# Showing dot product is commutative, please

• mr_coffee

#### 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!

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

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.

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!

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

Done!

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

$\vec{A} \times \vec{B} = AB\sin(\theta) [/tex] [itex] \vec{B} \times \vec{A} = BA\sin(\theta) [/tex] [itex] \vec{A} \times \vec{B} = \vec{B} \times \vec{A} [/tex] or am i just tired? No, it doesn't. [itex]\vec{A}\times\vec{B}$ does not equal $ABsin(\theta )$.

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

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

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

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

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.

Astronuc said:
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. $| \vec{A} \times \vec{B}| = AB\sin(\theta) = BA\sin(\theta) = | \vec{B} \times \vec{A}| [/tex] That holds true, its just the direction that switches. Theta is the angle from A to B. If you switch A and B, theta becomes negative theta. cos(theta) = cos(-theta), so the dot product, which uses cosine, is commutative. On the other hand, sin(-theta) = -sin(theta), so the cross product, which uses sine, is anticommutative. JoAuSc said: Theta is the angle from A to B. If you switch A and B, theta becomes negative theta. cos(theta) = cos(-theta), so the dot product, which uses cosine, is commutative. I agree with that argument... On the other hand, sin(-theta) = -sin(theta), so the cross product, which uses sine, is anticommutative. ...but not this one. Or maybe I just don't understand it. I agree that the cross product is anticommutative, but I do not see how it depends on the oddness of the sine function. When we say [itex]|\vec{A}\times\vec{B}|=ABsin(\theta)$, 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:

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

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