Show that the xy plane W=(x,y,0) in R3 is generated by (i) u=[120] and v=[010]...

Let ${\displaystyle (x,y,0)}$ be any element from W. For a and b in ${\displaystyle \mathbb{R}}$, we have
${\displaystyle (x,y,0)=a(1,2,0)+b(0,1,0)}$
${\displaystyle =(a,2a+b,0)}$
Comparing both side we get
${\displaystyle x=a}$ and ${\displaystyle y=2a+b}$
Now from ${\displaystyle y=2a+b}$
${\displaystyle b=y-2a}$
${\displaystyle =-y-2x}$
Therfore, for any ${\displaystyle (x,y,0)}$ in W, we can write
${\displaystyle (x,y,0)=x(1,2,0)+(y-2x)(0,1,0)}$
$\Rightarrow (x,y,0)\{(1,2,0),(0,1,0)\}$
Therefore, W is generated by u and v.
Let ${\displaystyle (x,y,0)}$ be any element from W. For a and b in ${\displaystyle \mathbb{R}}$, we have
${\displaystyle (x,y,0)=a(2,-1,0)+b(1,3,0)}$
${\displaystyle =(2a+b,-a+3b,0)}$
Comparing both side we get
${\displaystyle x=2a+b}$ ...(1)
and
${\displaystyle y=-a+3b}$ ...(2)
Now we solve equationn (1) and (2) to find the value of a and b. From (1) and (2), we get
${\displaystyle x+2y=2a+b-2a+6b=7b}$
${\displaystyle \Rightarrow b=\frac{x+2y}{7}}$
Now from (2)
${\displaystyle a=y+3b}$
${\displaystyle =y+3\left(\frac{x+2y}{7}\right)}$
${\displaystyle =y+\frac{3x+6y}{7}}$
${\displaystyle =\frac{7y+3x+6y}{7}}$
${\displaystyle =\frac{3x+13y}{7}}$
${\displaystyle \Rightarrow a=\frac{3x+13y}{7}}$
Therfore, for any ${\displaystyle (x,y,0)}$ in W, we can write
${\displaystyle (x,y,0)=a(2,-1,0)+b(1,3,0)}$
${\displaystyle =\frac{3x+13y}{7}(2,-1,0)+\frac{x+2y}{7}(1,3,0)}$
$\Rightarrow (x,y,0)\{(2,-1,0),(1,3,0)\}$
Therefore, W is generated by u and v.