理解 tikz 的基础变化

Question 1

我认为你只是忘了转置，即反转变换矩阵，这是所谓的“主动”和“被动”转变关键字“transpose”已经被 StefanH 提及，但是由于某种原因，他似乎忘记了实现它，或者他通过“转置基础”表示其他意思。

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}[scale=2]

% Definition and projection on the screen of basis A
\coordinate (Ax) at (1,0);%,0);
\coordinate (Ay) at (0,1);%,0);
\coordinate (Az) at (0,0);%,1);

% Draw basis A
\draw[blue,-latex] (0,0,0) -- (Ax) node[shift={(0.1,0,0)}] {$e_x$};
\draw[red,-latex] (0,0,0) -- (Ay) node[shift={(0,0.1,0)}] {$e_y$};
\draw[green,-latex] (0,0,0) -- (Az) node[shift={(0,0,0.1)}] {$e_z$};

% Define angles
\pgfmathsetmacro\alpha{70}
\pgfmathsetmacro\beta{110}

% Rotation matrix from basis A (canonical) and basis B (R = Rx(-\alpha)*Rz(-\beta))
\pgfmathsetmacro\Rxx{cos(\beta)}
\pgfmathsetmacro\Rxy{-cos(\alpha)*sin(\beta)}
\pgfmathsetmacro\Rxz{sin(\alpha)*sin(\beta)}
\pgfmathsetmacro\Ryx{sin(\beta)}
\pgfmathsetmacro\Ryy{cos(\alpha)*cos(\beta)}
\pgfmathsetmacro\Ryz{-sin(\alpha)*cos(\beta)}
\pgfmathsetmacro\Rzx{0}
\pgfmathsetmacro\Rzy{sin(\alpha)}
\pgfmathsetmacro\Rzz{cos(\alpha)}

% Point X in basis A
\pgfmathsetmacro\xa{1}
\pgfmathsetmacro\ya{1}
\pgfmathsetmacro\za{1}

% Change basis from natural tikz (with fake perspective) to A
\begin{scope}[shift={(0,0,0)},x={(Ax)},y={(Ay)},z={(Az)}]

% Draw X in A
\draw[ultra thick,blue,-Circle] (0,0,0) -- (\xa,\ya,\za) 
node[left=2pt,font=\tiny] {$X_a$};

\end{scope}

% Point X in basis B (Xb = R*Xa) 
% <- transposed this matrix by replacing \Rxy with \Ryx etc.
\pgfmathsetmacro\xb{\Rxx*\xa+\Rxy*\ya+\Rxz*\za}
\pgfmathsetmacro\yb{\Ryx*\xa+\Ryy*\ya+\Ryz*\za}
\pgfmathsetmacro\zb{\Rzx*\xa+\Rzy*\ya+\Rzz*\za}

% Projection of B on the screen
\coordinate (Bx) at (\Rxx,\Rxy);%,\Rxz);
\coordinate (By) at (\Ryx,\Ryy);%,\Ryz);
\coordinate (Bz) at (\Rzx,\Rzy);%,\Rzz);
\typeout{\Rxx,\Rxy;\Ryx,\Ryy;\Rzx,\Rzy}
% Change basis from A to B
\begin{scope}[shift={(0,0,0)},x={(Bx)},y={(By)},z={(Bz)}]

% Draw basis B
\draw[blue,-latex] (0,0,0) -- (1,0,0) node[shift={(0.1,0,0)}] {$e_{x'}$};
\draw[red,-latex] (0,0,0) -- (0,1,0) node[shift={(0,0.1,0)}] {$e_{y'}$};
\draw[green,-latex] (0,0,0) -- (0,0,1) node[shift={(0,0,0.1)}] {$e_{z'}$};

% Draw X in B
\draw[red,thick,-Circle] (0,0,0) -- (\xb,\yb,\zb)
 node[right,font=\tiny] {$X_b$};

\end{scope}
\end{tikzpicture}
\end{document}

Answer

我认为你只是忘了转置，即反转变换矩阵，这是所谓的“主动”和“被动”转变关键字“transpose”已经被 StefanH 提及，但是由于某种原因，他似乎忘记了实现它，或者他通过“转置基础”表示其他意思。

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}[scale=2]

% Definition and projection on the screen of basis A
\coordinate (Ax) at (1,0);%,0);
\coordinate (Ay) at (0,1);%,0);
\coordinate (Az) at (0,0);%,1);

% Draw basis A
\draw[blue,-latex] (0,0,0) -- (Ax) node[shift={(0.1,0,0)}] {$e_x$};
\draw[red,-latex] (0,0,0) -- (Ay) node[shift={(0,0.1,0)}] {$e_y$};
\draw[green,-latex] (0,0,0) -- (Az) node[shift={(0,0,0.1)}] {$e_z$};

% Define angles
\pgfmathsetmacro\alpha{70}
\pgfmathsetmacro\beta{110}

% Rotation matrix from basis A (canonical) and basis B (R = Rx(-\alpha)*Rz(-\beta))
\pgfmathsetmacro\Rxx{cos(\beta)}
\pgfmathsetmacro\Rxy{-cos(\alpha)*sin(\beta)}
\pgfmathsetmacro\Rxz{sin(\alpha)*sin(\beta)}
\pgfmathsetmacro\Ryx{sin(\beta)}
\pgfmathsetmacro\Ryy{cos(\alpha)*cos(\beta)}
\pgfmathsetmacro\Ryz{-sin(\alpha)*cos(\beta)}
\pgfmathsetmacro\Rzx{0}
\pgfmathsetmacro\Rzy{sin(\alpha)}
\pgfmathsetmacro\Rzz{cos(\alpha)}

% Point X in basis A
\pgfmathsetmacro\xa{1}
\pgfmathsetmacro\ya{1}
\pgfmathsetmacro\za{1}

% Change basis from natural tikz (with fake perspective) to A
\begin{scope}[shift={(0,0,0)},x={(Ax)},y={(Ay)},z={(Az)}]

% Draw X in A
\draw[ultra thick,blue,-Circle] (0,0,0) -- (\xa,\ya,\za) 
node[left=2pt,font=\tiny] {$X_a$};

\end{scope}

% Point X in basis B (Xb = R*Xa) 
% <- transposed this matrix by replacing \Rxy with \Ryx etc.
\pgfmathsetmacro\xb{\Rxx*\xa+\Rxy*\ya+\Rxz*\za}
\pgfmathsetmacro\yb{\Ryx*\xa+\Ryy*\ya+\Ryz*\za}
\pgfmathsetmacro\zb{\Rzx*\xa+\Rzy*\ya+\Rzz*\za}

% Projection of B on the screen
\coordinate (Bx) at (\Rxx,\Rxy);%,\Rxz);
\coordinate (By) at (\Ryx,\Ryy);%,\Ryz);
\coordinate (Bz) at (\Rzx,\Rzy);%,\Rzz);
\typeout{\Rxx,\Rxy;\Ryx,\Ryy;\Rzx,\Rzy}
% Change basis from A to B
\begin{scope}[shift={(0,0,0)},x={(Bx)},y={(By)},z={(Bz)}]

% Draw basis B
\draw[blue,-latex] (0,0,0) -- (1,0,0) node[shift={(0.1,0,0)}] {$e_{x'}$};
\draw[red,-latex] (0,0,0) -- (0,1,0) node[shift={(0,0.1,0)}] {$e_{y'}$};
\draw[green,-latex] (0,0,0) -- (0,0,1) node[shift={(0,0,0.1)}] {$e_{z'}$};

% Draw X in B
\draw[red,thick,-Circle] (0,0,0) -- (\xb,\yb,\zb)
 node[right,font=\tiny] {$X_b$};

\end{scope}
\end{tikzpicture}
\end{document}

Question 2

转置有问题。基础(Bx,By,Bz)或映射(\xb,\yb,\yc)被转置了，但我不确定是哪一个（老实说，我开始搞不清楚哪个是什么了：）。如果我使用

\pgfmathsetmacro\xc{\Rxx*\xa+\Rxy*\ya+\Rxz*\za}
\pgfmathsetmacro\yc{\Ryx*\xa+\Ryy*\ya+\Ryz*\za}
\pgfmathsetmacro\zc{\Rzx*\xa+\Rzy*\ya+\Rzz*\za}

并在范围内添加

\draw[red,fill=red,opacity=0.5] (\xc,\yc,\zc) circle (1pt);

我明白了

3D 坐标

代码中有些地方有两个坐标，有些地方有三个。如果我把它们都改成 3D，就会得到

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[scale=2]

  % Definition and projection on the screen of basis A
  \coordinate (Ax) at (1,0,0);
  \coordinate (Ay) at (0,1,0);
  \coordinate (Az) at (0,0,1);

  % Draw basis A
  \draw[blue,-latex] (0,0,0) -- (Ax) node[shift={(0.1,0,0)}] {$e_x$};
  \draw[red,-latex] (0,0,0) -- (Ay) node[shift={(0,0.1,0)}] {$e_y$};
  \draw[green,-latex] (0,0,0) -- (Az) node[shift={(0,0,0.1)}] {$e_z$};

  % Define angles
  \pgfmathsetmacro\alpha{70}
  \pgfmathsetmacro\beta{110}
  % Rotation matrix from basis A (canonical) and basis B (R = Rx(-\alpha)*Rz(-\beta))
  \pgfmathsetmacro\Rxx{cos(\beta)}
  \pgfmathsetmacro\Rxy{-cos(\alpha)*sin(\beta)}
  \pgfmathsetmacro\Rxz{sin(\alpha)*sin(\beta)}
  \pgfmathsetmacro\Ryx{sin(\beta)}
  \pgfmathsetmacro\Ryy{cos(\alpha)*cos(\beta)}
  \pgfmathsetmacro\Ryz{-sin(\alpha)*cos(\beta)}
  \pgfmathsetmacro\Rzx{0}
  \pgfmathsetmacro\Rzy{sin(\alpha)}
  \pgfmathsetmacro\Rzz{cos(\alpha)}

  % Point X in basis A
  \pgfmathsetmacro\xa{1}
  \pgfmathsetmacro\ya{1}
  \pgfmathsetmacro\za{1}

  % Change basis from natural tikz (with fake perspective) to A
  \begin{scope}[shift={(0,0,0)},x={(Ax)},y={(Ay)},z={(Az)}]
    % Draw X in A
    \draw (0,0,0) -- (\xa,\ya,\za) node {\tiny $\bullet$} node[right] {\tiny $X_a$};
  \end{scope}

  % Point X in basis B (Xb = R*Xa)
  \pgfmathsetmacro\xb{\Rxx*\xa+\Ryx*\ya+\Rzx*\za}
  \pgfmathsetmacro\yb{\Rxy*\xa+\Ryy*\ya+\Rzy*\za}
  \pgfmathsetmacro\zb{\Rxz*\xa+\Ryz*\ya+\Rzz*\za}

  \pgfmathsetmacro\xc{\Rxx*\xa+\Rxy*\ya+\Rxz*\za}
  \pgfmathsetmacro\yc{\Ryx*\xa+\Ryy*\ya+\Ryz*\za}
  \pgfmathsetmacro\zc{\Rzx*\xa+\Rzy*\ya+\Rzz*\za}

  % Projection of B on the screen
  \coordinate (Bx) at (\Rxx,\Rxy,\Rxz);
  \coordinate (By) at (\Ryx,\Ryy,\Ryz);
  \coordinate (Bz) at (\Rzx,\Rzy,\Rzz);

  % Change basis from A to B
  \begin{scope}[shift={(0,0,0)},x={(Bx)},y={(By)},z={(Bz)}]
    % Draw basis B
    \draw[blue,-latex] (0,0,0) -- (1,0,0) node[shift={(0.1,0,0)}] {$e_{x'}$};
    \draw[red,-latex] (0,0,0) -- (0,1,0) node[shift={(0,0.1,0)}] {$e_{y'}$};
    \draw[green,-latex] (0,0,0) -- (0,0,1) node[shift={(0,0,0.1)}] {$e_{z'}$};
    % Draw X in B
    \draw (0,0,0) -- (\xb,\yb,\zb) node {\tiny $\bullet$} node[right] {\tiny $X_b$};
    \draw[red,fill=red,opacity=0.5] (\xc,\yc,\zc) circle (1pt);
  \end{scope}
\end{tikzpicture}
\end{document}

Answer