# Vector image on another vector

1. Mar 1, 2013

### baby_1

Hello
I want to demonstrate a equation of Vector image on another vector (A on B)
$image=http://latex.codecogs.com/gif.latex?A_{B}=\frac{(\bar{A}.\bar{B}).\bar{B}}{|B|^{2}}&hash=7c1c60b3685b0f24b2ccf1c304bd7c8a$
So i go this steps
$image=http://latex.codecogs.com/gif.latex?A_{B}=ACos\Theta%20_{AB}&hash=ffeb05dabcc4a7dbcfd2e9c87c0aad60$
and as we know
$image=http://latex.codecogs.com/gif.latex?|\hat{a}_{B}|=1&hash=31210087498f86ce0e00402d33cf3c86$
so change the equation
$image=http://latex.codecogs.com/gif.latex?A_{B}=ACos\Theta%20_{AB}=|\bar{A}||\hat{a}_{B}|Cos\Theta%20_{AB}=\bar{A}.\hat{a}_{B}=|\bar{A}|.\hat{a}_{B}.\hat{a}_{B}=\frac{(A.\bar{B}).\bar{B}}{|B|^{2}}&hash=0f0504118a621e32f7fc3b28324fc96e$
but in my equation A is (vector scale) not a (Vector)!!!!
what is my problem?

2. Mar 1, 2013

### HallsofIvy

Staff Emeritus
You have $\overline{A}$, $A$, and $\left|\overline{A}\right|$. I understand that $\overline{A}$ is a vector and that $\left|\overline{A}\right|$ is its length but what is $A$?

3. Mar 1, 2013

### baby_1

A and |A| both of them are length of A vector (in electromagnetic books we assume that A=|A| for easy to write)

4. Mar 1, 2013

### pwsnafu

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

5. Mar 1, 2013

### baby_1

I rewrite the Steps.
main formula:
$image=http://latex.codecogs.com/gif.latex?A_{B}=\frac{%28\bar{A}.\bar{B}%29.\bar{B}}{|B|^{2}}&hash=e86b20fb27371d597d9b666860e0966b$

steps:
$image=http://latex.codecogs.com/gif.latex?A_{B}=|A|Cos\Theta%20_{AB}&hash=c63eefecfe7201bf3abf95a0c2223673$
as we know
$image=http://latex.codecogs.com/gif.latex?|\hat{a}_{B}|=1&hash=31210087498f86ce0e00402d33cf3c86$
and
$image=http://latex.codecogs.com/gif.latex?\bar{A}.\hat{a}_{B}=|\bar{A}||\hat{a}_{B}|Cos\Theta%20_{AB}&hash=54e78dea6b850d6b0b2e718150b68446$
so
$image=http://latex.codecogs.com/gif.latex?A_{B}=|\bar{A}|Cos\Theta%20_{AB}=|\bar{A}||\hat{a}_{B}|Cos\Theta%20_{AB}=\bar{A}.\hat{a}_{B}=|\bar{A}|.\hat{a}_{B}.\hat{a}_{B}=\frac{%28|\bar{A}|.\bar{B}%29.\bar{B}}{|B|^{2}}&hash=24881cd8c867fdf77990fbc0c02ca890$

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

6. Mar 2, 2013

### 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

7. Mar 3, 2013

### 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
$image=http://latex.codecogs.com/gif.latex?A_{B}=ACos\Theta%20_{AB}.\hat{a}_{B}&hash=f6497c7d973452af4b888b2bbcf31302$