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Let $V\subset \mathbb{R}^3$ be the subspace created by $\{(1,1,0),(0,2,0)\}$. Write the cartesian equations of $V$.

Since the two vectors are linearly independent they are a basis of the 2D-space $V$. They thus create a plane. Moreover, since $V$ is a vector space, the plane has to pass through the origin (else it would be an affine space). In order to write the plane equation in cartesian form, I need the vector normal to it. Thus I can use the cross product, since I know two vectors of the plane which are linealrly independent. \begin{equation} \textbf{v}_{n}=(1,1,0)\wedge(0,2,0)=\begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k}\\1&1&0\\0&2&0\end{vmatrix}=(0,0,2) \end{equation} Then the wanted equation is: $\pi: 2z=0$.

I'm not convinced at all though. Can you explain to me where I messed up and how to get the right answer? Thank you all

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    $z=0$ is indeed the (unique!) plane containing both vectors. It is quite easy to check that this is true, since the third coordinate of both is $0$. Hence, the answer is correct. – 2017-02-03

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The span of $(1,1,0)$ and $(0,2,0)$ consists of the $x$-$y$-plane, or in other words: all vectors with $z = 0$. So your equation is fine.

You can test if a linear combination $u$ is part of the plane, by checking if it fulfills the normal equation of the plane $n \cdot u = 0$: $$ n \cdot (s (1,1,0) + t (0,2,0)) = s ((0,0,2) \cdot (1,1,0)) + t ((0,0,2) \cdot (0,2,0)) = 0 \quad (s, t \in \mathbb{R}) $$

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    Ok, Thank you! The thing is that I tried and solve it in another way, precisely by finding the equation of the plane in parametric form, i.e. $x=t, y =t+2s, z=0$ from wich I get the cartesian eq. $x-y+2s=0$ but I was not sure at how to motivate $s=0$ – 2017-02-03
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    A Cartesian equation of the plane seems to be $a x + b y + c z + d = $, where $a,b,c,d$ are constants. Your equation $2 z = 0$ fits that scheme (while $z = 0$ would do as well). What you tried is having a vector $(t, t + 2s, 0)$ for some choice of $s$ and $t$. It describes the plane only as $\{ (t,t+2s,0) \mid s, t \in \mathbb{R}\}$. – 2017-02-03