Dot product calculates what exactly?

[mjt99]

I have a pretty general question about vectors. The scalar product of two vectors is a calculation of what exactly?

For example, if the units of two vectors are meters, the resultant dot product would be a meters squared. So if it's a measurement of area, what area exactly?

I'm very confused about this!

Thanks for the help.

 

[lukeseed]

The scalar product (or dot product) of two vectors is a scalar that is proportional to the length of the projection of the first vector onto the second vector. An example would be a vector A which is 2 units long on the positive x-axis and a vector B which is at a 45 degree angle to the x-axis in the positive y direction with a length of 1. B dot A is then the distance on the x-axis that vector B covers which is 1*cos45. Does that help?

 

[mgb_phys]

In physics it's the amount of a quantity in a particular direction.
So the dot product of force and direction = work.

 

[mjt99]

The scalar product (or dot product) of two vectors is a scalar that is proportional to the length of the projection of the first vector onto the second vector. An example would be a vector A which is 2 units long on the positive x-axis and a vector B which is at a 45 degree angle to the x-axis in the positive y direction with a length of 1. B dot A is then the distance on the x-axis that vector B covers which is 1*cos45. Does that help?

I'm still very confused on this topic. In your example above, wouldn't the dot product be:
(1)(2)*cos45

 

[mathman]

I'm still very confused on this topic. In your example above, wouldn't the dot product be:
(1)(2)*cos45

You are right. The projection idea is valid when both vectors are of unit length.

 

[mjt99]

Thanks to all for the replies. There's a couple of really good tutorials on this topic on YouTube as well.

 

[maze]

For example, if the units of two vectors are meters, the resultant dot product would be a meters squared. So if it's a measurement of area, what area exactly?

You have to be a little careful about what the "units" of a vector are. The length of a vector is in meters, but you can't really say that the vector itself has units of meters. Perhaps a better way to think about it is this:

u.v = (u₁i+u₂j+u₃k).(v₁i+v₂j+v₃k)
= u₁v₁i.i + u₁v₂i.j + u₁v₃i.k + u₂u₁j.i + u₂u₂j.j + u₂u₃j.k + u₃u₁k.i + u₃u₂k.j + u₃u₃k.k

where the coefficients u₁, u₂, etc are just numbers, and the units are included via the unit vectors i, j, k. In other words, i represents the vector of length 1 meter in the x-direction, j represents the vector of length 1 meter in the y-direction, etc. Then it is taken as a physical principle that i.i = 1 meter, i.j = 0 meter, etc, and the formula becomes

u.v = 1 meter * (u₁v₁ + u₂u₂ + u₃u₃)

 

[Ben Niehoff]

You have to be a little careful about what the "units" of a vector are. The length of a vector is in meters, but you can't really say that the vector itself has units of meters. Perhaps a better way to think about it is this:

u.v = (u₁i+u₂j+u₃k).(v₁i+v₂j+v₃k)
= u₁v₁i.i + u₁v₂i.j + u₁v₃i.k + u₂u₁j.i + u₂u₂j.j + u₂u₃j.k + u₃u₁k.i + u₃u₂k.j + u₃u₃k.k

where the coefficients u₁, u₂, etc are just numbers, and the units are included via the unit vectors i, j, k. In other words, i represents the vector of length 1 meter in the x-direction, j represents the vector of length 1 meter in the y-direction, etc. Then it is taken as a physical principle that i.i = 1 meter, i.j = 0 meter, etc, and the formula becomes

u.v = 1 meter * (u₁v₁ + u₂u₂ + u₃u₃)

I don't think this works as a general principle. If both of the vectors have units of meters, then the dot product should have units of square meters. It's just that these square meters don't happen to measure an area (in a meaningful sense, anyway).

It's a bit more clear what is happening if the two vectors have different units, such as force (kg m / s^2) and distance (m). Then the dot product has units of energy (kg m^2 / s^2), consistent with the definition W = Fd.

Note, however, that torque is also measured in (kg m^2 / s^2)! Just because two quantities are expressed in the same units, does not necessarily mean they are measuring the same thing.

The difference between torque and work is that work measures force times parallel distance, while torque measures force times perpendicular distance.

In the same way, area (m^2) is a measure of distance times perpendicular distance, while the dot product (also m^2) measures distance times parallel distance. It has the units of area, but it is not the same thing as area.

 

[maze]

hmm actually you are right it should be square meters. The reason why it is not a problem is that ||x|| = sqrt(x.x). The key being the square root.

 

[robphy]

the units are included via the unit vectors i, j, k. In other words, i represents the vector of length 1 meter in the x-direction, j represents the vector of length 1 meter in the y-direction, etc. Then it is taken as a physical principle that i.i = 1 meter, i.j = 0 meter, etc, and the formula becomes

The unit vectors are dimensionless.
Suppose \vec r carries units of meters (m),
then
\hat r \equiv \frac{\vec r}{| \vec r |} = \frac{\vec r}{\sqrt{ \vec r \cdot \vec r}}
carries units of \frac{ [m]} {\sqrt{ [m][m]} }= \frac{ [m]} {\sqrt{ [m]^2} }=1.

The components carry the units...
F_x = \vec F \cdot \hat i has whatever units \vec F are measured in.

 

[lukeseed]

You are correct. Typo ;0)

 

[tiny-tim]

Welcome to PF!

… if the units of two vectors are meters, the resultant dot product would be a meters squared. So if it's a measurement of area, what area exactly?

Hi mjt99! Welcome to PF!

Yes, the dot product has dimensions of area, but it is a scalar (meaning that dot products obey the scalar law of addition).

As you know, the magnitude of the cross product a x b obviously represents a genuine area, of the parallelogram with sides a and b.

To find an area represented by a.b, think of them as matrices … so one must be a row matrix, and the other a column matrix …

in other words, the dot product is a^Tb or b^Ta.

Then one of the vectors is in "transpose space", in which it represents not a vector but a 2-form (or, in n-dimensional space, an (n-1)-form), which is essentially a plane segment (hyperplane segment) of a particular size, perpendicular to the original vector …

then a.b is sort-of a^T x b, the area between the vector a and the plane segment b^T.

 

[robphy]

To find an area represented by a.b, think of them as matrices … so one must be a row matrix, and the other a column matrix …

in other words, the dot product is a^Tb or b^Ta.

Then one of the vectors is in "transpose space", in which it represents not a vector but a 2-form (or, in n-dimensional space, an (n-1)-form), which is essentially a plane segment (hyperplane segment) of a particular size, perpendicular to the original vector …

then a.b is sort-of a^T x b, the area between the vector a and the plane segment b^T.

I don't think this construction works generally.

Suppose we compute a^Ta... i don't see any meaningful area.

Actually, a^T is really a 1-form (here formed with the help of a metric),
which can be visualized by a family of parallel hyperplanes with spacing inversely-proportional to the magnitude of a^T. The normals of the hyperplanes are parallel to the vector a, in the case of a Euclidean metric.

The dot-product a^Tb would then be "the number of http://www.google.com/search?q="bongs+of+a+bell""" for the number of successive hyperplanes of the 1-form a^T pierced by the vector b.
(For a^Ta, we get the squared-magnitude of a: if a is vector of length 4, a^T is a family of hyperplanes of spacing 1/4... the vector a pierces 16 hyperplanes of a^T.)

 

[n!kofeyn]

The best explanation of the dot product I've read is from the book A Student's Guide to Maxwell's Equations by Daniel Fleisch. Read pages 6-9 to get a fantastically clear explanation behind the physical significance of the dot product. It doesn't require much background. Here is the http://books.google.com/books?id=I-...i=xzvWSZmIFI_EM4-ylPcC&client=safari#PPA6,M1".

If you're interested in working some of the problems from the book using the dot product in electricity and magnetism, the author's http://www4.wittenberg.edu/maxwell/index.html" contains solutions split up into staggered hints and then the full solution.

 

[robphy]

A useful way to think about the dot-product is as a measure of "overlap".
The "projection" makes this more precise.