Linear Algebra:P1 - Dim(V), linear independence In $\mathbb{R}^3 $ with $  \underline{N}=
\begin{pmatrix} 
   1\\ 
   2\\
   1
\end{pmatrix} $

and the subspace $V={\underline{N}^\bot}$.

(a1) Are the following vectors linearly independent?

(a2)Define the dimension of V.

(a3) Show that the following vectors are vectors of/in V.


(a1)

$\underline{x}_1=
\begin{pmatrix} 
   0\\ 
   1/2\\
   -1
\end{pmatrix}$ , 
$\underline{x}_2=
\begin{pmatrix} 
   1\\ 
   0\\
   -1
\end{pmatrix}$
 , 
$\underline{x}_3=
\begin{pmatrix} 
   1\\ 
   -1\\
   1
\end{pmatrix}$


$\begin{vmatrix}
 0   &  1  &  1\\
 1/2 &  0  & -1 \\
 -1  & -1  &  1
\end{vmatrix}=-1/2+1-1/2=0$

$\Rightarrow$ these vectors are linearly dependent.


(a2)

$\begin{vmatrix}
 0   & 1\\
 1/2 & 0
\end{vmatrix}=-1/2 \ne 0$

$\begin{vmatrix}
 1  & 1\\
 0 & -1
\end{vmatrix}=-1\ne 0$

$\Rightarrow  Rg(M)=Rg
\begin{pmatrix}
 0   &  1  &  1\\
 1/2 &  0  & -1 \\
 -1  & -1  &  1
\end{pmatrix}=2$ 

so that the Dimension of V is 2.

 
Is (a1) and (a2) done in this way correct?

For (a3)I don't know how to show that,
can I use the parametric definition of a plane:

$\underline{N}=\underline{x}_1+r\underline{x}_2+s\underline{x_3}$
 and if yes, does the order of the vectors matter?and how?
 A: For (a3), to prove the vectors are in $V$, take the dot product of these vectors with $N$ and the dot product should be zero.
A: a1 is correct. You calculated the determinant of the matrix containing the 3 vectors and since it is 0 you concluded correctly that they are linearly dependant. Another way to see it is to see that $2x_1 + (-1)x_2 + x_3 = 0$, which gives you a non-trivial linear combination of the 3 vectors giving 0, thus saying they are dependant.
Like I said in my comment I think you have the correct answer to a2 but with an incorrect calculation. Actually you calculated the dimension of the vector space spanned by $x_1, x_2, x_3$. But this might not be the same vector space as $V$ as far as you know. 
Probably, the goal of exercise a2 was for you to find that dimension of $V$ was 2 because it is the orthogonal complement of a 1 dimensional vector space in $\mathbb{R}^3$.
[Edit: and here I mean using some argument seen in the course / book. No calculations is needed if you have some theorems on dimensions of orthogonal vector spaces in your book / course.]
Finally, like I said in another of my comments, for a3 you just need to use the definition of orthogonal complement. Other answers show you how to.
A: Yes, (a1) and (a2) are correct, well, if I understand well, provided only that (a3) is true. 
However, we can also argue by simpler arguments:
For (a1), note that $\underline x_3=\underline x_2-2\cdot\underline x_1$, so these are linearly dependent. However, $\underline x_1$ and $\underline x_2$ are not parallel (then one would be a scalar times the other), so their rank must be at least $2$, so it is $2$.
For (a3), we need only to check that $\underline x_1$ and $\underline x_2$ are orthogonal to $\underline N$ (i.e. that they are in ${\underline N}^\perp$), then by the above equality,. it will also follow for $\underline x_3$. For this, nothing but the dot product is needed to be computed, and has to be zero:
$$\underline N\cdot\underline x_1=1\cdot 0+2\cdot 1/2+1\cdot(-1)=1-1=0\,,$$
and verify similarly that $\underline N\cdot\underline x_2=0$. Since, in 3d, the orthogonal of a nonzero vector has dimension $2$, we have that $\underline x_1,\underline x_2$ forms a basis of $\underline N^\perp$ (actually, any two vectors of $\underline x_1,\underline x_2,\underline x_3$ do so).
