refraction - Snell's law in vector form

Snell's law of refraction at the interface between 2 isotropic media is given by the equation: \begin{equation} n_1 \,\text{sin} \,\theta_1 = n_2 \, \text{sin}\,\theta_2 \end{equation} where $\theta_1$ is the angle of incidence and $\theta_2$ the angle of refraction. $n_1$ is the refractive index of the optical medium in front of the interface and $n_2$ is the refractive index of the optical medium behind the interface.

How can this be expressed in vector form: \begin{equation} n_1(\textbf{i} \times \textbf{n}) = n_2 (\textbf{t} \times \textbf{n}) \end{equation} where $\textbf{i}(i_\text{x}, i_\text{y}$ and $i_\text{z})$ and $\textbf{t}(t_\text{x}, t_\text{y}, t_\text{z})$ are the unit directional vector of the incident and transmitted ray respectively. $\textbf{n}(n_\text{x}, n_\text{y}, n_\text{n})$ is the unit normal vector to the interface between the two media pointing from medium 1 with refractive index $n_1$ into medium 2 with refractive index $n_2$.

Further more how can the Snell's law of refraction be expressed$^\text{1}$ in the following way? \begin{equation} t = \mu \textbf{i} + n\sqrt{1- \mu^2[1-(\textbf{ni})^2]} - \mu \textbf{n}(\textbf{ni}) \end{equation} Here $\mu = \dfrac{n_1}{n_2}$ and $\textbf{n}\textbf{i}= n_{\text{x}} i_{\text{x}} + n_{\text{y}}i_{\text{y}} + n_{\text{z}} i_{\text{z}}$ denotes the dot (scalar) product of vectors $\textbf{n}$ and $\textbf{i}$.

References:

1. Antonín Mikš and Pavel Novák, Determination of unit normal vectors of aspherical surfaces given unit directional vectors of incoming and outgoing rays: comment, 2012 Optical Society of America, page 1356