# Vector image on another vector

by baby_1
Tags: image, vector
 P: 77 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: 39,564 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: 77 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: 77 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 $A_{B}=ACos\Theta _{AB}.\hat{a}_{B}$