Question: > 
> $\mathbf{a}\times \mathbf{b}$ follows the right hand rule? Why not left hand rule? Why is it $a b \sin (x)$ times the perpendicular vector? Why is $\sin (x)$ used with the vectors but $\cos(x)$ is a scalar product?
> 
> 
> 

So why is cross product defined in the way that it is?
I am mainly interested in the right hand rule defintion too as it is out of reach?

Answer: 
As far as $\sin \theta$ and $\cos \theta$ are concerned,

Using the law of cosines, 

\begin{align}
 \|\overrightarrow{v\_2}\|^2 + \|\overrightarrow{v\_1}\|^2
 -2\|\overrightarrow{v\_2}\| \, \|\overrightarrow{v\_1}\| \cos \theta
 &= \|\overrightarrow{v\_2} - \overrightarrow{v\_1}\|^2
\\
 2x\_1 x\_2 + 2y\_1 y\_2 + 2z\_1 z\_2
 &= 2\|\overrightarrow{v\_2}\|\, \|\overrightarrow{v\_1}\| \cos \theta
\\
 \cos \theta &= \dfrac{x\_1 x\_2 + y\_1 y\_2 + z\_1 z\_2}
 {\|\overrightarrow{v\_2}\| \, \|\overrightarrow{v\_1}\|}
\\
\cos \theta &= \dfrac{\overrightarrow{v\_2} \circ \overrightarrow{v\_1}}
 {\|\overrightarrow{v\_2}\| \, \|\overrightarrow{v\_1}\|}
\end{align}

And so, $\overrightarrow{v\_1} \circ \overrightarrow{v\_2}
 = \|\overrightarrow{v\_2}\| \, \|\overrightarrow{v\_1}\| \cos \theta$

Note that $\sin \theta$ is non negative for all $0 \le \theta \le \pi$.

So

\begin{align}
 \sin^2 \theta &= 1 - \cos^2 \theta \\
 &= 1 - \dfrac{(x\_1 x\_2 + y\_1 y\_2 + z\_1 z\_2)^2}
 {\|\overrightarrow{v\_2}\|^2 \, \|\overrightarrow{v\_1}\|^2}
\\
 &= \dfrac{\|\overrightarrow{v\_2}\|^2 \, \|\overrightarrow{v\_1}\|^2
 - (x\_1 x\_2 + y\_1 y\_2 + z\_1 z\_2)^2}
 {\|\overrightarrow{v\_2}\|^2 \, \|\overrightarrow{v\_1}\|^2}
\\
 &= \dfrac{x\_2^2 y\_1^2 - 2x\_1 x\_2 y\_2 y\_1 + 
 x\_1^2 y\_2^2 + x\_2^2 z\_1^2 +x\_1^2 z\_2^2 - 2 x\_1 x\_2 z\_1 z\_2
 +y\_1^2 z\_2^2-2 y\_2 y\_1 z\_1 z\_2+y\_2^2 z\_1^2}
 {\|\overrightarrow{v\_2}\|^2 \, \|\overrightarrow{v\_1}\|^2}
\\
 &= \dfrac{(-y\_2 z\_1+y\_1 z\_2)^2+(x\_2 z\_1-x\_1 z\_2)^2+(-x\_2 y\_1+x\_1 y\_2)^2}
 {\|\overrightarrow{v\_2}\|^2 \, \|\overrightarrow{v\_1}\|^2}
\\
 &= \dfrac
 {\|\overrightarrow{v\_1} \times \overrightarrow{v\_2} \|^2}
 {\|\overrightarrow{v\_2}\|^2 \, \|\overrightarrow{v\_1}\|^2}
\end{align}

And we conclude, for purely formal reasons,
$\|\overrightarrow{v\_1} \times \overrightarrow{v\_2} \|
= \|\overrightarrow{v\_2}\| \, \|\overrightarrow{v\_1}\| \sin \theta$
As a complement to Bye\_World's illuminating answer, which defines $b \times\ c$ as the vector whose dot product with $a$ equals $\det(a, b, c)$, $\|b\|\|c\|\sin(\theta)$ is the surface of the parallelogram formed by vectors $b$ and $c$. If $a$ is a unit vector orthogonal to the $b$-and-$c$ plane, then $\det(a, b, c)$, which is the (signed) volume of the parallelepiped formed by the 3 vectors, equals the surface of the $b$-and-$c$ parallelogram. In this case, i.e.$\|a\|=1$ and angle from $a$ to $b$ (and to $c$) equal to positive $\pi/2$, one can picture $b \times c$ as $a$ times the surface of the parallelogram.