Vector image on another vector

baby_1

Hello
I want to demonstrate a equation of Vector image on another vector (A on B)
$A_{B}=\frac{(\bar{A}.\bar{B}).\bar{B}}{|B|^{2}}$
So i go this steps
$A_{B}=ACos\Theta _{AB}$
and as we know
$|\hat{a}_{B}|=1$
so change the equation
$A_{B}=ACos\Theta _{AB}=|\bar{A}||\hat{a}_{B}|Cos\Theta _{AB}=\bar{A}.\hat{a}_{B}=|\bar{A}|.\hat{a}_{B}.\hat{a}_{B}=\frac{(A.\bar{B}).\bar{B}}{|B|^{2}}$
but in my equation A is (vector scale) not a (Vector)!!!!
what is my problem?

HallsofIvy

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]?

baby_1

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)

pwsnafu

Then pick one and stick with it. Similarly, don't swap between upper and lower case.

baby_1

I rewrite the Steps.
main formula:
$A_{B}=\frac{(\bar{A}.\bar{B}).\bar{B}}{|B|^{2}}$

steps:
$A_{B}=|A|Cos\Theta _{AB}$
as we know
$|\hat{a}_{B}|=1$
and
$\bar{A}.\hat{a}_{B}=|\bar{A}||\hat{a}_{B}|Cos\Theta _{AB}$
so
$A_{B}=|\bar{A}|Cos\Theta _{AB}=|\bar{A}||\hat{a}_{B}|Cos\Theta _{AB}=\bar{A}.\hat{a}_{B}=|\bar{A}|.\hat{a}_{B}.\hat{a}_{B}=\frac{(|\bar{A}|.\bar{B}).\bar{B}}{|B|^{2}}$

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

*Ratch*

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

baby_1

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
$A_{B}=ACos\Theta _{AB}.\hat{a}_{B}$

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HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	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]?

baby_1	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)

pwsnafu	Vector image on another vector Then pick one and stick with it. Similarly, don't swap between upper and lower case.

Ratch	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