Showing dot product is commutative, need help please

  • Thread starter mr_coffee
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  • #1
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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) [tex]\dot[/tex] (bx, by, bz) = axbx + ayby + azbz;
a[tex]\dot[/tex]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[tex]\dot[/tex]a = (bx, by, bz) [tex]\dot[/tex] (ax,ay,az) = bxax + byay + bzaz;

b[tex]\dot[/tex]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!
 

Answers and Replies

  • #2
matt grime
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What is a.b, what is b.a? (in terms of components? No idea what those dashes are doing there)
 
  • #3
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a.b is the dot product, a and b are vectors. // thats just a symbol for comment in C++, out of habbit when I want to note somthing I put a //, sorry for the confusion.
 
  • #4
Tom Mattson
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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 [itex]\vec{A}\cdot\vec{B}=ABcos(\theta)[/itex].

Then:

[itex]\vec{A}\cdot\vec{B}=ABcos(\theta)[/itex]
[itex]\vec{A}\cdot\vec{B}=BAcos(\theta)[/itex]
[itex]\vec{A}\cdot\vec{B}=\vec{B}\cdot\vec{A}[/itex].

Done!
 
  • #5
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[itex]\vec{A}\cdot\vec{B}=ABcos(\theta)[/itex]
[itex]\vec{A}\cdot\vec{B}=BAcos(\theta)[/itex]
[itex]\vec{A}\cdot\vec{B}=\vec{B}\cdot\vec{A}[/itex].

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?
 
  • #6
Tom Mattson
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No, it doesn't. [itex]\vec{A}\times\vec{B}[/itex] does not equal [itex]ABsin(\theta )[/itex].
 
  • #7
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THanks for the help guys!! I do like your proof alot better then mine :)
 
  • #8
Astronuc
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[itex]\vec{B}\cdot\vec{A}=BAcos(\theta)[/itex] minor technicality

[itex] \vec{A} \times \vec{B} = - \vec{B} \times \vec{A} [/itex] the cross product is also known as "vector product".
 
  • #9
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Astronuc, are you saying his proof doesn't work or are you replying to whozum?
 
  • #10
Astronuc
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My comment about [itex]\vec{B}\cdot\vec{A}=BAcos(\theta)[/itex] is purely editorial. I believe Tom meant to right that.

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

and I was responding to Whozum also.
 
  • #11
Tom Mattson
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Astronuc said:
My comment about [itex]\vec{B}\cdot\vec{A}=BAcos(\theta)[/itex] is purely editorial. I believe Tom meant to right that.
Um...I did write that. :confused:

Another way to right the proof is [itex]\vec{A}\cdot\vec{B}=ABcos(\theta)=BAcos(\theta)=\vec{B}\cdot\vec{A}[/itex] 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. :smile:
 
  • #12
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[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.
 
  • #13
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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.
 
  • #14
Tom Mattson
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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)[/itex], that's actually a little sloppy. It's understood that [itex]\theta[/itex] is chosen so that [itex]sin(\theta)[/itex] is nonnegative, because the left side of that equation is nonnegative. A more precise way to say it would be:

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

or:

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