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对齐：在对齐问题中使用对齐

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对齐：在对齐问题中使用对齐

我希望前两部分看起来align好像是align环境中仅有的两个项目，其余部分在等号处对齐，并由一些空间隔开以表示不同的等式。

现在编译后的效果如下：

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因此，我想将前两个对齐如下：

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最后我想强调的是

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\documentclass{article}
\usepackage{mathtools}
\usepackage{amssymb}
\begin{document}
\begin{alignat*}{2}
    \vec{x}_1(t) &=
    \begin{pmatrix}
      x_1^{(1)}(t)\\
      x_2^{(1)}(t)
    \end{pmatrix} &&{}=
    \begin{pmatrix}
      X_1^{(1)}\cos(\omega_1t + \phi_1)\\
      r_1X_1^{(1)}\cos(\omega_1t + \phi_1)
    \end{pmatrix}\\
    \vec{x}_2(t) &=
    \begin{pmatrix}
      x_1^{(2)}(t)\\
      x_2^{(2)}(t)
    \end{pmatrix} &&{}=
    \begin{pmatrix}
      X_1^{(2)}\cos(\omega_2t + \phi_2)\\
      r_2X_1^{(2)}\cos(\omega_2t + \phi_2)
    \end{pmatrix}\\
    r_1 &=
    \begin{aligned}
      \frac{3k - m\omega_1^2}{2k} && \qquad
      X_1^{(1)} &&{}= \frac{1}{r_2 - r_1}\bigg[(r_2x_1(0) - x_2(0))^2 +
      \frac{(-r_2\dot{x}_1(0) + \dot{x}_2(0))^2}{\omega_1^2}\bigg]^{1/2}
    \end{aligned}\\
    r_2 &=
    \begin{aligned}
      \frac{3k - m\omega_2^2}{2k} && \qquad
      X_1^{(2)} &&{}= \frac{1}{r_2 - r_1}\bigg[(-r_1x_1(0) + x_2(0))^2 +
      \frac{(r_1\dot{x}_1(0) - \dot{x}_2(0))^2}{\omega_2^2}\bigg]^{1/2}
    \end{aligned}\\
    \phi_1 &=
    \begin{aligned}
      \arctan\bigg[\frac{-r_2\dot{x}_1(0) + \dot{x}_2(0)}
      {\omega_1^2(r_2x_1(0) - x_2(0))}\bigg] && \qquad
      \phi_2 &&{}= \arctan\bigg[\frac{r_1\dot{x}_1(0) - \dot{x}_2(0)}
      {\omega_2^2(-r_1x_1(0) + x_2(0))}\bigg]
    \end{aligned}
  \end{alignat*}
\end{document}

答案1

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\documentclass{article}
\usepackage{mathtools}
\usepackage{amssymb}
\begin{document}
\begin{alignat*}{2}
    \vec{x}_1(t) &=
    \begin{pmatrix}
      x_1^{(1)}(t)\\
      x_2^{(1)}(t)
    \end{pmatrix} =
    \mathrlap{\begin{pmatrix}
      X_1^{(1)}\cos(\omega_1t + \phi_1)\\
      r_1X_1^{(1)}\cos(\omega_1t + \phi_1)
    \end{pmatrix}}\\
    \vec{x}_2(t) &=
    \begin{pmatrix}
      x_1^{(2)}(t)\\
      x_2^{(2)}(t)
    \end{pmatrix} =
    \mathrlap{\begin{pmatrix}
      X_1^{(2)}\cos(\omega_2t + \phi_2)\\
      r_2X_1^{(2)}\cos(\omega_2t + \phi_2)
    \end{pmatrix}}\\
    r_1 &=
      \frac{3k - m\omega_1^2}{2k} &
      X_1^{(1)} &= \frac{1}{r_2 - r_1}\bigg[(r_2x_1(0) - x_2(0))^2 +
      \frac{(-r_2\dot{x}_1(0) + \dot{x}_2(0))^2}{\omega_1^2}\bigg]^{1/2}
\\
    r_2 &=
      \frac{3k - m\omega_2^2}{2k} &
      X_1^{(2)} &= \frac{1}{r_2 - r_1}\bigg[(-r_1x_1(0) + x_2(0))^2 +
      \frac{(r_1\dot{x}_1(0) - \dot{x}_2(0))^2}{\omega_2^2}\bigg]^{1/2}
\\
    \phi_1 &=
      \arctan\bigg[\frac{-r_2\dot{x}_1(0) + \dot{x}_2(0)}
      {\omega_1^2(r_2x_1(0) - x_2(0))}\bigg] &
\phi_2 &= \arctan\bigg[\frac{r_1\dot{x}_1(0) - \dot{x}_2(0)}
      {\omega_2^2(-r_1x_1(0) + x_2(0))}\bigg]
  \end{alignat*}
\end{document}

答案2

我不明白所提议的布局和随之而来的各种符号的对齐方式应该达到什么目的=。为了避免给人留下各种方程式以可能不是预期的方式相互关联的印象，我会简化布局，使用单一align*环境并使用\qquad语句在需要的地方插入一些水平空格。请注意，使用\biggl[而不是仅仅\bigg[遵循这两个\arctan语句很重要；如果省略l(mathopen) 说明符，LaTeX 将插入不适当的空格。

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath} % for "align*" and "pmatrix" environments
\usepackage{newpxtext}             % Palatino text font
\usepackage[euler-digits]{eulervm} % Euler math font
\begin{document}
\begin{align*}
    \vec{x}_1(t) &=
    \begin{pmatrix}
      x_1^{(1)}(t)\\
      x_2^{(1)}(t)
    \end{pmatrix} =
    \begin{pmatrix}
      X_1^{(1)}\cos(\omega_1t + \phi_1)\\
      r_1X_1^{(1)}\cos(\omega_1t + \phi_1)
    \end{pmatrix}\\
    \vec{x}_2(t) &=
    \begin{pmatrix}
      x_1^{(2)}(t)\\
      x_2^{(2)}(t)
    \end{pmatrix} =
    \begin{pmatrix}
      X_1^{(2)}\cos(\omega_2t + \phi_2)\\
      r_2X_1^{(2)}\cos(\omega_2t + \phi_2)
    \end{pmatrix}\\[2ex]  % additional vertical space
    r_1 &=
      \frac{3k - m\omega_1^2}{2k} \qquad
      X_1^{(1)} = \frac{1}{r_2 - r_1}\biggl[(r_2x_1(0) - x_2(0))^2 +
      \frac{(-r_2\dot{x}_1(0) + \dot{x}_2(0))^2}{\omega_1^2}\biggr]^{1/2}\\
    r_2 &=
      \frac{3k - m\omega_2^2}{2k}  \qquad
      X_1^{(2)} = \frac{1}{r_2 - r_1}\biggl[(-r_1x_1(0) + x_2(0))^2 +
      \frac{(r_1\dot{x}_1(0) - \dot{x}_2(0))^2}{\omega_2^2}\biggr]^{1/2}\\[2ex] % additional vertical space
    \phi_1 &=
      \arctan\biggl[\frac{-r_2\dot{x}_1(0) + \dot{x}_2(0)}
      {\omega_1^2(r_2x_1(0) - x_2(0))}\biggr] \qquad
      \phi_2 = \arctan\biggl[\frac{r_1\dot{x}_1(0) - \dot{x}_2(0)}
      {\omega_2^2(-r_1x_1(0) + x_2(0))}\biggr]
  \end{align*}
\end{document}

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