Cross Products

  • #1
I'm having trouble relating the cross product form |a||b|sin(theta) to its component form (a1b2 - a2b1) ... and so on... I know how to do this mathematically so please don't just suggest some proof that I can find in every textbook... The component form involves the solutions to equations using determinants I believe... I was wondering if anyone could get me going in the right direction as far as setting up a set of equations to solve in order to arrive at this component form... I know I have seen this somewhere but cannot find the right book... So I guess you could say I'm trying to setup the right question, in other words, is there a set of equations for 2 vectors in a plane that can be solved via determinants in order to arrive at this component form for the cross product? I'm having a little trouble stating the question even...
 

Answers and Replies

  • #2
I guess another way of what I am asking is - is there any way to arrive at the component form of the cross product without knowing it is equal to absin(theta)?
 
  • #3
radou
Homework Helper
3,115
6
Well, a x b = [tex]\det \left( \begin{array}{ccc}
\textbf{i} & \textbf{j} & \textbf{k} \\
a1 & a2 & a3 \\
b1 & b2 & b3 \end{array} \right)[/tex], unless you were referring to something else?

(Of course, a = a1i + a2j + a3k, etc.)
 
  • #4
well that's kind of what I'm asking... is there a way to arrive at that determinant form without knowing a x b equals |a||b|sin(theta)? So we want to construct a vector that is perpendicular to 2 vectors in a plane and has a magnitude that somehow expresses the amount of rotation.
 
  • #5
nicksauce
Science Advisor
Homework Helper
1,272
5
"well that's kind of what I'm asking... is there a way to arrive at that determinant form without knowing a x b equals |a||b|sin(theta)?"

Well that's kind of the definition for the cross-product. Nevertheless, maybe this might help a bit:

Start with the following definitions for a right-handed co-ordinate system:

[tex]\hat{x}\times\hat{x} = \hat{y}\times\hat{y} = \hat{z}\times{z} = 0[/tex]
[tex]\hat{x}\times\hat{y} = -\hat{y}\times\hat{x} = \hat{z}[/tex]
[tex]\hat{y}\times\hat{z} = -\hat{z}\times\hat{y} = \hat{x}[/tex]
[tex]\hat{z}\times\hat{x} = -\hat{x}\times\hat{z} = \hat{y}[/tex]

So if you write

[tex]A\times B = (A_x\hat{x} + A_y\hat{y} + A_z\hat{z}) \times (B_x\hat{x} + B_y\hat{y} + B_z\hat{z})[/tex]

(Ie: [tex]A_y\hat{y}\times B_x\hat{x}=A_yB_x(\hat{y}\times\hat{x}) = -A_yB_x\hat{z}[/tex])

Expand, regroup, and this will lead you to the determinant form.
 
Last edited:
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,833
956
well that's kind of what I'm asking... is there a way to arrive at that determinant form without knowing a x b equals |a||b|sin(theta)? So we want to construct a vector that is perpendicular to 2 vectors in a plane and has a magnitude that somehow expresses the amount of rotation.
If you don't use "length of a x b= |a||b|sin(theta)" (and the fact that a x b is perpendicular to be a and b with the "right hand rule"- it is not correct that a x b= |a||b|sin(theta)!) the what definition of cross product ARE you using?

Obviously, you have to have some definition before you can derive a formula!

nicksause is using, as a definition, that [itex]\vec{i}\times \vec{j}= \vec{k}[/itex], [itex]\vec{j}\times\vec{k}= \vec{i}[/itex], and [itex]\vec{k}\times\vec{i}= \vec{j}[/itex] together with requiring that the cross product be associative, distribute over vector addition, and be anti-commutative.
 
  • #7
nicksauce
Science Advisor
Homework Helper
1,272
5
nicksause is using, as a definition, that [itex]\vec{i}\times \vec{j}= \vec{k}[/itex], [itex]\vec{j}\times\vec{k}= \vec{i}[/itex], and [itex]\vec{k}\times\vec{i}= \vec{j}[/itex] together with requiring that the cross product be associative, distribute over vector addition, and be anti-commutative.
Right, I should have mentioned that.
 
  • #8
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
685
nicksause is using, as a definition, that [itex]\vec{i}\times \vec{j}= \vec{k}[/itex], [itex]\vec{j}\times\vec{k}= \vec{i}[/itex], and [itex]\vec{k}\times\vec{i}= \vec{j}[/itex]
That is the original quaternion-based definition of the cross product. Even the use of [itex]\vec{i}, \vec{j}, \vec{k}[/itex] as unit vectors comes straight from the quaternions. The determinant form is an easy mnemonic for some; I prefer the even/odd permutations of i,j,k (or whatever).

As an aside, the concept of vectors and vector spaces is a relatively recent invention (end of the 19th century). We are introduced to vectors in the first week of freshman physics and use vector-based calculations throughout. How did physicists do things, even very basic freshman-level physics things, before the invention of vectors and all that is associated with them?
 
  • #9
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
5,701
994
They probably used systems of equations expressed in some choice of coordinate system. I would imagine that there was more use of geometric and trigonometric arguments.
 

Related Threads on Cross Products

  • Last Post
Replies
6
Views
821
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
11
Views
909
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
2
Views
8K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
727
Replies
5
Views
629
  • Last Post
Replies
1
Views
711
Replies
2
Views
1K
Top