# Vector image on another vector

by baby_1
Tags: image, vector
 P: 38 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?
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,706 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 $A$?
 P: 38 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)
P: 764

## Vector image on another vector

Then pick one and stick with it. Similarly, don't swap between upper and lower case.
 P: 38 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?
 P: 315 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
 P: 38 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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