Cross product of 2 vectors of same magnitude

Click For Summary

Homework Help Overview

The discussion revolves around the cross product of two vectors, A and B, both having the same magnitude M and forming a 30-degree angle when joined at their tails. Participants are exploring how to express the cross product A x B in terms of M.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the magnitudes of the vectors and the sine of the angle between them. There are attempts to determine the value of the cross product based on the given angle, with some suggesting various potential outcomes.

Discussion Status

The discussion is active, with participants providing insights about the properties of the cross product and its dependence on the angle between the vectors. Some guidance has been offered regarding the implications of different angles on the result of the cross product.

Contextual Notes

There is a focus on the specific angle of 30 degrees and its impact on the calculation of the cross product. Participants are considering the implications of different configurations of the vectors, such as being parallel or perpendicular.

Okokya
Messages
2
Reaction score
0

Homework Statement


Vectors A and B both have magnitude M. Joined at the tails, they create a 30' angle. What is A x B in terms of M?

Homework Equations

The Attempt at a Solution


0? OR M^2? Sqrt(3)M/3?
 
Physics news on Phys.org
Remember that the cross product of two vectors is related to their magnitudes and the sine of the angle between them. Do you remember this?

If they were parallel, the answer would be zero, since the cross product gives you the perpendicular component of one vector along the other. If they were perpendicular, the answer would be M^2.
 
samnorris93 said:
Remember that the cross product of two vectors is related to their magnitudes and the sine of the angle between them. Do you remember this?

If they were parallel, the answer would be zero, since the cross product gives you the perpendicular component of one vector along the other. If they were perpendicular, the answer would be M^2.

So for theta 30', ab sin 30 = (M^2)/2 ?
 
Okokya said:
So for theta 30', ab sin 30 = (M^2)/2 ?

Yep!
 

Similar threads

Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
26
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
26
Views
3K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K