I have a problem which I could not resolve. Consider a known single frequency plane wave coming from an arbitrary $(\theta, \phi)$ direction where $(\theta ,\phi)$ are the usual spherical coordinates. We know the time difference that are made by the wave with three given points below.
$B\left(\lambda, \frac{\pi}{2}, \frac{60\pi}{180}\right); $ $C\left(\lambda,\frac{\pi}{2},\frac{180\pi}{180}\right) ;$ $D\left(\lambda,\frac{\pi}{2},\frac{300\pi}{180}\right) ;$
I want to find the $\theta$ and $\phi$ of the wave in terms of the time difference. We know the speed of propagation as well so $\lambda$ which is wavelength is known .
Any help would be highly appreciated.
Usually one would proceed as follows: Let $N$ be the unit vector pointing in the direction where the plane wave comes from, let $u_1:=(C-B)/\|C-B\|$, $u_2:=(D-B)/\|D-B\|$ and $u_3:=(u_1\times u_2)/\|u_1\times u_2\|$. Then $N=n_1 u_1+n_2 u_2+n_3 u_3$ for some real numbers $n_1,n_2,n_3$. We have $k_1:=N\cdot u_1=\frac{v(t_B-t_C)}{\|C-B\|}$ and $k_2:=N\cdot u_2=\frac{v(t_B-t_D)}{\|D-B\|}$, where $v$ is the speed. We set $s:=u_1\cdot u_2$ and can compute $$ n_1=\frac{k_1-s k_2}{1-s^2}\quad\text{and}\quad n_2=\frac{k_2-s k_1}{1-s^2}. $$ Finally compute $n_3=\pm\sqrt{1-\|n_1u_1+n_2 u_2\|^2}$ and then you can convert the vector $N$ into spherical coordinates.
Evidently, there are conditions on $v$, $t_B-t_C$ and $t_D-t_C$ that guarantee that the data comes from a wave. The conditions can be resumed in the following inequality on $k_1,k_2$: $$ k_1^2+k_2^2-2s k_1 k_2+s^2\le 1. $$ Note that if $u_1$ and $u_2$ are orthogonal, we are requiring that $k_1^2+k_2^2\le 1$. The inequality guarantees that $\|n_1u_1+n_2 u_2\|\le 1$.