# Homework Help: Need Help Finding the Angle Between Two Vectors

1. Apr 1, 2012

### hardygirl989

1. The problem statement, all variables and given/known data
Two vectors are given by A = 3i + j and = -1i + 2j.
A) Find A x B <---This is the cross product not multiplication.
B) Find the angle between A and B.

2. Relevant equations

|AxB|=|A||B|sin(theta)

3. The attempt at a solution

A) I used the determinant to find the cross product as 3(2)-1(-1)=7 <---This answer is correct

B) I need help with this problem...
I tried |AxB|=|A||B|sin(theta) to get θ= arcsin( |AxB| / ( |A||B| )
|A|=√(Ax^2+Ay^2)=3.16
|B|=√(Bx^2+Ay^2)=2.24
Thus, θ= arcsin(7/((3.16)(2.24)))=81.8 °.

Thank you.

2. Apr 1, 2012

### Staff: Mentor

Hi hardygirl989, Welcome to Physics Forums.

The problem is, there is an ambiguity introduced by the arcsin() function when the angle between the vectors exceeds 90 degrees. If you think about it, the sine of 90°-x is the same as the sine of 90°+x, as both angles "straddle" the y-axis. The arcsin function will only return one of the angles, and that would be the smaller of the two choices (in the first quadrant).

Note that the arccos() function doesn't have the same issue for angles in the 1st and 2nd quadrants. Maybe a dot product approach is in your future

Alternatively, find the angles associated with both vectors and take the difference.

3. Apr 1, 2012

### hardygirl989

Yes! I got the correct answer! Thank you! I was not even thinking about the ambiguity. :)

Dot product of A and B = |A||B|cos(theta)
so theta = arccos[Dot product of A and B / ( |A||B| ) ]
theta = arccos { [ (3)(-1)+(2)(1) ] / [ (3.16)(2.24) ] } = 98.13 degrees.