Confused about dot product multiplication

  • #1

NoahsArk

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TL;DR Summary
Dot Product Multiplication Meaning
I'm confused about what we are really measuring when taking the dot product of two vectors. When we say we are measure "how much one vector points in the direction of the other", that description is not clear. At first I thought it meant how much of a shadow one vector casts on another and I thought this was the same as finding the adjacent side of a triangle where we'd just multiply cosine by the hypotenuse. But I'm not sure what we are doing in using the dot product formula where we multiply corresponding vector points and then add the products together. E.g. take vectors A and B where A = (3,0) and B = (6,8). Had B been instead = to (3,4) then B would be like the hypotenuse of a right 3,4,5 triangle forming over A where A is the adjacent (base) side of the triangle. When B = (6,8) it's like we doubled the length of B from when it was (3,4). The dot product of A and B is now (3x6) + (0x8) = 18.

What exactly does 18 represent? B does not go 6 times further to the right then A did. It only went twice as far to the right. Why are we even multiplying the two together? Thanks
 
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  • #2
The dot product is an interesting vector operation. It can be used to get the projection of one vector upon another.

##|A|cos(\alpha)## is the projection of A on the vector B

similarly ##|B|cos(\alpha)## is the projection of B on the vector A

where ##\alpha## is the angle between vectors A and B

The dot product is defined as ##A \cdot B = |A||B|cos(\alpha)## where its now easy to see how you can get the projections from the dot product by dividing by either |A| or |B|.

Projections are useful in physics where you have a force vector A and some object which is constrained to move along B and so you want to find what force is acting on the object. A good example is a block sliding down a ramp with A being the vertical force of gravity pulling of the sliding block and with B pointing downward along the ramp which constrains the object.
 
  • #3
@jedishrfu thanks. I understand visually what we are doing when we add vectors or when we scale vectors because both of these operations can be shown visually on a graph. I do not understand, visually, what it is we are doing when we multiply vectors.
 
  • #4
NoahsArk said:
@jedishrfu thanks. I understand visually what we are doing when we add vectors or when we scale vectors because both of these operations can be shown visually on a graph. I do not understand, visually, what it is we are doing when we multiply vectors.
I'm not sure I have a visual model of the dot product. The dot product of unit vectors gives the cosine of the angle between them. The proof of that is quite important. That gives the two forms of the dot product:
$$\vec a \cdot \vec b = a_1b_1 +a_2b_2+a_3b_3 = |\vec a||\vec b|\cos \theta$$This is one of the most important concepts in all of mathematics. But, perhaps, it is quite abstract in its conception?
 
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  • #5
NoahsArk said:
When we say we are measure "how much one vector points in the direction of the other", that description is not clear
See if this helps. Consider 2 vectors ##\vec A## and ##\vec B## where ##A=2## and ##B=3##. (I'm using the convention that a vector's name with no arrow on top represents the vector's length.)

Their dot product (DP) has:
- a maximum value of 6 when the vectors point in the same direction;
- a value of 0 when the vectors are perpendicular;
- a minimum value of -6 when the vectors point in opposite directions.

So, knowing the possible range of the DP [-6, 6], the more positive the DP is, the more the vectors point in the same direction. This gives you a ‘feel’ for the relative directions - which may be helpful in some situations.

NoahsArk said:
I thought it meant how much of a shadow one vector casts on another and I thought this was the same as finding the adjacent side of a triangle where we'd just multiply cosine by the hypotenuse.
The DP's value is the length of one vector's 'shadow' multiplied by the other vector's length. For example...

The usual basic physics example of a DP is calculating work done (##W##) by a (conservative) force ##\vec F## for a given displacement ##\vec d##. Then ##W = \vec F \cdot \vec d##.

One way to understand this is to note that (for angle ##\alpha## between ##\vec F## and ##\vec d##), the component of ##\vec F## in the direction of ##\vec d## is ##F \cos \alpha## (that's the 'shadow's length').

It is only this component that does work, so ##(F \cos \alpha) \times d## is the work done and this is the DP, ##\vec F \cdot \vec d##.

(With a bit of calculus this can be extended to non-conservative forces.)

NoahsArk said:
But I'm not sure what we are doing in using the dot product formula where we multiply corresponding vector points and then add the products together.
It' not intuitively obvious that this approach should give these same result as using the 'cosine form'. The proof can be done with some manipulation of vectors and use of the cosine rule. E.g. see discussion here: https://math.stackexchange.com/questions/2380217/why-are-the-two-dot-product-definitions-equal

But there's no simple geometric intuition that I know of.
 
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  • #6
@PeroK you showed the two ways of finding the dot product which I have recently learned- i.e. either by multiplying the corresponding units and then adding the products or by multiplying the magnitude of the vectors and then multiplying by the cosine of the angle between them.

PeroK said:
The dot product of unit vectors gives the cosine of the angle between them.

I looked up why this is the case, and, if I'm not mistaken, it's only the case when we are dealing unit vectors. On a re-read, you did specify that we are dealing with unit vectors. I must be missing something, though. If all we are trying to do is find the cosine, there is the normal way introduced in trigonometry for finding the cosine. That leads back to the question about what the dot product is actually measuring?
 
  • #7
If you think of a 1d space there isn't much distinction between a vector and a signed number. So multiplication of ordinary numbers can be seen as some kind of operation on 1d vectors. Can we generalise that to n dimensional vectors? Possibly. The only constraint we can export from 1d space is that if the vectors point in the same direction the product should be ##|\vec A||\vec B|## and if they point in opposite directions it should be ##-|\vec A||\vec B|##. So one possible generalisation is ##|\vec A||\vec B|\cos\theta##, and that one turns out to be useful.

It should be noted that the inner product turns out to generalise well to curved spaces, including the curved spacetimes of general relativity. In fact, it's pretty much the way the metric tensor of such spaces provides a notion of angle.
 
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  • #8
Here’s a more detailed explanation of dot product that may help.

I feel you are seeing it but somehow refusing to see it because it’s so obvious maybe because you’re thinking of the dot product of two vectors has to be an area of some sort.

https://betterexplained.com/articles/vector-calculus-understanding-the-dot-product/

Also the dot has dimensions of area when used with units of measure but to get the projections of one vector on another you need to divide it by the length of the other vector making the result a projection with a length unit of measure.
 
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  • #9
Steve4Physics said:
The DP's value is the length of one vector's 'shadow' multiplied by the other vector's length.

I tried to draw an example of a situation where it would be useful to do this. Say you have Alice throwing a ball into the air at an angle of 60 degrees. The ball's path is vector A and has a magnitude of 8. Say Bob is standing at a distance of 12 meters across from Alice and the line between Alice and Bob is Vector B. The "shadow" cast by A onto B will be 4 (8 times cosine of 60 degrees). To get the dot product, though, we now need to multiply 4 times 12 to get 48, and 48 seems like an arbitrary number representing no piece of information that I can see a use for.

If we want to find the total distance that the ball traveled by the time it got to Bob we could divide 12 by the cosine of 60 degrees to get 24. Where the number 48 comes in I don't see.

I also thought about the Work example you gave and looked into it and found an example of someone pushing a box. The force they push the box with is one vector and the distance the box is moving is another. I didn't understand this example either because if you are pushing a box you will be pushing it straight in front of you so the direction of the force will be the same as the direction the box is moving in. Thanks
 
  • #10
Ibix said:
So multiplication of ordinary numbers can be seen as some kind of operation on 1d vectors. Can we generalise that to n dimensional vectors?
Take a simple example of two vectors A and B where A is a ball travelling and has the coordinates 3,0. Say B is another vector representing another ball traveling and also has coordinate of 3,0. The dot product here is 9, but what does the 9 even mean? Likewise, what would it even mean to say that the second ball traveled 9 units in the direction of the first ball (we know it only traveled a distance of 3 horizontal units). Or, is this just a bad example that doesn't illustrate the usefulness of the dot product?
 
  • #11
My favorite math animation fb feed hasn't a lot to say either ...



It is and it remains the definition of the cosine of an angle.
 
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  • #12
NoahsArk said:
The force they push the box with is one vector and the distance the box is moving is another. I didn't understand this example either because if you are pushing a box you will be pushing it straight in front of you so the direction of the force will be the same as the direction the box is moving in.
Maybe this illustration will help:
1707587314975.png
 
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  • #13
NoahsArk said:
I tried to draw an example of a situation where it would be useful to do this. Say you have Alice throwing a ball into the air at an angle of 60 degrees. The ball's path is vector A and has a magnitude of 8.
The ball's path is not a vector, it is a curve - in fact it's a parabola if air resistance is negligible.

The ball's position (relative to the start point say) can be represented by a vecctor (it's called a position vector) but this vector changes continuously in both size and direction as the ball follows its curved path.

The ball's velocity is a vector. For example it could initially be 8m/s at 60º to the horizntal. But the ball's velocity also changes in both magnitude and direction through the flight.

A vector product is not useful/relevant in this situation.

NoahsArk said:
Say Bob is standing at a distance of 12 meters across from Alice and the line between Alice and Bob is Vector B. The "shadow" cast by A onto B will be 4 (8 times cosine of 60 degrees). To get the dot product, though, we now need to multiply 4 times 12 to get 48, and 48 seems like an arbitrary number representing no piece of information that I can see a use for.
No idea what you are doing!

NoahsArk said:
If we want to find the total distance that the ball traveled by the time it got to Bob we could divide 12 by the cosine of 60 degrees to get 24. Where the number 48 comes in I don't see.
By total distance do you mean the arc length or the straight line distance?

But in neither case is the dot product relevant.

NoahsArk said:
I also thought about the Work example you gave and looked into it and found an example of someone pushing a box. The force they push the box with is one vector and the distance the box is moving is another. I didn't understand this example either because if you are pushing a box you will be pushing it straight in front of you so the direction of the force will be the same as the direction the box is moving in.
If you are pushing (or pulling) horizontally the angle is 0 and cos(0) =1.

But you can easily exert a non-horizontal force. E.g push on a small box with a force of 100N with your arms at (say) 60º to the horizontal. The horizontal component of the force you apply is 100cos(60º) = 50N. If you push the box 3m horizotnally, the work done by you is 50N x 3m =150J.

Edit: Aha, @Hill beat me to it using a nice diagram.
 
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  • #14
NoahsArk said:
@PeroK you showed the two ways of finding the dot product which I have recently learned. That leads back to the question about what the dot product is actually measuring?
Five points about the dot product.

You should find it useful. It crops up all the time in mathematics.

It has some important physical applications. For example, the work done by a force through a displacement.

It does not depend on your choice of x,y,z axes. The dot product is invariant.

It generalises to advanced mathematics and physics. You'll find it at the heart of linear algebra, differential geometry, General Relativity and Quantum Mechanics.

Don't try to avoid abstraction. "We do not use abstraction in order to be vague, but in order to be absolutely precise. "(Edsger Dijkstra, computer scientist)
 
  • #15
NoahsArk said:
Take a simple example of two vectors A and B where A is a ball travelling and has the coordinates 3,0. Say B is another vector representing another ball traveling and also has coordinate of 3,0. The dot product here is 9, but what does the 9 even mean?
You have taken the dot product of 2 equal vectors. This gives the square of each vector's length. So the length of each vector is √9 = 3.

In fact this a standard way of finding the length of a vector: you find the dot product of the vector with itself and then take the square root.

E.g. ##\vec A = <3,4>##. Find its length (magnitude), ##A##.
##\vec A \cdot \vec A = 3\times3 + 4\times4 =25##
##A = \sqrt {25} = 5##

But note you can't just take the dot product of 2 random vectors and expect it to mean something. There has to be a reason for choosing to use the dot product.

NoahsArk said:
Likewise, what would it even mean to say that the second ball traveled 9 units in the direction of the first ball (we know it only traveled a distance of 3 horizontal units). Or, is this just a bad example that doesn't illustrate the usefulness of the dot product?
Nothing travelled 9 units. You have forgotten to take the square root.

Remember, if you take a dot product of 2 (say) position vectors, you are effectively multiplying a distance by a distance. This gives an area. Taking the square root takes you back to something that's a distance.
 
  • #16
I will add this video as a link here.
 
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  • #17
NoahsArk said:
Take a simple example of two vectors A and B where A is a ball travelling and has the coordinates 3,0. Say B is another vector representing another ball traveling and also has coordinate of 3,0. The dot product here is 9, but what does the 9 even mean?
What does it mean to multiply two velocities? Why would you do that? A velocity with itself, on the way to ##m|\vec v|^2/2##, sure. Or you might want the component of my velocity parallel to yours, which would be ##\vec v\cdot\hat{\vec u}##. But what would you expect the ptoduct of two velocities to mean?

On the other hand you might be interested in the component of velocity perpendicular to a force ##\vec F##, which would be ##\vec v-\hat{\vec F}(\hat{\vec F}\cdot\vec v)##. There are lots of applications.
 
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  • #18
NoahsArk said:
When we say we are measure "how much one vector points in the direction of the other", that description is not clear.
Might be clearer with some simple examples.

Consider two vectors in the plane: ##\vec u =<1, 0>## and ##\vec v =<0, 1>##. These are unit vectors in the directions, respectively, of the x and y axes. These two vectors are perpendicular to each other, so clearly neither one points in the direction of the other. ##\vec u \cdot \vec v = 1 \cdot 0 + 1 \cdot 0 = 0##. Or if you like, ##\vec u \cdot \vec v = |\vec u||\vec v| \cos(\pi/2) = 1 \cdot 1 \cdot 0 = 0##.

OTOH, if there are two vectors whose directions are fairly close to each other, then the angle between the two vectors will be small (close to 0), so their dot product will be close to the product of the magnitudes of the two vectors.
 
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  • #19
OP I think you’re getting too caught up in the analogy between multiplication of regular numbers and vectors. There isn’t one necessarily.

The dot product is just a useful operation for projection. And while you contended that something to the effect of “why can’t we just use regular trigonometry instead?”…….this for me becomes hard to visualize in 3 dimensions.

Furthermore the dot product comes from the Law of Cosines.
 
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  • #20
With the force example it makes more sense to me I think. Before, I was multiplying two kinds of vectors, but it didn't make sense to multiply those kinds of vectors together. Since I don't have a background in physics, though, I wanted to clarify: Say a rope is pulling a box at an angle. The rope is latched onto the box from above and is going up at an angle into the hand of the person pulling. In the example I saw for this the box moves in a horizontal path, but would pulling the box this way cus it to either lift in the air or do a wheelie?
 
  • #21
NoahsArk said:
With the force example it makes more sense to me I think. Before, I was multiplying two kinds of vectors, but it didn't make sense to multiply those kinds of vectors together. Since I don't have a background in physics, though, I wanted to clarify: Say a rope is pulling a box at an angle. The rope is latched onto the box from above and is going up at an angle into the hand of the person pulling. In the example I saw for this the box moves in a horizontal path, but would pulling the box this way cus [you mean 'cause'!] it to either lift in the air or do a wheelie?
We're considering only the simple case - the box slides without tilting. In this case you can use the dot product to calculate the work done as described above.

If the block starts to tilt, it's a different problem. To avoid confusion, stick with the simple case.
 
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  • #22
@Steve4Physics are there any more mundane examples of how dot product multiplication works that you can think of which don't involve forces?
 
  • #23
NoahsArk said:
@Steve4Physics are there any more mundane examples of how dot product multiplication works that you can think of which don't involve forces?
Does finding a vector space basis in which every pair of vectors is perpendicular to each other and of unit length count? It is the core concept why we can even speak of angles and lengths!
 
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  • #24
jedishrfu said:
Here’s a more detailed explanation of dot product that may help.

I feel you are seeing it but somehow refusing to see it because it’s so obvious maybe because you’re thinking of the dot product of two vectors has to be an area of some sort.

https://betterexplained.com/articles/vector-calculus-understanding-the-dot-product/

Also the dot has dimensions of area when used with units of measure but to get the projections of one vector on another you need to divide it by the length of the other vector making the result a projection with a length unit of measure.
In the bannanas and oranges example, it made me re-think what can qualify as a vector itself. It seems the first vector in the bannas and oranges example consists of 2 coordinates- one is a rule for multiplying bananas, and the other is a rule for multiplying oranges. Can we have a vector where both coordinates represent some operation like that? It seems like this would make more sense if we were applying that vector to some other vector that consisted of a certain number of bananas and oranges. In the example, though, the second vector also consists of two coordinates which each represent a rule on how many times to multiply the two fruits.

Assume the first vector was (3,3) (meaning triple the number of bananas and triple the number of oranges), and the second vector was (2,4) (meaning 2 bananas and 4 oranges). The result of the dot product would be (3x2) + (3x4) = 18 where 18 consists of 6 bananas and 12 oranges. But can the result of a dot product, like 18 in this case, be comprised of different kinds of things (like bananas and oranges), or does the result of a dot product always have to represent the same information (like in the calculation of Work, the answer of the dot product was always a certain number of joules and not a number that represented joules plus something else like in the banana orange example). Thanks
 
  • #25
fresh_42 said:
Does finding a vector space basis in which every pair of vectors is perpendicular to each other and of unit length count? It is the core concept why we can even speak of angles and lengths!

I meant some kind of word problem example using a specific real world application to illustrate how the dot product works.
 
  • #26
NoahsArk said:
@Steve4Physics are there any more mundane examples of how dot product multiplication works that you can think of which don't involve forces?

I’m going to elaborate on what I think @fresh_42 is saying.

Finding an orthogonal basis for ANY vector space allows you to decompose (find a component representation for) for any vector in that space.

The abstract generalization of the “dot product” is called the “inner product” and one of the most useful applications of this refers to “Fourier series” where you try to decompose a complicated signal/waveform in terms of simpler signals.
 
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  • #27
NoahsArk said:
@Steve4Physics are there any more mundane examples of how dot product multiplication works that you can think of which don't involve forces?
Try a Google search! But a couple that spring to mind are...

If you are designing a computer game, you might want to know the angular separation between two objects as viewed by a player. If you know the objects' position-vectors relative to the player, you can work out the angle. This then allows you to display the objects the correct distance apart on the screen, allowing for how far away each is (i.e. to have the correct perspective).

There are various physics examples in electromagentism but you probably don't know enough physics yet.

Power = force x velocity in vector form is ##P=\vec F \cdot \vec v##. [Edit -just realised that has force.]
 
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  • #28
NoahsArk said:
I meant some kind of word problem example using a specific real world application to illustrate how the dot product works.
Find any application that involves the law of cosines, e.g. Pythagoras.
 
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  • #29
I could be wrong. But don't all intuitive examples involving both physics and dot product that are also simple, involve forces?
Steve4Physics said:
There are various physics examples in electromagentism but you probably don't know enough physics yet.

Power = force x velocity in vector form is ##P=\vec F \cdot \vec v##. [Edit -just realised that has force.]
 
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  • #30
Another example of dot products (integrated) is surface integrals for Gauss’ Law in electrostatics.
 
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  • #31
Given a gradient of a function, ##\vec \nabla f##, and a vector ##\vec v##, the dot product, ##\vec \nabla f \cdot \vec v##, is the directional derivative of ##f## in ##\vec v##.
 
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  • #32
Aurelius120 said:
I could be wrong. But don't all intuitive examples involving both physics and dot product that are also simple, involve forces?
You could be right. But presumably it depends upon what is classified as intuitive/simple.

For example Gauss's law is fairly intuituve/simple after studying basic electromagnetism. It involves the dot product of the electric field vector and the 'area vector'. (But in fact force is hidden in there - because the electric field vector depends on electric force.)

Edit: Aha. @PhDeezNutz beat me to it.
 
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  • #33
NoahsArk said:
I meant some kind of word problem example using a specific real world application to illustrate how the dot product works.
Suppose a company sells five different products, call them A, B, C, D, and E. Suppose also that during a month, there were, respectively 25, 30, 15, 20, and 40 units sold, with per-unit revenues of $100, $120, $85, $140, and $135.
If the number sold is ##\overrightarrow{N_{sold}} = <25, 30, 15, 20, 40>## and the per-unit revenue is ##\vec R = <100, 120, 85, 140, 135>##, then the total revenue is given by ##\overrightarrow{N_{sold}} \cdot \vec R##. The units of this number would be dollars.
 
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  • #34
NoahsArk said:
I meant some kind of word problem example using a specific real world application to illustrate how the dot product works.
There must be a parable that covers your situation. Mathematics is a mountain. You have to start climbing to make progress. You can only sit at the bottom looking up for so long. You must have made a decision at some point to attempt this mountain, so you ought to get on with it.
 
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