Equivalence of 2 definitions of the magnitude of the cross product?

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Discussion Overview

The discussion revolves around the equivalence of two definitions for calculating the magnitude of the cross product of vectors. Participants explore the mathematical steps involved, identify potential errors, and clarify concepts related to the sine of the angle between vectors. The scope includes technical reasoning and mathematical exploration.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines their approach to proving the equivalence of two methods for calculating the magnitude of the cross product but expresses uncertainty about a mistake in their calculations.
  • Another participant points out that the cross product is not arbitrary if certain coordinates are the same, suggesting a potential flaw in the initial setup.
  • A later reply emphasizes the importance of the sine of the angle between vectors and questions the introduction of cosine in the calculations, indicating that it may add confusion.
  • One participant proposes using the law of cosines to relate the sine and cosine of the angle, presenting a detailed mathematical derivation to support their reasoning.
  • Another participant acknowledges finding an error in their arithmetic and expresses gratitude for the assistance received, indicating progress in their understanding.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to proving the equivalence of the definitions, and multiple viewpoints on the calculations and methods remain present throughout the discussion.

Contextual Notes

Some participants note potential arithmetic errors and the need for careful attention to detail in calculations. The discussion also highlights the complexity of relating sine and cosine in the context of vector operations.

nomadreid
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TL;DR
The two definitions of the magnitude ||u x v|| of a cross product, one via coordinates, and the other with uv.sin A, are obviously equivalent, but my brute-force attempt hits a snag.
In order to prove that two of the means to calculate the magnitude of an arbitrary cross product are equivalent, I tried a direct approach, which I outline below. Any hint as to where I am going wrong would be appreciated.

So I hope this is the right section-- although this may look like HW, it isn't. It is my own fumbling.
Step 1A.png

Then
Step 2.png

Then,
Step 3.png


Alas, this doesn't come out. What am I doing wrong? Some silly mistake, no doubt, but I would be very grateful to have it pointed out.
 
Last edited:
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This isn't an arbitrary cross product if the second and third coordinates are the same.
 
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fresh_42 said:
This isn't an arbitrary cross product if the second and third coordinates are the same.
Oops. Typo edited, thanks. Apart from that?
 
nomadreid said:
Oops. Typo edited, thanks. Apart from that?
You carried that typo throughout the calculation. There is no 2B term in the cross product. The cross product is a Lie multiplication, hence the Jacobi identity holds, hence it has to be symmetric in (x,y,z) and the pairs (x,y), (y,z), (z,x) with a cyclic permutation.

Edit: Mistake corrected.
 
Last edited:
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Thanks very much for your reply, fresh_42. Let me start with your remark that there should be no 2B (my notation) in the result.
To find out where my mistake appears, I have broken Step 1 into four sub-steps for easy reference.

Steps 1a to 1c.png

Step 1d.png

After I have figured out where my mistake(s) is (are) here, I will comb my text for the typos o:) leading to the asymmetry you mentioned, as I don't see them at first glance. I will search again, but then I may turn to you again to specify. Thanks for your patience.
 
The essential point in your considerations is the sine of the angle between your vectors. How do you compute the sine without using the cross product? And turning it into a cosine doesn't answer the question, that's only adding confusion.

As I see it, you want to prove ##\sin \measuredangle (\vec{u},\vec{v})= \dfrac{|\vec{u}\times \vec{v}|}{|\vec{u}|\cdot |\vec{v}|}## where you have formulas for the right-hand side. So the question has to be: what is on the left? Do you use the law of cosines together with ##1=\cos^2\measuredangle (\vec{u},\vec{v})+\sin^2\measuredangle (\vec{u},\vec{v})## which is why you introduced the cosine? I guess, this can be done, even if it looks a bit like a nightmare of a calculation to me:
\begin{align*}
|(\vec{u}+\vec{v})^2|&=|\vec{u}|^2+|\vec{v}|^2-2|\vec{u}||\vec{v}|(1-\sin^2 \measuredangle (\vec{u},\vec{v}))\\
\sin^2 \measuredangle (\vec{u},\vec{v})&=1+\dfrac{|(\vec{u}+\vec{v})^2|-|\vec{u}|^2-|\vec{v}|^2}{2|\vec{u}||\vec{v}|}\\
&=\dfrac{|(\vec{u}+\vec{v})^2|-|\vec{u}|^2-|\vec{v}|^2+2|\vec{u}||\vec{v}|}{2|\vec{u}||\vec{v}|}\\
&=\dfrac{|(\vec{u}+\vec{v})^2|-(|\vec{u}|-|\vec{v}|)^2}{2|\vec{u}||\vec{v}|}\\
\sin \measuredangle (\vec{u},\vec{v})&=\sqrt{\left|\dfrac{|(\vec{u}+\vec{v})^2|-(|\vec{u}|-|\vec{v}|)^2}{2|\vec{u}||\vec{v}|}\right|}\\
\dfrac{|\vec{u}\times \vec{v}|}{|\vec{u}|\cdot |\vec{v}|}&\stackrel{!}{=}\sqrt{\left|\dfrac{|(\vec{u}+\vec{v})^2|-(|\vec{u}|-|\vec{v}|)^2}{2|\vec{u}||\vec{v}|}\right|}
\end{align*}
I didn't even use coordinates, yet, and I'm not sure whether I made a mistake or not. How do you plan to get through without any copy-and-paste errors?

If this wasn't your strategy, what else is it?
$$
$$
It doesn't answer your question, but you might be interested in this article:
https://arxiv.org/pdf/1205.5935
which has a nice introduction to the various products at its beginning.
 
Last edited:
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Thank you, fresh_42. First, I have downloaded the ArXiv article, which looks very useful. Second, yes, your calculations are essentially the direction I was going, although I was doing it more by brute force with the coordinates to show the two methods of calculating the cross product to be equivalent -- and in the meantime I found my error (which, as I suspected, was a silly arithmetic error). To be explicit, if you keep the Step 1 in my Posts #1 and #5 and then replace the other steps in the original Post #1 by the following Steps:
New Steps 2_3.png

New Step 4.png

one gets the desired
New Step 5.png

So, thanks again for the help.
 

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