Register to reply

Axb=-bxa, why?

by Starwatcher16
Tags: cross product, quaternions
Share this thread:
byee614
#2
Jun4-09, 01:22 PM
P: 9
a or b = 0?
berkeman
#3
Jun4-09, 01:22 PM
Mentor
berkeman's Avatar
P: 41,075
Quote Quote by Starwatcher16 View Post
axb=-bxa, why?
Because of the Right Hand Rule.

Starwatcher16
#4
Jun4-09, 01:58 PM
P: 53
Axb=-bxa, why?

Quote Quote by berkeman View Post
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.
Klockan3
#5
Jun4-09, 02:09 PM
P: 614
Quote Quote by Starwatcher16 View Post
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.
m00npirate
#6
Jun4-09, 02:09 PM
P: 50
compute it using determinants and you should see why
qntty
#7
Jun4-09, 04:39 PM
qntty's Avatar
P: 290
[tex]|A\times B| = |A||B|\sin{(\theta_B-\theta_A)}[/tex]

[tex]|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|[/tex]
AUMathTutor
#8
Jun4-09, 05:06 PM
P: 490
Uhhh, qntty...

I believe the magnitudes of the two vectors are the same.

;(((
0xDEADBEEF
#9
Jun4-09, 05:10 PM
P: 825
Why? Do the math with the definition of the cross product using the epsilon tensor which is antisymmetrical under the exchange of two indices:
[tex](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[/tex]
renaming of indices changing summation order:
[tex]&=& \sum_{j=1}^3 \sum_{k=1}^3 - \epsilon_{ijk}a_k b_j \\
&=& - (b \times a)_i[/tex]
AUMathTutor
#10
Jun4-09, 05:17 PM
P: 490
0xDEADBEEF:

Wow, that was helpful.
D H
#11
Jun4-09, 06:06 PM
Mentor
P: 15,167
None of the answers so far really answer the question. The real answer is that the cross product defined in the way it is defined is an extremely useful concept.

Mathematics has a dirty little secret. Mathematicians stumble in developing concepts just like everyone else. Eventually they stumble across a very concise and coherent representation. Mathematics is taught sans all of the stumbling around. The development of vector analysis took a lot of stumbling around throughout much of the 1800s.

One of the key developments was that of quaternions by Hamilton. Ever wonder why we use i,j,k to represent the unit vectors? Simple: In complex analysis, i is the square root of -1. Hamilton saw a way to extend complex numbers to 4 dimensions, using two new symbols j and k. The quaternions have real and imaginary parts, but now the imaginary part is written in terms of i, j, and k. As with complex numbers, i2=-1. Those extra symbols? Just as i2=-1, j2=k2=-1. However, j and k are something separate from i. For one thing, ijk=-1. One consequence: ij=k, jk=i, and ki=j. Another consequence is that multiplication in the quaternions is not commutative. Hamilton described his quaternions as comprising a scalar part and a vector part (the word vector is Hamilton's invention).

Heaviside saw Hamilton's vectors as being more useful than his quaternions. He tied vectors to things like the area of a parallelogram (see post #7), electrodynamics (Maxwell's equations are a lot easier in vectors than in quaternions), and a host of other things.

The Levi-Cevita symbol (post #9)? That came along later than the concept of the cross product and is also far too advanced a concept for someone struggling with basic vector analysis.


Register to reply