Columb law (vetorial) $$ \vec{E}(x,y,z) = \frac{e \vec{r}}{4 \pi \epsilon_0 r^3}, $$
Biot-Savart (vectorial + some calculus simplification from $ c = \frac{1}{\sqrt{\epsilon \mu}}$) $$ \vec{B}(x,y,z) = \frac{\mu_0 e (\vec{v} \times \vec{r}) }{4 \pi r^3} =
\frac{(\vec{v} \times \vec{r}) q}{c^2 (4 \pi \epsilon_0 r^3)} = \frac{\vec{v} \times \vec{E}}{c^2} $$
The correct way to treat electromagnetism interaction in relativity is to use electromagnetic tensor $F^{\mu\nu}$
$$ F^\mu_\nu = \left( \begin{matrix} 0 & -E_x/c & -E_y/c & -E_z/c \ -E_x/c & 0 & -B_z & B_y \ -E_y/c & B_z & 0 & -B_x \ -E_z/c & -B_y & B_x & 0 \ \end{matrix} \right) , $$
Four-velocity vector $$ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}, d\tau = \frac{dt}{\gamma}$$
$$ U^\mu = \left ( \begin{matrix} \gamma c \ \gamma v_x \ \gamma v_y \ \gamma v_z \ \end{matrix}\right ) , $$
Relativistic "force" law
$$ \frac{d(m_{0}U^\mu)}{d\tau} = - e F^\mu_\nu U^\nu $$
Some notations $$ \beta_x = \frac{v_x}{c} , \beta_x = \frac{v_y}{c} , \beta_z = \frac{v_z}{c} , \beta = \frac{v}{c}, v = \sqrt{v_x^2 + v_y^2 + v_z^2} $$
Lorentz Boost in arbitrary direction: $$ \Lambda^\mu_\nu(v) = \left ( \begin{matrix} \gamma & -\beta_x\gamma & -\beta_y\gamma & -\beta_z\gamma \ -\beta_x\gamma & 1 + (\gamma - 1)\beta_x^2/\beta^2 & (\gamma - 1)\beta_x\beta_y/\beta^2 & (\gamma - 1)\beta_x\beta_z/\beta^2 \ -\beta_y\gamma & (\gamma - 1)\beta_y\beta_z/\beta^2 & 1 + (\gamma - 1)\beta_y^2/\beta^2 & (\gamma - 1)\beta_y\beta_z/\beta^2 \ -\beta_z\gamma & (\gamma - 1)\beta_z\beta_x/\beta^2 & (\gamma - 1)\beta_z\beta_y/\beta^2 & 1 + (\gamma - 1)\beta_z^2/\beta^2 \ \end{matrix} \right ) $$ Lowering + uppering indexes $$ \Lambda_\mu^\nu(v) = \eta_{\mu\alpha}\eta^{\nu\beta} \Lambda^\alpha_\beta(v) $$ $$ \Lambda_\mu^\nu(v) = \left ( \begin{matrix} \gamma & \beta_x\gamma & \beta_y\gamma & \beta_z\gamma \ \beta_x\gamma & 1 + (\gamma - 1)\beta_x^2/\beta^2 & (\gamma - 1)\beta_x\beta_y/\beta^2 & (\gamma - 1)\beta_x\beta_z/\beta^2 \ \beta_y\gamma & (\gamma - 1)\beta_y\beta_z/\beta 804