Proving Vector Cross Product Properties in ℝ3?

[Jow]

If e1 and e2 are vectors in ℝ3 show that e1 x e2 = e3, e2 x e3 = e1 and e3 x e1 = e2. I have tried to prove this but I can't get it.

My attempt:
Step 1: [a1, a2, a3] x [b1, b2, b3] = [a2b3-a3b2, a3b1-a1b3, a1b2-a2b1]
Step 2: [b1, b2, b3] x [a2b3-b2a3, a3b1-a1b3, a1b2-a2b1] = [b2(a1b2-a2b1)-b3(a3b1-a1b3), b3(a2b3-a3b2)-b1(a1b2-a2b1), b1(a3b1-a1b3)-b2(a2b3-b2a3)] ... nothing cancels and I do not end up with [a1, a2, a3], which I should shouldn't I?
I also try this with actual numbers but it still doesn't work. Am I doing something completely wrong? I understand, geometrically, why this should happen because the cross product is orthogonal to the two vectors. Am I doing something wrong or is my textbook wrong? (I am pretty sure the former is the right answer to that question).

 

[Dick]

If e1 and e2 are vectors in ℝ3 show that e1 x e2 = e3, e2 x e3 = e1 and e3 x e1 = e2. I have tried to prove this but I can't get it.

My attempt:
Step 1: [a1, a2, a3] x [b1, b2, b3] = [a2b3-a3b2, a3b1-a1b3, a1b2-a2b1]
Step 2: [b1, b2, b3] x [a2b3-b2a3, a3b1-a1b3, a1b2-a2b1] = [b2(a1b2-a2b1)-b3(a3b1-a1b3), b3(a2b3-a3b2)-b1(a1b2-a2b1), b1(a3b1-a1b3)-b2(a2b3-b2a3)] ... nothing cancels and I do not end up with [a1, a2, a3], which I should shouldn't I?
I also try this with actual numbers but it still doesn't work. Am I doing something completely wrong? I understand, geometrically, why this should happen because the cross product is orthogonal to the two vectors. Am I doing something wrong or is my textbook wrong? (I am pretty sure the former is the right answer to that question).

I think they mean e1=[1,0,0], e2=[0,1,0] and e3=[0,0,1]. They are special vectors. It's not true for any three vectors.

 

[Jow]

Oh good. I thought I was missing something really important. I must have just misread the question. Thanks.

 

[HallsofIvy]

In fact, one way of defining the "cross product" is to assert that e_1\times e_2= e_3, e_2\times e_3= e_1, e_3\times e_1= e_2, that the product is "bilinear" and anti-commutative.