Prove: vectors u•v=u1•v1+u2+v2 from a• b=|a|•|b|•cos D

Answer 1

u•v=u1•v1+u2+v2 from a• b=|a|•|b|•cos D
|u| = sqrt(u1^2 + u2^2)
|v| = sqrt (v1^2 + v2^2)
so
|u| |v| = sqrt(u1^2 + u2^2) * sqrt (v1^2 + v2^2)
then for the angle
cos (alpha-beta) = cos alpha cos beta + sin alpha sin beta
= v1/sqrt(v1^2+v2^2)*u1/sqrt(u1^2+u2^2)
+ v2/sqrt(v1^2+v2^2)*u1/sqrt(u1^2+u2^2)
so
v1* u1 - v2*u1
whew

Answer 2

are u1,u2,v1,v2 the components of u and v?
If so, then I think you must mean u•v=u1•v1+u2•v2
But that is poor notation, since one • means vector dot product, and the other • means scalar multiplication.

First, it is each to show that
i•i = 1 and j•j=1
since |i|=|j|=1 and cosθ=1 since θ=0
also, i•j=0 since cosθ=0

Now you can see that
u•v = (u1 i + u2 j)•(v1 i + v2 j)
Now just expand that out. All the i•i and j•j are 1, and i•j = j•i = 0

Answer 3

I mean:
so
v1* u1 + v2*u1
whew again