Is the Dot Product of Two Vector Pairs Always Commutative?

Click For Summary
SUMMARY

The discussion confirms that the dot product of two vector pairs is not always commutative when considering multiple pairs. Specifically, the equation (a*b)(c*d) does not equal (a*c)(b*d) despite the individual dot products being commutative. A counterexample using vectors a = [1,0], b = [1,0], c = [1,1], and d = [1,2] demonstrates this, yielding different scalar results for both expressions. The mathematical reasoning provided emphasizes the importance of understanding the properties of dot products and their limitations.

PREREQUISITES
  • Understanding of vector operations, specifically dot products.
  • Familiarity with vector notation and properties.
  • Basic knowledge of trigonometry, particularly cosine functions.
  • Ability to construct and analyze counterexamples in mathematical proofs.
NEXT STEPS
  • Study the properties of dot products in vector algebra.
  • Learn about vector magnitudes and angles, including how to calculate them.
  • Explore counterexample techniques in mathematical proofs.
  • Investigate the implications of non-commutativity in vector operations.
USEFUL FOR

Students in physics or mathematics, educators teaching vector algebra, and anyone seeking to deepen their understanding of vector operations and their properties.

digipony
Messages
35
Reaction score
0

Homework Statement

The Attempt at a Solution




I am working a physics problem and want to make sure I'm not making a mistake in the math. Here is my math inquiry:

Say you have (a*b)(c*d) where * indicates the dot product, and a,b,c, and d are all vectors. Can you say that (a*b)(c*d) = (a*c)(b*d) since the dot product is commutative and gives you a scalar? Thanks!
 
Physics news on Phys.org
digipony said:

Homework Statement

The Attempt at a Solution




I am working a physics problem and want to make sure I'm not making a mistake in the math. Here is my math inquiry:

Say you have (a*b)(c*d) where * indicates the dot product, and a,b,c, and d are all vectors. Can you say that (a*b)(c*d) = (a*c)(b*d) since the dot product is commutative and gives you a scalar? Thanks!

No. You can't. Neither of those properties says you can swap vectors between two different dot products.
 
If you're ever unsure of these things, always try to find a counterexample. Take a = [1,0], b = [1,0], c = [1,1] and d = [1,2]. Then,

(a*b)(c*d) = (1)(1 + 2) = 3

but

(a*c)(b*d) = (1)(1) = 1
 
Sorry, accidentally double posted.
 
Dick said:
No. You can't. Neither of those properties says you can swap vectors between two different dot products.

Darn- it made the problem so easy, but I guess that was a sign that I was probably doing something that I should not be doing. Thank you for explaining why I could not shuffle the vectors around. :)

Karnage1993 said:
If you're ever unsure of these things, always try to find a counterexample. Take a = [1,0], b = [1,0], c = [1,1] and d = [1,2]. Then,

(a*b)(c*d) = (1)(1 + 2) = 3

but

(a*c)(b*d) = (1)(1) = 1
I did try an example, however it worked out. But I guess it wasn't that good of an example since it was a fluke that it worked out. Thanks
 
digipony said:

Homework Statement

The Attempt at a Solution




I am working a physics problem and want to make sure I'm not making a mistake in the math. Here is my math inquiry:

Say you have (a*b)(c*d) where * indicates the dot product, and a,b,c, and d are all vectors. Can you say that (a*b)(c*d) = (a*c)(b*d) since the dot product is commutative and gives you a scalar? Thanks!

No, as others have already pointed out.

You can see this another way: a*b = |a||b| cos(a,b) and c*d = |c||d|cos(c,d), where |a|, |b| are the magnitudes of a and b and (a,b) is the angle between a and b, and similarly for c and d. In general, we do not have cos(a,c).cos(b,d) equal to cos(a,b).cos(c,d), so the two expressions are generally different.
 

Similar threads

Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K