Unknown angle between two vectors

  • Thread starter student34
  • Start date
  • #1
346
3

Homework Statement


Scalar product of vectors AB = -6; vector product of AB = 9. Find the angle between vectors A and B.


Homework Equations


Scalar product: cos(θ)*AB
Vector product: sin(θ)*AB
(sin(θ))/cos(θ) = tanθ

The Attempt at a Solution


AB = -6/cos(θ)
AB = 9/sin(θ)
9/sin(θ) = -6/cos(θ)
9/-6 = (sin(θ))/cos(θ)
3/-2 = tan(θ)
tan^-1(3/-2) = -56° which I take to mean 56° from each other.
But their answer is 124° which is -56° + 180°. How can this be when the whole time we are dealing with one angle between two vectors?
 

Answers and Replies

  • #2
374
0
3/-2 = tan(θ)
tan θ = negative value means θ can only be 90<θ<180 or 270<θ<360.


I have a question also , why the vector product which is a vector can equal to 9? is that a magnitude of AxB?
Thank you
 
  • #3
346
3
tan θ = negative value means θ can only be 90<θ<180 or 270<θ<360.
Ohhhh, I see, thanks.

I have a question also , why the vector product which is a vector can equal to 9? is that a magnitude of AxB?
Thank you
Yes, I should have mentioned that.
 

Related Threads on Unknown angle between two vectors

Replies
3
Views
8K
Replies
11
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
7
Views
220
Top