Evaluating Scalar Field in Spherical Coordinates

[bowlbase]

Homework Statement

Evaluate the scalar field $f(r, \theta, \phi)= \mid 2\hat{r}+3\hat{\phi} \mid$ in spherical coords.

Homework Equations

Law of Cosines?

$\mid \vec{A} + \vec{B} \mid = \sqrt{A^2+B^2+2ABCos(\theta)}$

The Attempt at a Solution

I'm not sure the law of cosines is exactly what I'm suppose to use but so far it is the only thing that I've found that seems to fit the way the problem is presented.

If this is the correct way then:
$\mid 2\hat{r}+3\hat{\phi} \mid=\sqrt{2^2+3^2+12cos(\theta)}$


Am I doing this correctly?

 

[pasmith]

Homework Statement

Evaluate the scalar field $f(r, \theta, \phi)= \mid 2\hat{r}+3\hat{\phi} \mid$ in spherical coords.

Homework Equations

Law of Cosines?

$\mid \vec{A} + \vec{B} \mid = \sqrt{A^2+B^2+2ABCos(\theta)}$

The \theta which occurs in this expression is not the spherical coordinate \theta. This is obviously going to be a source of confusion, so you need to find a different letter for the angle between \vec A and \vec B, such as \alpha:
<br /> \|\vec{A} + \vec{B}\| = \sqrt{A^2 + B^2 + 2AB\cos \alpha}<br />

However I think the intended method is to start from
<br /> \|2 \hat r + 3 \hat \phi\|^2 = (2 \hat r + 3 \hat \phi) \cdot (2 \hat r + 3 \hat \phi)<br />

The Attempt at a Solution

I'm not sure the law of cosines is exactly what I'm suppose to use but so far it is the only thing that I've found that seems to fit the way the problem is presented.

If this is the correct way then:
$\mid 2\hat{r}+3\hat{\phi} \mid=\sqrt{2^2+3^2+12cos(\alpha)}$


Am I doing this correctly?

Now all you need is the angle \alpha between \hat r and \hat \phi.

 

[bowlbase]

I made a mistake. The question should be: $\mid 2\hat{r} +3\hat{\theta} \mid$

 

[bowlbase]

I'm not sure how I would go about finding the angle between the two vectors in spherical. I could probably switch them to Cartesian but is there a simpler way via spherical?

 

[vanhees71]

Spherical coordinates are orthogonal coordinates! Thus \hat{r} and \hat{\phi} are orthogonal with unit norm...

 

[bowlbase]

So just 90° then?