Help with caption from Kittel: Can we get phase angle difference from dot product? Can someone help me understand how the phase angle difference is just the dot product of k and r? Also, how do we get that the difference in phase angle is 2pir*sin(phi)/lambda?
 A: Generally, the local phase angle $\alpha$ of a wave described by
$$\psi=e^{i\vec k\cdot \vec r}=e^{i\alpha(\vec r)}$$
is of course
$$\alpha(\vec r)=\vec k\cdot \vec r$$
That is just the very definition of phase $\alpha$ of a plane wave (in stationary description, i.e. temporal phase $e^{-i\omega t}$ has been divided out of the wave).
Due to that, all the wave fronts (the lines that look like "stair steps" perpendicular to the peripheral rays of the beam in Kittel's figure) have the same phase. In mathematical terms:
$$\vec k\cdot \vec r=const.$$
defines a plane (or in this 2D figure, a line).
From that it is clear that the phase difference between $r_O$ and $r_{dV}$ is given by the quotient of the respective phase factors:
$$e^{i\delta\alpha} = \frac{e^{i\vec k \cdot \vec r_{dV}}}{e^{i\vec k \cdot \vec r_O}}=e^{i\vec k \cdot (\vec r_{dV}-\vec r_O)}=e^{i\vec k \cdot \vec r}$$
and hence the phase difference is
$$\delta\alpha=\vec k \cdot \vec r$$
Again, this is just the definition of the term "phase difference".
As to your second question: suppose that $\chi$ was the angle opposite to $\varphi$ next to the other leg of the right (in the sense of orthogonal) triangle. Then $\chi$ is the angle that corresponds to the dot product:
$$\vec k\cdot \vec r = kr\cos\chi=kr\cos (\pi/2-\varphi)=kr\sin \varphi$$
Since the angular wave number is defined by
$$k=2\pi/\lambda$$
(you want to have a phase difference of $2\pi$ if $r=\lambda$ in $e^{ikr}$, i.e. exactly one period of the wave after a distance of one wavelength), you finally get
$$\vec k\cdot \vec r=\frac{2\pi r}{\lambda}\sin \varphi$$
By the way: the same formulae are also useful for single or double slit diffraction.
