Show that Dot-Product is Distributive

  • Thread starter Saladsamurai
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In summary, the dot product is distributive when the vectors are coplanar, and the general case is also 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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  • #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
 
  • #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]?
 
  • #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]
 
  • #5
Well the intermediate step would rely on the definition. Plus it's easy to generalize.
 
  • #6
qntty said:
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
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
Saladsamurai said:
[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
Saladsamurai said:
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
qntty said:
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
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
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
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
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
Hint: [itex]|\vec{B}+\vec{C}|=[/itex]___?
 
  • #16
gabbagabbahey said:
Hint: [itex]|\vec{B}+\vec{C}|=[/itex]___?

I don't know without using components :confused:
 
  • #17
Saladsamurai said:
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
gabbagabbahey said:
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
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
Photo1-2.jpg
Hmmm. I don't see how an angle inside my triangle is pi-gamma.

Maybe I'm cooked ...
 
  • #21
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
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
Saladsamurai said:
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
gabbagabbahey said:
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
 

1. What is the definition of a dot-product?

The dot-product, also known as the scalar product, is a mathematical operation that takes two vectors and returns a scalar value. It is calculated by multiplying the corresponding components of the two vectors and summing the results.

2. How is the dot-product used in vector algebra?

The dot-product is used to find the angle between two vectors, determine if the vectors are orthogonal (perpendicular), and to calculate the projection of one vector onto another.

3. What does it mean for the dot-product to be distributive?

When we say that the dot-product is distributive, it means that it follows the distributive property of multiplication. This means that the dot-product of two vectors multiplied by a third vector is equal to the dot-product of each individual vector multiplied by the third vector. In other words, (a + b) · c = a · c + b · c.

4. How can we prove that the dot-product is distributive?

The dot-product is distributive by definition. However, to prove it mathematically, we can use the properties of vector algebra. We can expand the dot-product of (a + b) · c to (a · c) + (b · c) and show that it is equal to a · c + b · c, which follows the distributive property.

5. Why is it important to understand the distributive property of the dot-product?

Understanding the distributive property of the dot-product is crucial in vector algebra as it allows us to simplify complex expressions and solve problems more efficiently. It also helps us to better understand the relationship between vectors and how they interact with each other in mathematical calculations.

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