Linear algebra - Coordinate Systems I'm preparing for an upcoming Linear Algebra exam, and I have come across a question that goes as follows: Let U = {(s, s − t, 2s + 3t)}, where s and t are any real numbers. Find the coordinates of x = (3, 4, 3) relative to the basis B if x is in U . Sketch the set U in the xyz-coordinate system.
It seems that in order to solve this problem, I'll have to find the basis B first! how do I find that as well?
The teacher hae barely covered the coordinate systems and said she will less likely include anything from the section on the exam, but I still want to be safe. The book isn't of much help. It explains the topic but doesn't give any examples.
Another part of the question ask about proving that U is a subspace of R3, but I was able to figure that one out on my own. I'd appreciate if someone could show me how to go about solving the question above.
 A: What is probably meant here for $B$ (you could also choose a different basis) is taking $s=1, t=0$ for the first basis vector and $t=1, s=0$ for the second one. Then you have $U=\{s(1,1,2)+t(0,-1,3)|s,t\in \mathbb{R}\}$ (This also gives an easy proof that this is a vector space). Now to present $x$ in this particular basis, you' have to solve the linear system of equation that is lurking in the exercise, namely $s=3, s-t=4, 2s+3t=3$.
A: This might be an answer, depending on how one interprets the phrase "basis $B$", which is undefined in the question as stated:
Note that $(s, s - t, 2s + 3t) = s(1, 1, 2) + t(0, -1, 3)$.  Taking $s = 1$, $t = 0$ shows that $(1, 1, 2) \in U$.  Likewise, taking $s = 0$, $t = 1$ shows $(0, -1, 3) \in U$ as well.  Incidentally, the vectors $(1, 1, 2)$ and $(0, -1, 3)$ are clearly linearly independent; to see this in detail, note that if  $(s, s - t, 2s + 3t) = s(1, 1, 2) + t(0, -1, 3) = (0,0,0)$, then we must obviously have $s = 0$, whence $s - t = -t = 0$ as well.  Assuming $B$ refers to the basis $(1, 1, 2)$, $(0, -1, 3)$ of $U$, it is easy to work out the values of $s$ and $t$ corresponding to $x$:  setting $(s, s - t, 2s + 3t) = (3, 4, 3)$, we see that we must have
$s = 3$ whence $t = -1$ follows from $s - t = 4$.  These check against $2s + 3t = 3$, as the reader may easily verify.  The desired coordinates for $x$ in the basis $B$ are thus $(3, -1)$.
Think that about covers it, if my assumption about $B$ is correct.  
Can't provide a graphic, but one is easily constructed noting that the vectors $(1, 1, 2)$, $(0, -1, 3)$ span $U$ in $R^3$ (the "$xyz$" coordinate system).
