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.(adsbygoogle = window.adsbygoogle || []).push({});

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 lets 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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# Homework Help: Dot product of basis vectors in orthogonal coordinate systems

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