Dot product of basis vectors in orthogonal coordinate systems

In summary: I apologize for any confusion that may have arisen.In summary, the problem was that quasar987 was not able to find the angle between \hat r and \hat R in cylindrical coordinates, and was looking for help. HallsofIvy was able to find the angle by finding the scalar projection of \hat R on the x-y plane.
  • #1
FrogPad
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I'm doing a series of questions right now that is basically dealing with the dot and cross products of the basis vectors for cartesian, cylindrical, and spherical coordinate systems.

I am stuck on [itex] \hat R \cdot \hat r [/itex] right now.

I'll try to explain my work, and the problem I am running into as well as I can. It's somewhat difficult to do since I can't draw pictures, but...

Ok, first the notation being used is:

Cylindrical: [itex] (r,\phi,z) [/itex]
Spherical: [itex] (R, \phi, \theta ) [/itex]

First, the dot product of any two unit vectors, say [itex] \hat A [/itex] and [itex] \hat B [/itex] will be: [itex] \hat A \cdot \hat B = \cos \theta_{AB} [/itex]

So my trouble is finding the angle between [itex] \hat r [/itex] and [itex] \hat R [/itex]. I know that [itex] \phi [/itex] does not affect this angle, so it really comes down to [itex] \theta [/itex].

So let's say [itex] \theta = 0[/itex], then [itex] \theta_{rR} [/itex] would be 90 degrees right?

Now say [itex] \theta = 45\,\,deg [/itex], then [itex] \theta_{rR} [/itex] would be 45 degrees right?

Now say [itex] \theta = 135 \,\,deg [/itex], then does [itex] \theta_{rR} [/itex] equal 45 degrees?

I'm not sure how to deal with measuring the angle between the two vectors when [itex] \theta [/itex] causes [itex] \hat R [/itex] to point below the x-y plane.

I hope this makes sense. Any help would be nice. Thanks :)
 
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  • #2
It's probably easier to go at this problem from the point of view [itex]\vec{A}\cdot \vec{B}=A_xB_x+A_yB_y[/itex] rather than [itex]\vec{A}\cdot \vec{B}=ABcos(\theta_{AB})[/itex]

In the partcular case [tex]\hat{R}\cdot \hat{r}[/tex], try expressing [tex]\hat{R}[/tex] in cylindrical coordinates*. If you can do that, then the dot product will be obviously just the r-component of [tex]\hat{R}[/tex]!

*If you struggle too long on this, it is because you're not using the easiest approach.. I'll give you a hint if you need one.
 
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  • #3
The [itex]\hat r[/itex], in cylindrical coordinates, is just the projection of the [itex]\hat R[/itex], in spherical coordinates, projected onto the xy-plane. Since [itex]\phi[/itex] is the "co-latitude", the angle [itex]\hat R[/itex] makes with the z-axis, the angle you want, between [itex]\hat R[/itex] and [itex]\hat r[/itex] is the complement of that [itex]\frac{\pi}{2}-\phi[/itex].
 
  • #4
Thanks quasar987 for pointing me in the right direction. I didn't really struggle on the problem. I actually posted it, went up to the store and got some food, ate it and fell asleep :smile: I had every intention of working on it last night, but sleep got the best of me. Good looking out though :approve:

I ended up solving it (I think?) by finding the scalar projection of [itex] \hat R [/itex] on the x-y plane in the direction of [itex] \hat r [/itex], and the [itex] \hat z [/itex] component along the cylinder with another projection in the direction of [itex] \hat z [/itex].

I ended up with,

[tex] \hat R = \hat r \sin \theta + \hat z \cos \theta [/tex]
(in cylindrical coordinates)

Thus, the dot product [itex] \hat R \cdot \hat r = \sin \theta [/itex]

HallsofIvy: I appreciate the help :smile: I think what you are saying is a confirmation of what I found. Does everything look ok?
 
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  • #5
[itex]\sin\theta[/itex] is my answer too. (Or in cartesian coordinates, [itex]\sqrt{x^2+y^2}/\sqrt{x^2+y^2+z^2}[/itex])

Though if you want an honest answer to "Does everything look ok?", I'd answer that when you wrote...

[tex] \hat R = \hat r \sin \theta + \hat z \cos \theta [/tex]
(in cylindrical coordinates)

I'm not sure this is correct. In my understanding of the term "coordinates", what you wrote is [itex] \hat R [/itex] in spherical coordinates in the [itex]\hat{r},\hat{\phi},\hat{z}[/itex] (cylindrical) basis.

But I could easily be wrong! Anyone can comment on that?
 
  • #6
quasar987 said:
I'm not sure this is correct. In my understanding of the term "coordinates", what you wrote is LaTeX graphic is being generated. Reload this page in a moment. in spherical coordinates in the LaTeX graphic is being generated. Reload this page in a moment. (cylindrical) basis.

I'm very glad you expressed the concern in my wording. I see where you are coming from. It is definitely not accurate for me to say "in cylindrical coordinates" while the variable [itex] \theta [/itex] is in use for the basis vector.
 

What is the dot product of basis vectors in orthogonal coordinate systems?

The dot product of basis vectors in orthogonal coordinate systems is a mathematical operation that involves multiplying two vectors together and taking the sum of their products. It is commonly used in vector calculus and linear algebra to measure the angle between two vectors or to find the projection of one vector onto another.

How is the dot product calculated in orthogonal coordinate systems?

In orthogonal coordinate systems, the dot product is calculated by multiplying the corresponding components of the two vectors and then summing them up. For example, if the two vectors are A = (a1, a2, a3) and B = (b1, b2, b3), then the dot product is given by A ⋅ B = (a1 * b1) + (a2 * b2) + (a3 * b3).

What is the significance of the dot product in orthogonal coordinate systems?

The dot product has several important applications in orthogonal coordinate systems. It can be used to determine the angle between two vectors, the projection of one vector onto another, and the length of a vector. It is also used in many physical and engineering problems, such as calculating work and energy.

Can the dot product be negative in orthogonal coordinate systems?

Yes, the dot product can be negative in orthogonal coordinate systems. This occurs when the angle between the two vectors is greater than 90 degrees. In this case, the dot product represents the component of one vector that is perpendicular to the other vector.

How is the dot product related to the concept of orthogonality?

The dot product is closely related to the concept of orthogonality, which refers to two vectors being perpendicular to each other. In orthogonal coordinate systems, the dot product of two orthogonal vectors is equal to 0. This means that the two vectors are perpendicular to each other and do not share any common direction.

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