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

by LearninDaMath
Tags: abcostheta, product, scalar, vectors
 P: 300 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.
 P: 260 How does this work for you: http://en.wikipedia.org/wiki/Dot_pro...interpretation
P: 300
 Quote by elfmotat How does this work for you: http://en.wikipedia.org/wiki/Dot_pro...interpretation
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.

P: 300

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

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.

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