Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Showing dot product is commutative, need help please

  1. Sep 12, 2005 #1
    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!
     
  2. jcsd
  3. Sep 12, 2005 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    What is a.b, what is b.a? (in terms of components? No idea what those dashes are doing there)
     
  4. Sep 12, 2005 #3
    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.
     
  5. Sep 12, 2005 #4

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    You need either this form:

    or this one:

    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!
     
  6. Sep 13, 2005 #5
    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?
     
  7. Sep 13, 2005 #6

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No, it doesn't. [itex]\vec{A}\times\vec{B}[/itex] does not equal [itex]ABsin(\theta )[/itex].
     
  8. Sep 13, 2005 #7
    THanks for the help guys!! I do like your proof alot better then mine :)
     
  9. Sep 13, 2005 #8

    Astronuc

    User Avatar
    Staff Emeritus
    Science Advisor

    [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".
     
  10. Sep 13, 2005 #9
    Astronuc, are you saying his proof doesn't work or are you replying to whozum?
     
  11. Sep 13, 2005 #10

    Astronuc

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  12. Sep 13, 2005 #11

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Um...I did write that. :confused:

    That is exactly the same as my proof. You just wrote it horizontally instead of vertically. :smile:
     
  13. Sep 13, 2005 #12
    [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.
     
  14. Sep 13, 2005 #13
    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.
     
  15. Sep 13, 2005 #14

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I agree with that argument...

    ...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].
     
    Last edited: Sep 13, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook