Why is cross product defined in the way that it is? 
$\mathbf{a}\times \mathbf{b}$ 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?

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,
$\|\overrightarrow{v_1} \times \overrightarrow{v_2} \|
= \|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\| \sin \theta$
