dot product (scalar product) of 2 vectors: ABcos[itex]\theta[/itex]by LearninDaMath Tags: abcostheta, product, scalar, vectors 

#1
Jan312, 07:14 PM

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. 



#2
Jan312, 07:34 PM

P: 260

How does this work for you: http://en.wikipedia.org/wiki/Dot_pro...interpretation




#3
Jan312, 07:56 PM

P: 300





#4
Jan912, 07:22 PM

P: 300

dot product (scalar product) of 2 vectors: ABcos[itex]\theta[/itex]
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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