Comparing a geometric definition of cross product to the "usual" one Could you help me with my little problem?
Given this definition of cross product:
1) $a \times b$ is perpendicular to $a$ and $b$, whenever $ a,b$ are linearly independent
2) basis $a, \ b, \ a \times b$ is positively oriented
3) $\|a \times b\| = \|a\| \cdot \|b\| \cdot \sin( \angle(a, b))$ 
How can I prove that it's equivalent with the determinant definition of cross product (we put the basis vectors $e_1, e_2, e_3$ in the first row and the coordinates of $a$ and $b$ in the second and third rows)?
 A: I don't think I've completely understood your question but here is how I would answer it
By definition, if the basis vectors are $\{\vec{i},\vec{j},\vec{k} \}$, the cross product of $\vec{a}=(a_1,a_2,a_3)$ and $\vec{b}=(b_1,b_2,b_3)$ (expressed in those basis vectors) is 
$$\vec{a}\times\vec{b}=\left| \begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \end{array}\right| $$
For the first one, just change the basis vectors to $\{\vec{e_1},\vec{e_2},\vec{e_3}\}$ such as the two vectors $\vec{a}$ and $\vec{b}$ are in the same plane given by the two vectors $\vec{e_1}$ and $\vec{e_2}$. Then they can be expressed as follows : 
$$\vec{a}=(a'_1,a'_2,0) \quad \text{and} \quad \vec{b}=(b'_1,b'_2,0)$$
$$\vec{a}\times\vec{b}=\left| \begin{array}{ccc}
\vec{e_1} & \vec{e_2} & \vec{e_3} \\
a'_1 & a'_2 & 0 \\
b'_1 & b'_2 & 0 \end{array}\right| = 
\left| \begin{array}{cc}
a'_1 & a'_2 \\
b'_1 & b'_2 \end{array}\right| \cdot \vec{e_3} $$
which is perpendicular to the plane and thus to $\vec{a}$ and $\vec{b}$.
For the second one, choose your basis vectors such as $\vec{a} = a''_1\cdot\vec{e_1}$ with $a''_1>0$. Then we have 
$$\vec{a}=(a''_1,0,0) \quad \text{and} \quad \vec{b}=(b''_1,b''_2,0)$$
$$\left| \begin{array}{cc}
a''_1 & a_0 \\
b''_1 & b''_2 \end{array}\right|=a''_1\cdot b''_2$$
As $a''_1$ is by definition positive, only the sign of $b''_2$ matters... And if you make a quick drawing of the situation, you directly see that the sign of
$$\vec{a}\times\vec{b}=\left| \begin{array}{ccc}
\vec{e_1} & \vec{e_2} & \vec{e_3} \\
a''_1 & 0 & 0 \\
b''_1 & b''_2 & 0 \end{array}\right| = 
\left| \begin{array}{cc}
a''_1 & 0 \\
b''_1 & b''_2 \end{array}\right| \cdot \vec{e_3} = a''_1 \cdot b''_2 \cdot \vec{e_3} $$ 
is positive if they are positively oriented, and negative if not. As $\vec{e_3}$ is positively oriented in $\{\vec{e_1},\vec{e_2},\vec{e_3}\}$, that's it.
For the third one I'm not really sure, but as $$|\vec{a}.\vec{b}|=||\vec{a}||\cdot||\vec{b}||\cdot|\cos(\phi)| = |a''_1\cdot b''_1|$$ and that $$||\vec{a}\times\vec{b}||=|a''_1\cdot b''_2| $$
I would do 
$$\begin{align}
(|a''_1\cdot b''_1|)^2+(|a''_1\cdot b''_2|)^2 &= (a''_1)^2\cdot [(b''_1)^2+(b''_2)^2] \\
||\vec{a}||^2\cdot||\vec{b}||^2\cdot|\cos(\phi)|^2 + (|a''_1\cdot b''_2|)^2 &= ||\vec{a}||^2\cdot||\vec{b}||^2 \\
(|a''_1\cdot b''_2|)^2 &= ||\vec{a}||^2\cdot||\vec{b}||^2 - ||\vec{a}||^2\cdot||\vec{b}||^2\cdot|\cos(\phi)|^2 \\
||\vec{a}\times\vec{b}||^2&= ||\vec{a}||^2\cdot||\vec{b}||^2\cdot|\sin(\phi)|^2 \\
||\vec{a}\times\vec{b}||&= \pm ||\vec{a}||\cdot||\vec{b}||\cdot|\sin(\phi)|
\end{align}$$
Then as previously, the sign of $a''_1 \cdot b''_2$ show you the sign of the cross product. And justifies the $\sin\phi$
$$||\vec{a}\times\vec{b}||= ||\vec{a}||\cdot||\vec{b}||\cdot\sin(\phi)$$
