Let $\mathit f$ be the mapping from $\Bbb Q$$^2$ to $\Bbb Q$$^3$ defined by:
$ \begin{pmatrix} x \\ y \\ \end{pmatrix} $ $\rightarrow$ (x+y)$ \begin{pmatrix} 1\\ 1\\ 1\\ \end{pmatrix} $
Let $e_1, e_2, e_3$ be the canonical basis vectors of $\Bbb Q$$^3$, and let $e'_1:= e_1, e'_2:= e_2, e'_3:= e_1 + e_2 + e_3.$
(A) $\mathit f$($ \begin{pmatrix} 1\\ 1\\ \end{pmatrix} $) coordinated to ($e_1, e_2, e_3$), equals $ \begin{pmatrix} 1\\ 1\\ 1\\ \end{pmatrix} $
(B) $\mathit f$($ \begin{pmatrix} 1\\ 1\\ \end{pmatrix} $) coordinated to ($e_1, e_2, e_3$), equals $ \begin{pmatrix} 0\\ 0\\ 2\\ \end{pmatrix} $
(C) The vector from $\Bbb Q$$^3$ with coordinates $ \begin{pmatrix} 1\\ 1\\ 1\\ \end{pmatrix} $ regarding ($e_1, e_2, e_3$) is included in the image of $\mathit f$.
I think only (B) is true but could someone please explain? I understand why (A) isn't true but don't really understand (B) and (C).
(A) is not true because $$f(1,1)=(2,2,2)=2e_1+2e_2+2e_3,$$ so in coordinates $(e_1,e_2,e_3)$, $f(1,1)$ is $(2,2,2)$, not $(1,1,1)$.
(B) is true because $$f(1,1)=(2,2,2)=0e_1'+0e_2'+2e_3',$$ so in coordinates $(e_1',e_2',e_3')$, $f(1,1)$ is $(0,0,2)$.
(C) is true because, for example, $$f(1,0)=(1,1,1)=1e_1+1e_2+1e_3.$$ In fact the image of $f$ is the subspace generated by $(1,1,1)$.