From geometric algebra, I know the definition of a bivector. Now suppose we have three 2-d vectors $a$, $b$, $c$. What is the geometrical intuition behind the following formula? (The dot product of a vector and a bivector)
$$ c\cdot(a\wedge b) $$
If I find the answer to this question, I will understand what is happening in 3-d space, when $c$ is not in the same plane of $a$ and $b$. Or what the value of $(a\wedge b) \cdot c$ is and its relation to the specified formula.
In 2D, let $a \wedge b = B i$, where $i = e_1 e_2$ is the pseudoscalar for the plane, and $B$ is the magnitude of the bivector (i.e. $B = (a \wedge b)/i$). Then
$c \cdot (a \wedge b) = c i B$.
The vector $c$ is scaled by the magnitude of the bivector $a \wedge b$, and also rotated 90 degrees. The sense of the rotation is from $e_1$ towards $e_2$ if $B > 0$ (counterclockwise with $e_1$ to the right, and $e_2$ up.)
In dimensions greater than 2, let
$c = c_\parallel + c_\perp$
where $c_\parallel \in \textrm{span}\{a, b\}$, and $c_\parallel \cdot c_\perp = 0$. None of the components of $c$ that lie outside of the plane $a \wedge b$ contribute to the vector-bivector dot product, leaving
$c \cdot (a \wedge b) = c_\parallel \cdot (a \wedge b).$
This is now a 2D relationship, as above. We have is a projection of $c$ into the plane spanned by $a$ and $b$. That projection is by the magnitude of $a \wedge b$, and then rotated by 90 degrees.