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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:

$$ \iint_{V} f(x,y) dx\ dy = \iint_{Q} f(\Phi(u,v) \Bigg| \frac{\partial{\Phi}}{\partial{u}} \times \frac{\partial{\Phi}}{\partial{v}} \Bigg| $$

$$ \Phi(u,v)=(2u \cos v,\ \ u \sin v) $$

DiscreteMath
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  • Point for you to ponder on - the cross product of two vectors gives a third vector, perpendicular to the original two vectors. Could you define a (meaningful) cross product in R^2 that would imitate that property? Moreover, there are similar threads to this one in MSE, I recommend you to try reading some of them :) – GSofer Mar 22 '19 at 20:29
  • mean is in th last integral listed above... I have a parameterization defined over R^2... wondering what this degenerates to when taking R^3 formula and applying it to R^2 case... – DiscreteMath Mar 22 '19 at 20:30
  • for example, maybe we pretend there's a 3 dimension there when we need it to perform the cross product? – DiscreteMath Mar 22 '19 at 20:31
  • You can consider the "cross product" of a single vector $(a,b)$ to be the vector $(-b,a)$ perpendicular to it. – Théophile Mar 22 '19 at 20:32
  • Alternatively, you could consider the two vectors in $\Bbb R^2$ to have $z$-co-ordinate $0$, so that their "cross product" would be $(0,0,k)$ for some $k$. But it really depends on what you're looking for. – Théophile Mar 22 '19 at 20:33

2 Answers2

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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, $\binom{a}{b}\land\binom{c}{d}=ad-bc$.

J.G.
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The cross product in 2 dimensions is a scalar give my a 2x2 determinant:

$$ (a, b) \times (c, d) = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = ad - cb $$

The cross product in 3 dimensions is a vector given by the 3x3 determinant:

$$ (a, b, c) \times (d, e, f) = \begin{vmatrix} e_x & e_y & e_z \\ a & b & c \\ d & e & f \\ \end{vmatrix} $$

E. A. Abbott describes a 2D cross product nicely in his mathematical fantasy book "Flatland":

Flatland describes life and customs of people in a 2-D world: in this universe vectors can be summed together and projected, areas are calculated, rotations are clock-wise or counter clock-wise, reflection is possible... but cross product does not exist; otherwise, 2-D inhabitants should have great fantasy to imagine a 3rd dimension to contain a vector orthogonal to their plane.

By the way, in 2-D a single scalar number is sufficient to describe a force’s moment.

$$ M(\vec{r}, \vec{F}) = r_1 F_2 - r_2 F_2 $$

With such a definition, this operation respects all algebraic properties of cross product, but the result is a scalar.

flatland

DiscreteMath
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