Axb=-bxa, why?

by Starwatcher16
Tags: cross product, quaternions
 P: 53 axb=-bxa, why?
 P: 9 a or b = 0?
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P: 40,649
 Quote by Starwatcher16 axb=-bxa, why?
Because of the Right Hand Rule.

P: 53
Axb=-bxa, why?

 Quote by berkeman Because of the Right Hand Rule.
I have two vectors perpendicular to each other in the xy plane, if I take their cross product, I get another vector in the z plane.

I don't understand why one should be going +z, as opposed to the other way.
P: 614
 Quote by Starwatcher16 I have two vectors perpendicular to each other in the xy plane, if I take their cross product, I get another vector in the z plane. I don't understand why one should be going +z, as opposed to the other way.
Because there is no logical way they should go so we just defined AxB as -BxA and it follows the right hand rule in a right handed coordinate system. The cross product is a vector perpendicular to both the crossed vectors and its length is the area of the parallelogram you get, the deal is that either up or down works for this definition so we just have to define either AxB or BxA as up and then the other down, which is why AxB=-BxA.
 P: 50 compute it using determinants and you should see why
 P: 290 $$|A\times B| = |A||B|\sin{(\theta_B-\theta_A)}$$ $$|B\times A| = |A||B|\sin{(\theta_A-\theta_B)}=|A||B|\sin{-(\theta_B-\theta_A)}=-|A||B|\sin{(\theta_B-\theta_A)} = - |A\times B|$$
 P: 490 Uhhh, qntty... I believe the magnitudes of the two vectors are the same. ;(((
 P: 824 Why? Do the math with the definition of the cross product using the epsilon tensor which is antisymmetrical under the exchange of two indices: $$(a \times b)_i &=& \sum_{j=1}^3 \sum_{k=1}^3 \epsilon_{ijk}a_jb_k \\ &=& \sum_{j=1}^3 \sum_{k=1}^3 - \epsilon_{ikj}a_jb_k$$ renaming of indices changing summation order: $$&=& \sum_{j=1}^3 \sum_{k=1}^3 - \epsilon_{ijk}a_k b_j \\ &=& - (b \times a)_i$$