# Finding the angle between two vectors

1. Sep 23, 2011

### hyde2042

1. The problem statement, all variables and given/known data

A force F =( 6 i - 2 j ) N acts on a particle that under
goes a displacement D r = ( 3 i + j )m. Find (a) the work done
by the force on the particle and (b) the angle between F
and D r .

2. Relevant equations

I've found the work to be about 16N. My problem is finding the angle.

The equation thebook gives is cos(inv)* (Products of vectors A and B) / (A)(i^2+j^2)*(B)(i^2+j^2)

3. The attempt at a solution

I used cas(inv)*((VectorA * VectorB) / sqrt(6^2-2^2)(3^2+1^2))

Which came out to 26 degrees. The back of the book says 36.9 degrees for the answer. I don't think I'm missing anything. All your help is appreciated. Thank you.

2. Sep 23, 2011

### MetalManuel

The easiest way that i can think of to find the angle between two vectors is the dot product.

Remember $\vec{A}\cdot\vec{B}=|A||B|Cos(\theta)$

3. Sep 23, 2011

### hyde2042

That's what I used, but I still got 26.5 degrees instead of the 36.9.

I got part a correct (finding the Force on the object) so I don't think I did any previous calculations incorrectly for the numbers I'm using now.

4. Sep 23, 2011

### SammyS

Staff Emeritus
The magnitude of vector A is √( 62 + (-2)2 ) = √( 36 + 4 ) .

5. Sep 24, 2011

### hyde2042

Holy crap.. thank you.