Showing dot product is commutative, please

[mr_coffee]

Hello everyone. I'm trying to show ab = ba. //communative property of dot product
This is what I have, is it enough to show this?
ab = (ax, ay, az) $$\dot$$ (bx, by, bz) = axbx + ayby + azbz;
a$$\dot$$b = (axi + ayj) (bxi + byj); //note: i and j are unit vectors
= axbx(i)(i) + ayby(j)(j) + axby(i)(j) + aybx(j)(i)
= axbx+ayby

b$$\dot$$a = (bx, by, bz) $$\dot$$ (ax,ay,az) = bxax + byay + bzaz;

b$$\dot$$a = (bxi + byj) (axi + ayj)
= bxax(i)(i) + byay(j)(j) + bxay(i)(j) + byax(j)(i)
= bxax + byay

Is this enough to show the dot product is communative?
Any help would be great if it isn't!

 

[matt grime]

What is a.b, what is b.a? (in terms of components? No idea what those dashes are doing there)

 

[mr_coffee]

a.b is the dot product, a and b are vectors. // that's just a symbol for comment in C++, out of habbit when I want to note somthing I put a //, sorry for the confusion.

 

[quantumdude]

You've got it, but I'm not sure of why you did it twice.

You need either this form:

ab = (ax, ay, az) (bx, by, bz) = axbx + ayby + azbz,

or this one:

ab = (axi + ayj) (bxi + byj),

but not both.

An even quicker way would be to define the dot product as $\vec{A}\cdot\vec{B}=ABcos(\theta)$.

Then:

$\vec{A}\cdot\vec{B}=ABcos(\theta)$
$\vec{A}\cdot\vec{B}=BAcos(\theta)$
$\vec{A}\cdot\vec{B}=\vec{B}\cdot\vec{A}$.

Done!

 

[whozum]

$\vec{A}\cdot\vec{B}=ABcos(\theta)$
$\vec{A}\cdot\vec{B}=BAcos(\theta)$
$\vec{A}\cdot\vec{B}=\vec{B}\cdot\vec{A}$.

Done!

Doesn't this imply that cross product is commutative then?

[itex] \vec{A} \times \vec{B} = AB\sin(\theta) [/tex]
[itex] \vec{B} \times \vec{A} = BA\sin(\theta) [/tex]
[itex] \vec{A} \times \vec{B} = \vec{B} \times \vec{A} [/tex]

or am i just tired?

 

[quantumdude]

No, it doesn't. $\vec{A}\times\vec{B}$ does not equal $ABsin(\theta )$.

 

[mr_coffee]

THanks for the help guys! I do like your proof a lot better then mine :)

 

[Astronuc]

$\vec{B}\cdot\vec{A}=BAcos(\theta)$ minor technicality

$\vec{A} \times \vec{B} = - \vec{B} \times \vec{A}$ the cross product is also known as "vector product".

 

[mr_coffee]

Astronuc, are you saying his proof doesn't work or are you replying to whozum?

 

[Astronuc]

My comment about $\vec{B}\cdot\vec{A}=BAcos(\theta)$ is purely editorial. I believe Tom meant to right that.

Another way to right the proof is $\vec{A}\cdot\vec{B}=ABcos(\theta)=BAcos(\theta)=\vec{B}\cdot\vec{A}$ using the commutative rule of scalars, A*B = B*A.

and I was responding to Whozum also.

 

[quantumdude]

My comment about $\vec{B}\cdot\vec{A}=BAcos(\theta)$ is purely editorial. I believe Tom meant to right that.

Um...I did write that.

Another way to right the proof is $\vec{A}\cdot\vec{B}=ABcos(\theta)=BAcos(\theta)=\vec{B}\cdot\vec{A}$ using the commutative rule of scalars, A*B = B*A.

That is exactly the same as my proof. You just wrote it horizontally instead of vertically.

 

[whozum]

[itex] | \vec{A} \times \vec{B}| = AB\sin(\theta) = BA\sin(\theta) = | \vec{B} \times \vec{A}| [/tex]

That holds true, its just the direction that switches.

 

[JoAuSc]

Theta is the angle from A to B. If you switch A and B, theta becomes negative theta. cos(theta) = cos(-theta), so the dot product, which uses cosine, is commutative. On the other hand, sin(-theta) = -sin(theta), so the cross product, which uses sine, is anticommutative.

 

[quantumdude]

Theta is the angle from A to B. If you switch A and B, theta becomes negative theta. cos(theta) = cos(-theta), so the dot product, which uses cosine, is commutative.

I agree with that argument...

On the other hand, sin(-theta) = -sin(theta), so the cross product, which uses sine, is anticommutative.

...but not this one.

Or maybe I just don't understand it. I agree that the cross product is anticommutative, but I do not see how it depends on the oddness of the sine function.

When we say $|\vec{A}\times\vec{B}|=ABsin(\theta)$, that's actually a little sloppy. It's understood that $\theta$ is chosen so that $sin(\theta)$ is nonnegative, because the left side of that equation is nonnegative. A more precise way to say it would be:

$|\vec{A}\times\vec{B}|=AB|sin(\theta )|$

or:

$|\vec{A}\times\vec{B}|=ABsin(\theta )$, $0\leq\theta\leq\frac{\pi}{2}$.