The math book i'm using states that the cross product for two vectors is defined over $R^3$:
$$u = (a,b,c)$$
$$v = (d,e,f)$$
is:
$$u \times v = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & b & c \\ d & e & f \\ \end{vmatrix} $$
and the direction of the resultant is determined by curling fingers from vector v to u with thumb pointing in direction of the cross product of u x v.
Out of curiosity, what's the cross product if u and v are defined over $R^2$ instead of $R^3$ instead:
$$u = (a,b)$$
$$v = (d,e)$$
Is there a "degenerate" case for the cross product of $R^2$ instead $R^3$? like this is some type of 2x2 determinant instead?
for instance if had a parameterization:
$$\Phi(u,\ v) = (\ f(u),\ \ g(v)\ )$$
and needed to calculate in $R^2$:
$$ D = \Bigg| \frac{\partial{\Phi}}{\partial{u}} \times \frac{\partial{\Phi}}{\partial{v}} \Bigg| $$
There are plenty of examples in the book for calculating the determinate D in $R^3$ but none at all for $R^2$ case.
As in:
$$ \iint_{V} f(x,y) dx\ dy = \iint_{Q} f(\Phi(u,v) \Bigg| \frac{\partial{\Phi}}{\partial{u}} \times \frac{\partial{\Phi}}{\partial{v}} \Bigg| $$
$$ \Phi(u,v)=(2u \cos v,\ \ u \sin v) $$
In $n$ dimensions, the Levi-Civita symbol has $n$ indices, two of which contract with those of two vectors whose wedge product is sought in the geometry's exterior algebra. This obtains an object with $n-2$ indices, so whereas we get a familiar vector if $n=3$, we get a scalar if $n=2$. In particular, $\binom{a}{b}\land\binom{c}{d}=ad-bc$.