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The 3+1-Lorentz transformation is \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot} \mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,t \tag{01a}\label{01a}\\ c\,t^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,t\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathbf{x}}{c}\right) \tag{01b}\label{01b}\\ \gamma_{\mathrm u} & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{01c}\label{01c} \end{align} and in differential form \begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathrm d\mathbf{x}\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \left(\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,\mathrm dt \tag{02a}\label{02a}\\ c\, \mathrm dt^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,\mathrm dt\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c}\right) \tag{02b}\label{02b} \end{align}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

For time-like vectors

Could you check in \eqref{02a} what would be the quantity $\mathrm d\mathbf{x}^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{03}\label{03} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert <c \tag{04}\label{04} \end{equation}

For space-like vectors

Could you check in \eqref{02b} what would be the quantity $\mathrm d t^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{c^2}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert^2}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{05}\label{05} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert >c \tag{06}\label{06} \end{equation}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

ADDENDUM

1. From \eqref{02a} and \eqref{03} if \begin{equation} \mathbf{X}\boldsymbol{=} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}\\ \\ x_4 \end{bmatrix}\,\quad \texttt{with } \left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{>}0 \tag{07}\label{07} \end{equation} is a time-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity \begin{equation} \boxed{\:\:\mathbf{u}\boldsymbol{=}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{08}\label{08} \end{equation} its space component is zero \begin{equation} \mathbf{X}'\boldsymbol{=} \begin{bmatrix} x'_1\\ x'_2\\ x'_3\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}'\\ \\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \boldsymbol{0}\\ \\ x'_4 \end{bmatrix} \,\quad \texttt{with } x'^{2}_4\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3 \tag{09}\label{09} \end{equation}

2. From \eqref{02b} and \eqref{05} if \begin{equation} \mathbf{X}\boldsymbol{=} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}\\ \\ x_4 \end{bmatrix}\,\quad \texttt{with } \left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{<}0 \tag{10}\label{10} \end{equation} is a space-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity \begin{equation} \boxed{\:\:\mathbf{u}\boldsymbol{=}\left\Vert\dfrac{\mathbf{x}}{x_4}\right\Vert^{\boldsymbol{-}2}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{11}\label{11} \end{equation} its time component is zero \begin{equation} \mathbf{X}'\boldsymbol{=} \begin{bmatrix} x'_1\\ x'_2\\ x'_3\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \vphantom{x'_1}\\ \mathbf{x}'\\ \vphantom{x'_3}\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \vphantom{x'_1}\\ \mathbf{x}'\\ \vphantom{x'_3}\\ 0\vphantom{x'_4} \end{bmatrix} \,\quad \texttt{with } \Vert\mathbf{x}'\Vert^2\boldsymbol{=}x^2_1\boldsymbol{+}x^2_2\boldsymbol{+}x^2_3\boldsymbol{-}x^2_4 \tag{12}\label{12} \end{equation}

The 3+1-Lorentz transformation is \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot} \mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,t \tag{01a}\label{01a}\\ c\,t^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,t\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathbf{x}}{c}\right) \tag{01b}\label{01b}\\ \gamma_{\mathrm u} & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{01c}\label{01c} \end{align} and in differential form \begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathrm d\mathbf{x}\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \left(\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,\mathrm dt \tag{02a}\label{02a}\\ c\, \mathrm dt^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,\mathrm dt\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c}\right) \tag{02b}\label{02b} \end{align}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

For time-like vectors

Could you check in \eqref{02a} what would be the quantity $\mathrm d\mathbf{x}^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{03}\label{03} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert <c \tag{04}\label{04} \end{equation}

For space-like vectors

Could you check in \eqref{02b} what would be the quantity $\mathrm d t^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{c^2}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert^2}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{05}\label{05} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert >c \tag{06}\label{06} \end{equation}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

ADDENDUM

1. From \eqref{02a} and \eqref{03} if \begin{equation} \mathbf{X}\boldsymbol{=} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}\\ \\ x_4 \end{bmatrix}\,\quad \texttt{with } \left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{>}0 \tag{07}\label{07} \end{equation} is a time-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity \begin{equation} \boxed{\:\:\mathbf{u}\boldsymbol{=}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{08}\label{08} \end{equation} its space component is zero \begin{equation} \mathbf{X}'\boldsymbol{=} \begin{bmatrix} x'_1\\ x'_2\\ x'_3\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}'\\ \\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \boldsymbol{0}\\ \\ x'_4 \end{bmatrix} \,\quad \texttt{with } x'^{2}_4\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3 \tag{09}\label{09} \end{equation}

2. From \eqref{02b} and \eqref{05} if \begin{equation} \mathbf{X}\boldsymbol{=} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}\\ \\ x_4 \end{bmatrix}\,\quad \texttt{with } \left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{<}0 \tag{10}\label{10} \end{equation} is a space-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity \begin{equation} \boxed{\:\:\mathbf{u}\boldsymbol{=}\left\Vert\dfrac{\mathbf{x}}{x_4}\right\Vert^{\boldsymbol{-}2}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left\Vert\dfrac{\mathbf{x}}{x_4}\right\Vert^{\boldsymbol{-}2}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{11}\label{11} \end{equation} its time component is zero \begin{equation} \mathbf{X}'\boldsymbol{=} \begin{bmatrix} x'_1\\ x'_2\\ x'_3\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \vphantom{x'_1}\\ \mathbf{x}'\\ \vphantom{x'_3}\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \vphantom{x'_1}\\ \mathbf{x}'\\ \vphantom{x'_3}\\ 0\vphantom{x'_4} \end{bmatrix} \,\quad \texttt{with } \Vert\mathbf{x}'\Vert^2\boldsymbol{=}x^2_1\boldsymbol{+}x^2_2\boldsymbol{+}x^2_3\boldsymbol{-}x^2_4 \tag{12}\label{12} \end{equation}

The 3+1-Lorentz transformation is \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot} \mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,t \tag{01a}\label{01a}\\ c\,t^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,t\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathbf{x}}{c}\right) \tag{01b}\label{01b}\\ \gamma_{\mathrm u} & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{01c}\label{01c} \end{align} and in differential form \begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathrm d\mathbf{x}\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \left(\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,\mathrm dt \tag{02a}\label{02a}\\ c\, \mathrm dt^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,\mathrm dt\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c}\right) \tag{02b}\label{02b} \end{align}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

For time-like vectors

Could you check in \eqref{02a} what would be the quantity $\mathrm d\mathbf{x}^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{03}\label{03} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert <c \tag{04}\label{04} \end{equation}

For space-like vectors

Could you check in \eqref{02b} what would be the quantity $\mathrm d t^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{c^2}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert^2}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{05}\label{05} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert >c \tag{06}\label{06} \end{equation}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

ADDENDUM

1. From \eqref{02a} and \eqref{03} if \begin{equation} \mathbf{X}\boldsymbol{=} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}\\ \\ x_4 \end{bmatrix}\,\quad \texttt{with } \left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{>}0 \tag{07}\label{07} \end{equation} is a time-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity \begin{equation} \boxed{\:\:\mathbf{u}\boldsymbol{=}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{08}\label{08} \end{equation} its space component is zero \begin{equation} \mathbf{X}'\boldsymbol{=} \begin{bmatrix} x'_1\\ x'_2\\ x'_3\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}'\\ \\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \boldsymbol{0}\\ \\ x'_4 \end{bmatrix} \,\quad \texttt{with } x'^{2}_4\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3 \tag{09}\label{09} \end{equation}

2. From \eqref{02b} and \eqref{05} if \begin{equation} \mathbf{X}\boldsymbol{=} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}\\ \\ x_4 \end{bmatrix}\,\quad \texttt{with } \left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{<}0 \tag{10}\label{10} \end{equation} is a space-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity \begin{equation} \boxed{\:\:\mathbf{u}\boldsymbol{=}\left\Vert\dfrac{\mathbf{x}}{x_4}\right\Vert^{\boldsymbol{-}2}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{11}\label{11} \end{equation} its time component is zero \begin{equation} \mathbf{X}'\boldsymbol{=} \begin{bmatrix} x'_1\\ x'_2\\ x'_3\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}'\\ \\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}'\\ \\ 0 \end{bmatrix} \,\quad \texttt{with } \Vert\mathbf{x}'\Vert^2\boldsymbol{=}x^2_1\boldsymbol{+}x^2_2\boldsymbol{+}x^2_3\boldsymbol{-}x^2_4 \tag{12}\label{12} \end{equation}

The 3+1-Lorentz transformation is \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot} \mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,t \tag{01a}\label{01a}\\ c\,t^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,t\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathbf{x}}{c}\right) \tag{01b}\label{01b}\\ \gamma_{\mathrm u} & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{01c}\label{01c} \end{align} and in differential form \begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathrm d\mathbf{x}\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \left(\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,\mathrm dt \tag{02a}\label{02a}\\ c\, \mathrm dt^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,\mathrm dt\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c}\right) \tag{02b}\label{02b} \end{align}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

For time-like vectors

Could you check in \eqref{02a} what would be the quantity $\mathrm d\mathbf{x}^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{03}\label{03} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert <c \tag{04}\label{04} \end{equation}

For space-like vectors

Could you check in \eqref{02b} what would be the quantity $\mathrm d t^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{c^2}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert^2}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{05}\label{05} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert >c \tag{06}\label{06} \end{equation}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

ADDENDUM

1. From \eqref{02a} and \eqref{03} if \begin{equation} \mathbf{X}\boldsymbol{=} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}\\ \\ x_4 \end{bmatrix}\,\quad \texttt{with } \left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{>}0 \tag{07}\label{07} \end{equation} is a time-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity \begin{equation} \boxed{\:\:\mathbf{u}\boldsymbol{=}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{08}\label{08} \end{equation} its space component is zero \begin{equation} \mathbf{X}'\boldsymbol{=} \begin{bmatrix} x'_1\\ x'_2\\ x'_3\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}'\\ \\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \boldsymbol{0}\\ \\ x'_4 \end{bmatrix} \,\quad \texttt{with } x'^{2}_4\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3 \tag{09}\label{09} \end{equation}

2. From \eqref{02b} and \eqref{05} if \begin{equation} \mathbf{X}\boldsymbol{=} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}\\ \\ x_4 \end{bmatrix}\,\quad \texttt{with } \left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{<}0 \tag{10}\label{10} \end{equation} is a space-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity \begin{equation} \boxed{\:\:\mathbf{u}\boldsymbol{=}\left\Vert\dfrac{\mathbf{x}}{x_4}\right\Vert^{\boldsymbol{-}2}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{11}\label{11} \end{equation} its time component is zero \begin{equation} \mathbf{X}'\boldsymbol{=} \begin{bmatrix} x'_1\\ x'_2\\ x'_3\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \vphantom{x'_1}\\ \mathbf{x}'\\ \vphantom{x'_3}\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \vphantom{x'_1}\\ \mathbf{x}'\\ \vphantom{x'_3}\\ 0\vphantom{x'_4} \end{bmatrix} \,\quad \texttt{with } \Vert\mathbf{x}'\Vert^2\boldsymbol{=}x^2_1\boldsymbol{+}x^2_2\boldsymbol{+}x^2_3\boldsymbol{-}x^2_4 \tag{12}\label{12} \end{equation}

The 3+1-Lorentz transformation is \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot} \mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,t \tag{01a}\label{01a}\\ c\,t^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,t\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathbf{x}}{c}\right) \tag{01b}\label{01b}\\ \gamma_{\mathrm u} & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{01c}\label{01c} \end{align} and in differential form \begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathrm d\mathbf{x}\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \left(\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,\mathrm dt \tag{02a}\label{02a}\\ c\, \mathrm dt^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,\mathrm dt\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c}\right) \tag{02b}\label{02b} \end{align}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

For time-like vectors

Could you check in \eqref{02a} what would be the quantity $\mathrm d\mathbf{x}^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{03}\label{03} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert <c \tag{04}\label{04} \end{equation}

For space-like vectors

Could you check in \eqref{02b} what would be the quantity $\mathrm d t^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{c^2}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert^2}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{05}\label{05} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert >c \tag{06}\label{06} \end{equation}

The 3+1-Lorentz transformation is \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot} \mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,t \tag{01a}\label{01a}\\ c\,t^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,t\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathbf{x}}{c}\right) \tag{01b}\label{01b}\\ \gamma_{\mathrm u} & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{01c}\label{01c} \end{align} and in differential form \begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathrm d\mathbf{x}\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \left(\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,\mathrm dt \tag{02a}\label{02a}\\ c\, \mathrm dt^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,\mathrm dt\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c}\right) \tag{02b}\label{02b} \end{align}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

For time-like vectors

Could you check in \eqref{02a} what would be the quantity $\mathrm d\mathbf{x}^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{03}\label{03} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert <c \tag{04}\label{04} \end{equation}

For space-like vectors

Could you check in \eqref{02b} what would be the quantity $\mathrm d t^{\boldsymbol{\prime}}$ if in this same equation you replace \begin{equation} \mathbf{u}\boldsymbol{\longrightarrow}\dfrac{c^2}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert^2}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt} \tag{05}\label{05} \end{equation} under the assumption \begin{equation} \left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert >c \tag{06}\label{06} \end{equation}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

ADDENDUM

1. From \eqref{02a} and \eqref{03} if \begin{equation} \mathbf{X}\boldsymbol{=} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}\\ \\ x_4 \end{bmatrix}\,\quad \texttt{with } \left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{>}0 \tag{07}\label{07} \end{equation} is a time-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity \begin{equation} \boxed{\:\:\mathbf{u}\boldsymbol{=}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{08}\label{08} \end{equation} its space component is zero \begin{equation} \mathbf{X}'\boldsymbol{=} \begin{bmatrix} x'_1\\ x'_2\\ x'_3\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}'\\ \\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \boldsymbol{0}\\ \\ x'_4 \end{bmatrix} \,\quad \texttt{with } x'^{2}_4\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3 \tag{09}\label{09} \end{equation}

2. From \eqref{02b} and \eqref{05} if \begin{equation} \mathbf{X}\boldsymbol{=} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}\\ \\ x_4 \end{bmatrix}\,\quad \texttt{with } \left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{<}0 \tag{10}\label{10} \end{equation} is a space-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity \begin{equation} \boxed{\:\:\mathbf{u}\boldsymbol{=}\left\Vert\dfrac{\mathbf{x}}{x_4}\right\Vert^{\boldsymbol{-}2}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{11}\label{11} \end{equation} its time component is zero \begin{equation} \mathbf{X}'\boldsymbol{=} \begin{bmatrix} x'_1\\ x'_2\\ x'_3\\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}'\\ \\ x'_4 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \\ \mathbf{x}'\\ \\ 0 \end{bmatrix} \,\quad \texttt{with } \Vert\mathbf{x}'\Vert^2\boldsymbol{=}x^2_1\boldsymbol{+}x^2_2\boldsymbol{+}x^2_3\boldsymbol{-}x^2_4 \tag{12}\label{12} \end{equation}