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

  • Context: Undergrad 
  • Thread starter Thread starter nomadreid
  • Start date Start date
  • Tags Tags
    Cross product
Click For Summary
SUMMARY

The forum discussion centers on proving the equivalence of two definitions for the magnitude of the cross product of vectors. The user initially encounters errors due to a persistent typo and misunderstanding of the cross product's properties, specifically regarding the absence of a 2B term. The discussion emphasizes the importance of correctly applying the sine of the angle between vectors and the law of cosines in the calculations. Ultimately, the user corrects their arithmetic mistake and acknowledges the guidance received from other forum members.

PREREQUISITES
  • Understanding of vector operations, specifically cross products
  • Familiarity with trigonometric identities, particularly sine and cosine
  • Knowledge of the law of cosines in vector mathematics
  • Basic proficiency in mathematical proof techniques
NEXT STEPS
  • Study the properties of the cross product in vector algebra
  • Learn about the Jacobi identity and its implications in Lie algebra
  • Explore the law of cosines and its application in vector calculations
  • Review mathematical proof strategies for establishing equivalences in definitions
USEFUL FOR

Mathematicians, physics students, and anyone involved in vector calculus or linear algebra who seeks to deepen their understanding of vector operations and their properties.

nomadreid
Gold Member
Messages
1,762
Reaction score
248
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:
Physics news on Phys.org
This isn't an arbitrary cross product if the second and third coordinates are the same.
 
  • Like
Likes   Reactions: nomadreid
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:
  • Like
Likes   Reactions: nomadreid
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:
  • Like
Likes   Reactions: nomadreid
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.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
21
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 59 ·
2
Replies
59
Views
9K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 17 ·
Replies
17
Views
11K