Proof of vector property in space

[chwala]

TL;DR

See attached

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?

 

[wrobel]

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"

 

[chwala]

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,

$c(\vec u⋅\vec v)=(c⋅\vec v)⋅\vec u$

 

[wrobel]

$$c(u,v)=(cu,v),\quad (u,v)=(v,u)\Longrightarrow c(u,v)=c(v,u)=(cv,u)=(u,cv)$$

 

[jedishrfu]

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.

 

[chwala]

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...

 

[wrobel]

...

$(c\vec v_1, c\vec v_2, c\vec v_3)⋅(\vec u_1,\vec u_2,\vec u_3)$

what's this?

 

[chwala]

Amended

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,

$c(\vec u⋅\vec v)=(c⋅\vec v)⋅\vec u$

 

[Structure seeker]

Yes that's completely fine due to your proof above it of the commutative property.

 

[Steve4Physics]

@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.