# Dot product (scalar product) of 2 vectors: ABcos$\theta$

1. Jan 3, 2012

### LearninDaMath

How, precisely, do you get/derive the Bcosθ term?

Is it simply [Cosθ=A/B] --> [BCosθ = A] ? It can't be that simple because then how is the extra length of vector A fit into [*A*Bcosθ]? I feel pretty confused as to what is going on here. To summerize, A x B = [ABcosθ] makes little sense.. and i think that is because for one reason, i'm not sure how Bcosθ is derived. And the other reason is because I don't know where they get the A variable either. Just can't seem to see what's going on here.

Last edited: Jan 3, 2012
2. Jan 3, 2012

3. Jan 3, 2012

### LearninDaMath

Re: dot product (scalar product) of 2 vectors: ABcosθ

Wow, that looks a little more complex than I thought it would be. I thought it was something much more straight forward. I'll have to do some further self study for a little while and get back to you on what I was able to figure out. At worst, I hope to at least come back with a more specific question. At best, I'll be able to figure it out. Thanks.

4. Jan 9, 2012

### LearninDaMath

Figured it out, this makes perfect sense now. However I have another vector question, but unrelated to this one, so I'll start a new thread for it.