How Do You Calculate the Angle Between Two Vectors?

  • Thread starter Thread starter azila
  • Start date Start date
  • Tags Tags
    Stuck
Click For Summary

Homework Help Overview

The discussion revolves around calculating the angle between two vectors A and B, given their magnitudes and the vector product. The original poster presents their approach and calculations, expressing uncertainty about the correctness of their method.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use the relationship between the vector product and the sine of the angle to find the angle between the vectors. They question whether their calculations and the value of the vector product are correct.

Discussion Status

Participants are exploring different aspects of the problem, with some suggesting the need to consider the absolute value of the vector product. There is acknowledgment of a potential misunderstanding regarding the use of sine versus arcsine in the calculations.

Contextual Notes

There is a mention of uncertainty regarding the correctness of the vector product value and the implications of using different trigonometric functions in the calculations.

azila
Messages
67
Reaction score
0
Easy problem that I am stuck on...

Homework Statement


Two vectors A and B have magnitude of A = 3.02 B=3.00. Their vector product is AXB = -4.90k + 2.01i. What is the angle between vector A and vector B?


Homework Equations


since I know the vector product of A and B, I was thinking that Cx = 2.01 and Cz = -4.90. I would then add them together to get -2.89. Then could i use the equation C = ABsin(theta). That would be like

-2.89 = (3.02)(3.00)sin(theta)
-.319=sin(theta)
-18.6 = theta
I don't know if this is the right path to follow but this is what I could come up with.


The Attempt at a Solution



THANKS for any help.
 
Physics news on Phys.org
it's
[tex]\left| c\right| =\left| a\right| \,\left| b\right| \,sin\left( p\right)[/tex]
where c = a x b
 
so am I solving it incorrectly because i am using the same equation, its just that I don't know if the value of c that I have is correct. Also the answer I get is not right. So, what should I change? thanks for the help though.
 
get the absolute value of c.
c^2=x^2+y^2+z^2
 
Thank You So Much! I Realized What I Was Doing Wrong Was That I Was Taking The Sin When I Should Have Been Taking The Arcsin. Thanks Again!
 

Similar threads

Replies
1
Views
1K
Replies
5
Views
910
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 18 ·
Replies
18
Views
1K
Replies
26
Views
4K
Replies
1
Views
2K
Replies
3
Views
1K
Replies
46
Views
7K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K