Is there any continuous surjective map from $\mathbb{R}^3$ to $\mathbb{R}^3$ s.t. for every $x$ in $\mathbb{R}^3$ $f(x) \cdot x= 0$? Is there any continuous surjective map from $\mathbb{R}^3$ to $\mathbb{R}^3$ such that for every $x$ in $\mathbb{R}^3$,  $f(x)  \cdot x=0$?
What about i replace $\mathbb{R}^3$ to $\mathbb{R}^4$ ?
 A: For  $\mathbb{R}^3$ the answer is "yes".
Let us say that a continuous map $f : M \to \mathbb{R}^3$ defined on a subset $M \subset \mathbb{R}^3$ is admissible if $f(x) \cdot x = 0$ for all $x \in M$.
Define the spheric shell  $S^2(r_1,r_2) = \lbrace p \in \mathbb{R}^3 \mid r_1 \le \lVert p \rVert \le r_2 \rbrace$ and $S^2(r) = S^2(r,r)$ = sphere with radius $r$. Note that $S^2(0,r)$ = closed ball with center $0$ and radius $r$.
The reason for the affirmative answer is that we can find a cover $P_i$, $i \in \mathbb{N}$, of $\mathbb{R}^3$ such that each $P_i$ is contained in the image of an admissible map defined on a suitable spheric shell $S_i$.
If we have admissible $f_i : S^2(r_i) \to  \mathbb{R}^3$, then
$$L(f_1,f_2) : S^2(r_1,r_2) \to \mathbb{R}^3, L(f_1,f_2)(p) = \frac{r_2 - \lVert p \rVert}{r_2 - r_1}f_1(r_1 \frac{p}{\lVert p \rVert}) + \frac{\lVert p \rVert - r_1}{r_2 - r_1}f_2(r_2 \frac{p}{\lVert p \rVert})$$
is easily seen to be admissible. In other words, each admissible map defined on the boundary of $S^2(r_1,r_2)$ has an admissible extension to $S^2(r_1,r_2)$.
Let $H_z^{\pm} = \lbrace (x,y,z) \mid sign(z) = \pm 1 \rbrace$ the open halfspaces above and below the $x$-$y$-plane. $H_x^{\pm}, H_y^{\pm}$ are defined similarly. 
Now let $S$ denote $S^2$ with north and south pole $(0,0,\pm 1)$ removed. Define a function $\varphi : S \to S^2$ as follows: For each $x \in S$ let $G_x$ be the great circle through $x$ and north and south pole. Let $\varphi(x)$ be the unit tangent vector at the manifold $G_x$ in the point $x$ going in downward direction. $\varphi$ is admissible, but cannot be extended to $S^2$. Define $\phi_z^- : S^2 \to  \mathbb{R}^3$, $\phi_1(0,0,\pm 1) = 0$, $\phi_z^-(x,y,z) = (1-z^2)\varphi(x,y,z)$ for $z \in (-1,1)$. Then $\phi_z^-$ is  admissible. For each $p = (x,y,z) \in S^2$ with $z < 0$ the line through $p$ and $0$ contains a non-zero point of $K_z^- = \phi_z^-(S^2)$.
Define $K_z^-(n) = \lbrace t p \mid p \in K_z^-, t \in [n-1,n] \rbrace$ for $n \in \mathbb{N}$. Then $\bigcup_{n=1}^\infty K_z^-(n) = H_z^- \cup \lbrace 0 \rbrace$.
Similarly we obtain an admissible $\phi_z^+$ and sets $K_z^+(n) \subset H_z^+ \cup \lbrace 0 \rbrace$ with analogous properties. Do the same to obtain the obvious $\phi_x^\pm, \phi_y^\pm$ and  $K_x^\pm(n), K_y^\pm(n)$.
On $S^2(12(n-1),12(n-1)+1)$ define $\phi_z^-(n)(p) = (\lVert p \rVert - 11(n-1)) \phi_z^-(\frac{p}{\lVert p \rVert})$. This is an admissible map such that $K_z^-(n) \subset \phi_z^-(n)(S^2(12(n-1),12(n-1)+1))$. Note that for $n = 1$ we obtain $S^2(12(n-1),12(n-1)+1) = S^2(0,1)$ = closed ball with center $0$ and radius $1$.
On the spheric shells $S^2(12(n-1) + 2,12(n-1)+ 3)$ define $\phi_z^+(n)$ similarly based on $\phi_z^+$. Do the same on $S^2(12(n-1) + 4,12(n-1)+ 5)$ based on $\phi_x^-$, on $S^2(12(n-1) + 6,12(n-1)+ 7)$ based on $\phi_x^+$, on $S^2(12(n-1) + 8,12(n-1)+ 9)$ based on $\phi_y^-$, on $S^2(12(n-1) + 10,12(n-1) + 11)$ based on $\phi_y^+$.
This gives an admissible map $f: \bigcup_{k=1}^\infty S^2(2k-2,2k-1) \to \mathbb{R}^3$ which is surjective by construction.
On the intermediate spheric shells $S^2(2k-1,2k)$, $k \in \mathbb{N}$, use admissible extensions to get an admissible surjective map $F : \mathbb{R}^3 \to \mathbb{R}^3$.
