I Proof of vector property in space

  • I
  • Thread starter Thread starter chwala
  • Start date Start date
  • Tags Tags
    Proof
chwala
Gold Member
Messages
2,825
Reaction score
413
TL;DR Summary
See attached
1689350162836.png

My interest is on the associative property; is there anything wrong of showing and concluding proof by;

##c(\vec u⋅\vec v)=(c⋅\vec v)⋅\vec u.##

or are we restricted in the prose?
 
Last edited:
Physics news on Phys.org
Your equality contains three different types of multiplications: scalar by scalar; scalar by vector and a vector by vector. I do not think that it makes sense to call this "associative property"
 
wrobel said:
Your equality contains three different types of multiplications: scalar by scalar; scalar by vector and a vector by vector. I do not think that it makes sense to call this "associative property"
Did you understand my question? I am going through the pdf notes and my question is specifically on: the associative property.

##c(\vec u⋅\vec v)=(c⋅\vec u)⋅\vec v##

as indicated on my pdf notes...

Unless, you are of the opinion that the notes are not correct.

i am asking if the proof could be allowed to take this route

...

##c(\vec u⋅\vec v)=(cv_1, cv_2, cv_3)⋅(u_1,u_2, u_3)##

...

to end up with,

##c(\vec u⋅\vec v)=(c⋅\vec v)⋅\vec u##
 
Last edited:
$$c(u,v)=(cu,v),\quad (u,v)=(v,u)\Longrightarrow c(u,v)=c(v,u)=(cv,u)=(u,cv)$$
 
What is c a scalar or a vector?

On the LHS you use c as a scalar multiplied by the scalar inner product of U and V

But on the RHS, you've dotted c with the U vector.

Since there is no hat on c then it must be a scalar and the RHS operator usage is confusingly wrong.
 
  • Like
Likes Steve4Physics
jedishrfu said:
What is c a scalar or a vector?

On the LHS you use c as a scalar multiplied by the scalar inner product of U and V

But on the RHS, you've dotted c with the U vector.

Since there is no hat on c then it must be a scalar and the RHS operator usage is confusingly wrong.
I put arrows on ##u## and ##v## to indicate that they are vectors...if that is not correct then you may guide me. I did not get it right on latex, sorry for that confusion and i ought to have clearly stated that ##c## is a scalar and ##u,v## are vectors... now can we focus on my question? Thanks...
 
chwala said:
...

##(c\vec v_1, c\vec v_2, c\vec v_3)⋅(\vec u_1,\vec u_2,\vec u_3)##
what's this?
 
Amended
wrobel said:
what's this?
Amended. The correct position is as follows:

i am asking if the proof could be allowed to take this route

...

##c(\vec u⋅\vec v)=(cv_1, cv_2, cv_3)⋅(u_1,u_2, u_3)##

...

to end up with,

##c(\vec u⋅\vec v)=(c⋅\vec v)⋅\vec u##
 
Yes that's completely fine due to your proof above it of the commutative property.
 
  • #10
@chwala, for information, part of the confusion is that you have written expressions containing ##c⋅\vec u## and ##c⋅\vec v##. But these are wrong.

##c## is a scalar.
##⋅## in this context is the dot (inner) product of two vectors.
##\vec u## and ##\vec v## are vectors.

##c⋅\vec u## should be ##c\vec u##
##c⋅\vec v## should be ##c\vec v##

For example, in Post #8 you wrote "##c(\vec u⋅\vec v)=(c⋅\vec v)⋅\vec u##". This is wrong and should be written as ##c(\vec u⋅\vec v)=(c\vec v)⋅\vec u##.

Edit: @jedishrfu already pointed out (Post #5) this problem.
 
  • Like
  • Informative
Likes jedishrfu, Mark44 and chwala

Similar threads

Back
Top