Unknown angle between two vectors

[student34]

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?

 

[Outrageous]

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

 

[student34]

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.