Originally Posted by

**Bingk** I think this might help.

Your vector e = [1,1] can be written as i + j. A generalization is if your vector e = [a,b] then it can be written as ai + bj.

Your vector e is written as if it was in 2D. But, you can write this as if it was in 3D (or more).

So, in 3D, your e would be [1,1,0] or i + j + 0k. In general, if your e = [a,b] then it would be [a,b,0] in 3D or ai +bj +0k.

Now, as mentioned before, i,j, and k are unit vectors pointing in the positive x-,y-, and z-axis respectively. This means that your i,j, and k already satisfy the right-hand rule. For emphasis, i,j, and k are vectors, meaning that i = [1,0,0], j = [0,1,0], and k = [0,0,1]. I emphasize this because this basically answers your question, e x k is actually e x [0,0,1] (in this case [1,1,0] x [0,0,1]), and this also shows that the i,j, and k were chosen. If you go to more advanced math, you can actually define your own i,j, and k (which will most likely roughly satisfy the right hand rule). Sometimes the i,j,k are referred to as the components of your vector.

So, your top row is basically your components (which has to be listed in the order i,j,k, otherwise you mess around with the orientation).

Since you are doing exk, your second row is the components of e (listed in the i,j,k order), and your third row is the components of k (again, listed in i,j,k order).

If you were told to do kxe, then your second row would be the components of k, and the third row would be the components of e, and this would be the same as ex(-k) (i.e. e x [0,0,-1])

A general example:

If s = [a,b,c] and t = [d,e,f] then

s x t =

i j k

a b c

d e f

and t x s =

i j k

d e f

a b c

That's how you would set up the determinant