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Show that Dot-Product is Distributive

  • #1
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PLEASE Skip to post #14 after reading problem statement; I am trying to solve this without using components

Homework Statement


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?
 
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Answers and Replies

  • #2
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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
 
  • #3
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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]?
 
  • #4
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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]
 
  • #5
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Well the intermediate step would rely on the definition. Plus it's easy to generalize.
 
  • #6
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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]?

[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?
 
  • #7
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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
 
  • #8
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[tex](\vec{A} \cdot \vec{B})\cdot\vec{C}[/tex]
Don't you mean [tex](\vec{A} + \vec{B})\cdot\vec{C}[/tex]?:confused:
 
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  • #9
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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
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.
 
  • #10
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Don't you mean [tex](\vec{A} + \vec{B})\cdot\vec{C}[/tex]?:confused:
Sorry, I was working on another problem at the same time :smile:

One minute....
 
  • #11
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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:
 
  • #12
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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]
 
  • #13
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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:
 
  • #14
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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?
 
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  • #15
gabbagabbahey
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Hint: [itex]|\vec{B}+\vec{C}|=[/itex]___?
 
  • #16
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Hint: [itex]|\vec{B}+\vec{C}|=[/itex]___?
I don't know without using components :confused:
 
  • #17
gabbagabbahey
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I don't know without using components :confused:
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.
 
  • #18
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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.
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.
 
  • #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:
 
  • #20
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Photo1-2.jpg



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

Maybe I'm cooked ...
 
  • #21
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What if I drop a perpendicular from the tip of C to my extension of B.. then wouldn't the angle between B+C and C just be gamma?
 
  • #22
gabbagabbahey
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Basic Trig....look at the large angle adjacent to gamma...if gamma were zero you would simply have a straight line (pointing in the direction of B) and that angle would be 180 degrees or pi radians, in general that angle is (180 deg- gamma)
 
  • #23
gabbagabbahey
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What if I drop a perpendicular from the tip of C to my extension of B.. then wouldn't the angle between B+C and C just be gamma?
No, because B+C is not parallel to B
 
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  • #24
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Basic Trig....look at the large angle adjacent to gamma...if gamma were zero you would simply have a straight line (pointing in the direction of B) and that angle would be 180 degrees or pi radians, in general that angle is (180 deg- gamma)
Oh man :blushing: When I read your post, you said pi - gamma, but I was thinking pi/2 - gamma...pi = 180 ... my brain = pudding
 

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