plane is a copy of $\mathbb{R}^2$ which means it is not bounded. Here $u+v,ku$ lie in the same plane since there are both linear combinations of $u,v$. Recall that vector spaces are closed under addition and scalar multiplication. The fact that you want the plane to go through the origin is because subspaces of $\mathbb{R}^n$ contain the origin. 
A: Planes are infinite in their extents. If we're given a vector $\mathbf{u}$, then the vector $k \mathbf{u}$ is obtained just by scaling the length of $\mathbf{u}$, but keeping the same direction. So, if the vector $\mathbf{u}$ lies in a plane $W$, then any vector $k \mathbf{u}$ will also lie in this same plane, no matter how large $k$ is.
The picture in your book is misleading -- it shows the plane $W$ as a bounded parallelogram, which is wrong. I guess it's a somewhat understandable mistake -- it's difficult to draw infinite things on the page of a book. But they could have drawn a region with a fuzzy border to indicate that it's infinite.
A: You can also approach it algebraically. 
A plane in $W\subset\textbf{R}^{3}$ which passes through the origin has general equation given by
\begin{align*}
ax + by + cz = 0
\end{align*}
Thus if the vectors $\textbf{u} = (u_{1},u_{2},u_{3})\in W$ and $\textbf{v} = (v_{1},v_{2},v_{3})\in W$, then their coordinates must satisfy
\begin{align*}
\begin{cases}
au_{1} + bu_{2} + cu_{3} = 0\\
av_{1} + bv_{2} + cv{3} = 0
\end{cases}\Longrightarrow a(u_{1} + v_{1}) + b(u_{2} + v_{2}) + c(u_{3} + v_{3}) = 0 \Rightarrow \textbf{u}+\textbf{v}\in W
\end{align*}
A similar reasoning applies to $k\textbf{u}$.
