Cross Product of Vectors a & b: Why 0?

Click For Summary

Discussion Overview

The discussion revolves around the cross product of two vectors, a <1,-1,0> and b <-1,1,0>, and why the result is the zero vector. Participants explore the implications of co-linearity, the right-hand rule, and the geometric interpretation of the cross product.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that the cross product results in the zero vector because vectors a and b are co-linear.
  • Another participant explains the mathematical definition of the cross product, emphasizing that since the angle between co-linear vectors is 180 degrees, the sine of that angle is zero, leading to a zero product.
  • Questions arise regarding the applicability of the right-hand rule for co-linear vectors, with some participants asserting that it does not apply since co-linear vectors do not define a plane.
  • One participant expresses confusion about the right-hand rule, stating that their thumb points out of the screen when applying it, leading to a discussion about the ambiguity of direction in co-linear cases.
  • Another participant introduces the concept of the area of the parallelogram formed by the vectors, stating that co-linear vectors result in a degenerate case with zero area, hence a zero cross product.
  • There is a discussion about the infinite number of planes that can be formed with co-linear vectors, raising questions about how to apply the right-hand rule in such scenarios.
  • Some participants clarify that the cross product is typically addressed in vector calculus rather than linear algebra.

Areas of Agreement / Disagreement

Participants generally agree that the cross product of co-linear vectors is zero, but there is disagreement regarding the application of the right-hand rule and the interpretation of direction in this context. The discussion remains unresolved regarding the right-hand rule's applicability to co-linear vectors.

Contextual Notes

Participants express uncertainty about the definitions and rules surrounding the right-hand rule and the cross product, particularly in relation to co-linear vectors. There are also references to the mathematical properties of the cross product that may not be fully understood by all participants.

BoundByAxioms
Messages
98
Reaction score
0
I have not yet taken Linear Algebra, so perhaps that is why I don't have an answer to this question. But:


Given a vector a <1,-1,0> and a vector b <-1,1,0>, I would suspect (by the right hand rule) that taking a X b, I would get a vector pointing out of the computer screen (assuming that the computer screen is the xy plane). However, when I take the actual cross product, I get:

i j k
1 -1 0
-1 1 0

(0-0)i-(0-0)j+(1-1)k, giving me the vector 0i+0j+0k. Why is this?
 
Physics news on Phys.org
Because b=-a = (-1)*a. In other words, a and b are co-linear and hence their cross product is zero.
 
Just to add onto what the previous poster said, the cross product between two vectors a and b is defined as a*b =|a||b|sinx(n^), where x is the angle between the vectors and n^ is a unit vector, n. Since they are collinear, the angle between them is 180 degrees. So, sin180 = 0, making the whole product 0.
 
So the right hand rule does not apply to vectors that are co-linear? I suppose that that was my main question.
 
You know the following rule for cross product: -aXb = -(aXb) = aX(-b).

Now this reduces it to -(aXa). What is aXa?
 
BoundByAxioms said:
So the right hand rule does not apply to vectors that are co-linear? I suppose that that was my main question.
The right-hand rule doesn't even make sense for vectors that are co-linear. Take your example, [1,-1,0]\times[-1,1,0] What makes you think the result would be coming out of the screen? Why not along [1,1,0], or for that matter, in any direction of the form [a,a,b]? The problem is that two co-linear vectors do not define a plane. Fortunately, the cross product of a pair of co-linear vectors is identically zero, so the fact that the direction is not well-defined is not a problem.
 
D H said:
The right-hand rule doesn't even make sense for vectors that are co-linear. Take your example, [1,-1,0]\times[-1,1,0] What makes you think the result would be coming out of the screen? Why not along [1,1,0], or for that matter, in any direction of the form [a,a,b]? The problem is that two co-linear vectors do not define a plane. Fortunately, the cross product of a pair of co-linear vectors is identically zero, so the fact that the direction is not well-defined is not a problem.


Well, I guess I'm not seeing this. I can't see how the right hand rule doesn't work. I put my hand at 315 degrees for vector a, and then I curl it until it points in the direction of vector b, at 135 degrees. And my thumb is pointing out of the computer screen.
 
Defennder said:
You know the following rule for cross product: -aXb = -(aXb) = aX(-b).

Now this reduces it to -(aXa). What is aXa?

aXa = 0i+0j+0k. I don't know those rules though, could you explain why they make sense?
 
If a and b are not colinear, you can only rotate a through one plane to put it on top of b.

If a and b *are* colinear, there is an infinite number of planes that you can rotate a through to put it onto b: any plane containing both vectors.

So how do you know which plane to put your "hand" in when applying the rule? You seem to think it should be the xy plane, but there is no good reason for this. It could just as well be any other plane containing the two vectors.
 
  • #10
One way of thinking about the cross product is that |a x b| is the area of the parallelogram whose sides are a and b. If a and b are colinear, we no longer have a parallelogram but a degenerate case of a parallelogram, i.e. a line segment. Since the area of a line segment is zero, |a x b| = 0 if a and b are colinear. Thus a x b = 0-> (where 0-> denotes the zero vector)
 
  • #11
BoundbyAxioms, the zero vector has an arbitrary direction. It is undefined (the direction).
 
  • #12
BoundByAxioms said:
Well, I guess I'm not seeing this. I can't see how the right hand rule doesn't work. I put my hand at 315 degrees for vector a, and then I curl it until it points in the direction of vector b, at 135 degrees. And my thumb is pointing out of the computer screen.

No, not necessarily. Since those vectors are diametrically opposite, you can turn the fingers of your right hand from one to the other either way, with the thumb pointing in either direction. Try it!
 
  • #13
Actually, there are infinite possible directions for the thumb to be pointing!
 
  • #14
uman said:
If a and b are not colinear, you can only rotate a through one plane to put it on top of b.

If a and b *are* colinear, there is an infinite number of planes that you can rotate a through to put it onto b: any plane containing both vectors.

So how do you know which plane to put your "hand" in when applying the rule? You seem to think it should be the xy plane, but there is no good reason for this. It could just as well be any other plane containing the two vectors.

Ok now I get it, that's what was confusing me. Thank you for your help, everyone.
 
  • #15
By the way, the cross product of vectors is a matter of vector calculus and is not normally dealt with in Linear Algebra.
 
  • #16
HallsofIvy said:
By the way, the cross product of vectors is a matter of vector calculus and is not normally dealt with in Linear Algebra.

Ah, ok. Good to know. I know cross products from my vector calculus class, not quite sure why I thought I should ask it in a linear algebra forum.
 

Similar threads

High School What is the cross product of two vectors?
  • · Replies 32 ·
2
Replies
32
Views
4K
Graduate Reason out the cross product (for the moment): a skew symmetric form
  • · Replies 14 ·
Replies
14
Views
4K
Undergrad Question about the vector cross product in spherical or cylindrical coordinates
  • · Replies 5 ·
Replies
5
Views
5K
Undergrad The vector to which a dual vector corresponds
  • · Replies 13 ·
Replies
13
Views
4K
Undergrad Relation Between Cross Product and Infinitesimal Rotations
  • · Replies 22 ·
Replies
22
Views
5K
Undergrad Rotational invariance of cross product matrix operator
  • · Replies 7 ·
Replies
7
Views
3K
Undergrad Visualization of complex vectors and dot product
  • · Replies 14 ·
Replies
14
Views
4K
High School Confused about dot product multiplication
  • · Replies 33 ·
2
Replies
33
Views
5K
Undergrad Understanding the cross product and rotations
  • · Replies 11 ·
Replies
11
Views
4K
Undergrad Vector Cross Product: Understanding the Perpendicular Result
  • · Replies 11 ·
Replies
11
Views
2K