Check if such a vector $x$ exist Given:
Matrix $A$ of dimension $m \times n$
Vector $b$ of dimension $m \times 1$ 
I want to know an algorithm to check if there exist a vector $x $ of dimension $n \times 1$ such that both the following constraints are satsfied.
1) $Ax \geq 0$  (meaning all entries of the vector $Ax$ are non negative).
2) $(Ax)^Tb < 0 $ 
 A: By Farkas' lemma, either there exists a vector $x\in\mathbb{R}^n$ satisfying your conditions, i.e. $Ax \ge 0$ and $(A^\top b)^\top x < 0$, or there exists a vector $y\in\mathbb{R}^m$ satisfying $A^\top y = A^\top b$ and $y\ge 0$ (but only one of these vectors exists). Note that you can formulate the searches for $x$ and $y$ as linear programs: $x^* \in \arg\min\{0 : Ax\ge 0, ~ (A^\top b)^\top x < 0\}$ and $y^* \in\arg\min\{0 : A^\top y = A^\top b, ~ y\ge 0\}$. You can solve these linear programs in Matlab using the linprog command or CVX. For for a given $A$ and $b$, one of these linear programs will be feasible with optimal value zero, and one of them will be infeasible with optimal value $+\infty$.
Now, aside from numerically solving the aforementioned linear programs, we can gain some analytical insight as follows. Consider the constraint set $\mathcal{Y}=\{y\in\mathbb{R}^m : A^\top y = A^\top b, ~ y\ge 0\}$. If $b\ge 0$, then certainly $b\in\mathcal{Y}$, and therefore by Farkas' lemma we conclude that there does not exist a vector $x\in\mathbb{R}^n$ satisfying $Ax\ge 0$ and $(A^\top b)^\top x < 0$. Therefore, looking at whether $b\ge 0$ provides you an easy check. In the case that $b\ngeq 0$, we resort to searching for $y\ge 0$ such that $A^\top(y-b)=0$. This amounts to finding $z\in\mathcal{N}(A^\top)$ such that $z+b\ge 0$. Let $\{z_1,z_2,\dots,z_q\}\subseteq\mathbb{R}^m$ denote a basis for the null space of $A^\top$. Then our search through the nullspace amounts to finding coefficients $\alpha_i$, $i\in\{1,2,\dots,q\}$, such that
\begin{equation*}
b + \sum_{i=1}^q \alpha_i z_i = b + \begin{bmatrix}z_1 & z_2 & \cdots & z_q \end{bmatrix}\begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_q\end{bmatrix} = b+Z\alpha \ge 0,
\end{equation*}
where we define $Z$ and $\alpha$ in the obvious way. Therefore, we've reduced our search to a single affine inequality. If you can find $\alpha\in\mathbb{R}^q$ satisfying $b+Z\alpha \ge 0$ with $Z$ being a matrix defined by a basis of $\mathcal{N}(A^\top)$, then $y=b+Z\alpha$ satisfies $y\in\mathcal{Y}$, indicating that there is no solution to your constraints on $x$.
I hope this gives you a few different useful ways of looking at your feasibility problem!
