I wanted to express angle between two 3D vectors pointing in arbitrary direction say $\vec{r}$ and $\vec{R}$. If I take the z axis along any other direction (other than the direction of $\vec{R}$ and
$\vec{r}$). Both vectors will have polar coordinates. $$r,\theta,\phi$$ and $$R', \theta' , \phi'$$ respectively for both $\vec{r}$ and $\vec{R}$. Now how do I express the angle between the vectors in terms of theses polar angles. Can some one suggest some good references to understand better about these concepts ?
Vectors – Expressing Angle Between Two Vectors in 3D Using Spherical Polar Coordinates
analytic geometryspherical trigonometryspherical-geometryvector analysisvectors
Related Solutions
If $a - \sqrt{2} b$ is a unit vector, then the inner product \begin{equation*} \langle a - \sqrt{2}b, a - \sqrt{2}b \rangle = ||a - \sqrt{2} b||^{2} = 1. \end{equation*}
My suggestion would be to work out the inner product, first using bilinearity, and then using the relationship between the inner product of two vectors, their magnitudes, and their angles.
So we have $AB=(-3,5,-1)$ and $OA=(2,2,3)$. Now use, for example, the formula
$$\frac{a\cdot b}{|a||b|}=\cos\theta,$$ where $\theta$ is the angle between $a$ and $b$.
Edit: I think the confusion here is what we really mean by angle between two vectors. Since vectors "float" in space, we can imagine placing them so that their tails touch. Then the angle between these two physical interpretations of vectors is defined as the angle between $a$ and $b$ as vectors. There is no ambiguity about where the vector is pointing because we place them tail-to-tail so both arrows always point away.
Here is a pretty bad picture I made in photoshop. Even though b "points away" from the angle, it doesn't really affect anything.
Best Answer
Using the dot product, the angle between $\vec{r}$ and $\vec{R}$ is
$\cos^{-1} \left( \frac{\vec{r}.\vec{R}}{|\vec{r}||\vec{R}|}\right) = \cos^{-1} \left( \frac{\vec{r}.\vec{R}}{rR'}\right)$
Calculting the dot product of $\vec{r}$ and $\vec{R}$ is simplest if you convert them to Cartesian co-ordinates first:
$\vec{r} = (r \sin \theta \cos \phi, r \sin \theta \sin \phi, r \cos \theta)$
$\vec{R} = (R' \sin \theta' \cos \phi', R' \sin \theta' \sin \phi', R' \cos \theta')$