Vector image on another vector

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baby_1
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Hello
I want to demonstrate a equation of Vector image on another vector (A on B)
gif.latex?A_{B}=\frac{(\bar{A}.\bar{B}).\bar{B}}{|B|^{2}}.gif

So i go this steps
gif.gif

and as we know
gif.gif

so change the equation
a}_{B}|Cos\Theta%20_{AB}=\bar{A}.\hat{a}_{B}=|\bar{A}|.\hat{a}_{B}.\hat{a}_{B}=\frac{(A.\bar{B}).gif

but in my equation A is (vector scale) not a (Vector)!
what is my problem?
 
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You have [itex]\overline{A}[/itex], [itex]A[/itex], and [itex]\left|\overline{A}\right|[/itex]. I understand that [itex]\overline{A}[/itex] is a vector and that [itex]\left|\overline{A}\right|[/itex] is its length but what is [itex]A[/itex]?
 
Thanks for your replay
A and |A| both of them are length of A vector (in electromagnetic books we assume that A=|A| for easy to write)
 
I rewrite the Steps.
main formula:
gif.latex?A_{B}=\frac{%28\bar{A}.\bar{B}%29.\bar{B}}{|B|^{2}}.gif


steps:
gif.gif

as we know
gif.gif

and
gif.latex?\bar{A}.gif

so
heta%20_{AB}=\bar{A}.\hat{a}_{B}=|\bar{A}|.\hat{a}_{B}.\hat{a}_{B}=\frac{%28|\bar{A}|.\bar{B}%29.gif


but in my equation |A| is (vector length) not a (Vector)!
what is my problem?
 
Oh Baby,

What are you doing? You are taking the dot product of vector A and vector B which results in a scalar. Then you are taking the dot product of that scalar with vector B. That makes no sense.

I think you want the dot product of vector A and vector B/|B|. B/|B| is the unit vector in the B direction.

Ratch
 
Thanks friends
i found my probelm
Image of Vector of A on B is a Vector and we should define the direct of it when write the image equation
gif.latex?A_{B}=ACos\Theta%20_{AB}.gif