Finding the Cross Product in R2: How is it Different from R3?

[Radfire]

How do you find the cross product when you are given two vectors in R2? i know how to do it for R3

 

[Radfire]

heres the two vectors btw, not that its really important since the general rules for finding the cross product should be the same for all vectors

Two vectors are given by a= 3.0 iˆ + 5.0 ˆj and
b = 2.0 iˆ + 4.0 ˆj

 

[dacruick]

any two vectors given to you in R3 creates a plane. You can then rotate the whole system so that the two vectors now lie in the xy plane. the cross product of that will be in the z direction. So what I'm trying tell you is that the cross product vector is still in the R3 plane. Just treat it like its in R3.

 

[fzero]

The cross product in 2d is a scalar, not a vector.

$$\vec{u}\times \vec{v} = \det(\vec{u}\vec{v}) = \det\begin{pmatrix} u_x & v_x \\ u_y & v_y \end{pmatrix} = u_x v_y - u_y v_x.$$

 

[Radfire]

cheers guys, sorted

 

[HallsofIvy]

Essentially, you take the "z" coordinate of each vector to be 0. I "kind of, sort of" disagree with fzero. I would say that the cross product of two vectors in a two dimensional plane, is a vector but, since the cross product of two vectors is perpendicular to both, the cross product of two vectors in the xy-plane will NOT be in that plane. It will be perpendicular to the plane. Of course, then only the length is important which is the number fzero gives.