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Show that Dot-Product is Distributive

  1. May 2, 2009 #1
    PLEASE Skip to post #14 after reading problem statement; I am trying to solve this without using components

    1. The problem statement, all variables and given/known data
    Griffith's E&M problem 1.1. I feel good about my life.

    Using the definition [itex]\vec{A}\cdot\vec{B} =AB\cos\theta[/itex] show that the dot product is distributive when

    (a) the 3 vectors are coplanar
    (b)the general case



    Okay then for part (a) I have started like this:

    Let A B & C be 3 coplanar vectors. Let [itex]\theta[/itex] be the angle between A & B; let [itex]\phi[/itex] be between B & C and let [itex]\alpha[/itex] be between A & C

    then

    [tex]\vec{A}\cdot(\vec{B}+\vec{C})=|\vec{A}||\vec{B+C}|\cos\gamma[/tex]

    where gamma is the angle between A and (B+C)

    ...now I am a little confused, i want to write that this implies

    [tex]\vec{A}\cdot(\vec{B}+\vec{C})=(AB+AC)\cos\gamma[/tex]

    but I am not sure if that is correct. And if it is, where to go from here?
     
    Last edited: May 3, 2009
  2. jcsd
  3. May 2, 2009 #2
    I also drew this and came up with the four relationships seen in the box. I am not sure if this helps me :smile:
    Picture1-30.png
     
  4. May 2, 2009 #3
    Let [tex]\vec{A} = (a_x,a_y)[/tex] etc. What is [tex]\vec{A} \cdot \vec{B}[/tex] in terms of [tex]a_x, a_y, b_x [/tex] and [tex] b_y[/tex]?
     
  5. May 2, 2009 #4
    I thought of using components, but the problem specifically asks to use the definition provided.

    [itex]
    \vec{A}\cdot\vec{B} =AB\cos\theta[/itex]
     
  6. May 3, 2009 #5
    Well the intermediate step would rely on the definition. Plus it's easy to generalize.
     
  7. May 3, 2009 #6

    [tex]\vec{A} \cdot \vec{B}=(a_xb_x+a_yb_y)[/tex]

    [tex]\Rightarrow (\vec{A} \cdot \vec{B})\cdot\vec{C}=(a_xb_xc_x+a_yb_yc_x)[/tex]

    Right?
     
  8. May 3, 2009 #7
    Not sure what this intermediate step is? I could do the same for [itex] \vec{A} \cdot (\vec{B}\cdot\vec{C})[/itex] and show that they are equal w/out using the definition
     
  9. May 3, 2009 #8
    Don't you mean [tex](\vec{A} + \vec{B})\cdot\vec{C}[/tex]?:confused:
     
    Last edited: May 3, 2009
  10. May 3, 2009 #9
    The intermediate step is to say that [tex]\vec{A} \cdot \vec{B}=(a_xb_x+a_yb_y)[/tex] which is easier to work with in this case. You don't need to say [itex] \vec{A} \cdot (\vec{B}\cdot\vec{C})[/itex] is equal to anything and in fact you can't because the dot product is only for vectors.
     
  11. May 3, 2009 #10
    Sorry, I was working on another problem at the same time :smile:

    One minute....
     
  12. May 3, 2009 #11
    okay...I still don't see how this:

    [tex](\vec{A} + \vec{B})\cdot\vec{C}=(a_x+b_x)c_x+(a_y+b_y)c_y=a_xc_x+b_xc_x+a_yc_y+b_yc_y=\vec{A}\cdot\vec{B}+\vec{B}\cdot\vec{C}[/tex]

    uses the definition of dot-product [itex]\vec{A}\cdot\vec{B} =AB\cos\theta[/itex]

    Sorry :confused:
     
  13. May 3, 2009 #12
    To prove that

    [tex]\vec{A}\cdot\vec{B}=a_xb_x+a_yb_y[/tex]

    you have to use the definition. If the angle of [tex]\vec{A}[/tex] is [tex]\alpha [/tex] and the angle of [tex]\vec{B}[/tex] is [tex]\beta[/tex] then


    [tex]\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos{(\alpha-\beta)}[/tex]

    [tex]=|\vec{A}||\vec{B}|(\cos{\alpha}\cos{\beta}+sin{\alpha}\sin{\beta})[/tex]

    [tex]=|\vec{A}|\cos{\alpha}|\vec{B}|\cos\beta+|\vec{A}|sin{\alpha}|\vec{B}|\sin{\beta}[/tex]

    [tex]=a_xb_x+a_yb_y[/tex]
     
  14. May 3, 2009 #13
    This is an interesting approach. Thanks qntty!

    The only thing is that the question is asked (in the text) before the component-wise definition of the Dot-Product is even given.

    This leads me to wonder if there is a way to show that the dot-product is distributive without ever going into components.

    I think that my original approach might lead somewhere. You have given me an idea to modify it though. I like how you used an absolute reference frame when naming your angles, instead of the way I named them (i.e., wrt the vectors themselves).

    I will play around with this a little, but it is really difficult for me to see a way to do this w/out components coming into play.

    Thanks again :smile:
     
  15. May 3, 2009 #14
    Okay, scratch that last post. I am going to stick with my original way of labeling angles (i.e., wrt the vectors themselves, not the reference frame)

    Here is my new diagram:

    Picture3-13.png

    Now, since I already know that the dot product distributes then I know that I am shooting for

    [tex](\vec{A} + \vec{B})\cdot\vec{C}=\vec{A}\cdot\vec{B}+\vec{ B}\cdot\vec{C}[/tex]

    So the end result should be

    something=[itex]|A||B|\cos\beta+|B||C|\cos\gamma[/itex]

    From my diagram I have

    [tex](\vec{A} + \vec{B})\cdot\vec{C}=|A||B+C|\cos\phi[/tex]

    This leads me to believe that I must write [itex]\phi[/itex] in terms of [itex]\beta[/itex] and [itex]\gamma[/itex]


    From my diagram, I also have the relationships:

    [itex]\phi=\beta+s[/itex]
    [itex]\phi=(\beta+\gamma)-r[/itex]

    Beta and gamma are 'known' and r, s, and phi are unknown....

    Seems like I am one equation short of doing something. :smile:

    Any ideas?
     
    Last edited: May 3, 2009
  16. May 3, 2009 #15

    gabbagabbahey

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    Hint: [itex]|\vec{B}+\vec{C}|=[/itex]___?
     
  17. May 3, 2009 #16
    I don't know without using components :confused:
     
  18. May 3, 2009 #17

    gabbagabbahey

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    Try drawing the tail of the vector C at the head of the vector B and draw in the vector B+C. Draw all the angles in and use the law of cosines.
     
  19. May 3, 2009 #18
    If I do it this way, I think that I will need to go back to defining the angle wrt a datum instead of from vector to vector... other wise, i do not know what angles to write in.
     
  20. May 3, 2009 #19

    gabbagabbahey

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    Extend a dotted line along the direction of B and label [itex]\gamma[/itex] the same way as before...you should quickly see that one of the angles in your triangle is [itex]\pi-\gamma[/itex]:wink:
     
  21. May 3, 2009 #20
    Photo1-2.jpg


    Hmmm. I don't see how an angle inside my triangle is pi-gamma.

    Maybe I'm cooked ...
     
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