do the null space vectors need to be an orthogonal complement to the rowspace vector, isn't it enough to be linearly independent? say for example I have two 3 dimensional vectors that are linearly independent then I would be be able to form a 2 dimensional subspace which is a plane in the 3 dimension. the null space in my understanding is the missing vectors that are needed to represent the whole 3 dimension and conventionally is it formed by finding the orthogonal complement for both vectors in my subspace which is for sure linearly independent. 
my question is, do the nullspace vectors need to be orthogonal, I am under the assumption that any vector that is linearly independent is good enough to represnt the whole dimension ? if that is not the case, why is it conventional for them to be orthogonal is it for simplicity?
 A: Your observations can be formalized as follows:
Let $A$ be a $m \times n$ matrix over a field $K$ (e.g. $K = \mathbb{R}$ for geometric intuition). For every $1 \leq i \leq m$ let $A_i = (A_{i1}, \dotsc, A_{in})^T \in K^n$ be the transposed of the $i$-th row of $A$ and let
$$
 R = \mathrm{span}_K\{A_1, \dotsc, A_m\} \subseteq K^n.
$$
For $x,y \in K^n$ let $x \cdot y = \sum_{i=1}^n x_i y_i$ and call $x$ and $y$ orthogonal if $x \cdot y = 0$. For any subspace $U \subseteq K^n$ let
$$
 U^\perp = \{x \in K^n \mid \text{$x \cdot y = 0$ for every $y \in U$}\}
$$
be the orthogonal complement of $U$.
Then $(A \cdot x)_i= \sum_{j=1}^n A_{ij} x_j$ for all $x \in K^n$ and $1 \leq i \leq m$. Therefore $x \in \ker A$ if and only if $\sum_{j=1}^n A_{ij} x_j = 0$ for every $1 \leq i \leq m$. Notice that $\sum_{j=1}^n A_{ij} x_j = A_i \cdot x$. Thus $x \in \ker A$ if and only if $A_i \cdot x = 0$ for every $1 \leq i \leq m$, i.e. if $x$ is orthogonal to each row of $A$. It then follows that $x$ is already orthogonal to every $y \in R$ because
$$
 \left( \sum_{i=1}^m \lambda_i A_i \right) \cdot x
 = \sum_{i=1}^m \lambda \underbrace{A_i \cdot x}_{=0}
 = 0.
$$
So we have $\ker A = R^\perp$. In this way is makes sense to say that the nullspace $\ker A$ is precisely the orthogonal complement $R^\perp$ of the subspace $R$ spanned by the rows of $A$.
(These who know the dual space will recognize this as a special case of the equality $\mathrm{im}(f^*) = \ker(f)^\perp$ where $f \colon V \to W$ is some linear map and $f^* \colon W^* \to V^*$ is the induced linear map.)
