如何制作右手三和弦

Question

我想表示两个向量 $\vec a,\vec b$ 的叉积。我想制作一个右手三角。如何沿 $\hat n$ 绘制一个螺旋线以及如何用 $\theta$ 标记角度？

 \documentclass[border=10pt]{standalone}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}[line width=.8pt,x={(1,0)},  z={(-0.5,-0.5)}]
\coordinate (O)  at (0,0,0);
\coordinate (Ay) at (0,3,0);
% Draw the axes
\foreach \c/\l/\p in {{4.5,0,0}/\vec{b}/right, {3,-3,0}/\vec{a}/below left}{
 \draw[->] (O) -- +(\c) node[\p] {$\l$};
}
\draw[->] (O) -- node[pos=0.7,above left] {$\hat{n}$} (Ay);
\end{tikzpicture}
\end{document}

编辑：

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,quotes}
\begin{document}

\begin{tikzpicture}
\draw (0,0) coordinate (O) -- (0,2)
 \draw (3,0) coordinate (A) -- (0,0) coordinate (B)
-- (2,-2) coordinate (C)
pic [fill=black!50] {angle = C--B--A}
pic [draw,->,red,thick,angle radius=1cm] {angle = C--B--A};
\end{tikzpicture}
\end{document}

Answer 1

我发现了一个参数化方程螺旋维基百科。tikz从头开始处理三维图形，但我相信额外的库可以简化绘制更复杂图形的过程。

这是您要尝试实现的某个目标的示例。从下面的代码开始。3d view让您更改向量的方向。

\documentclass{standalone}
\usepackage[svgnames]{xcolor}
\usepackage{tikz}
\usetikzlibrary {perspective,arrows.meta}

\begin{document}
\begin{tikzpicture}[
    3d view={135}{30},   % "isometric view" with azimuth=45
    axes/.style = {-Latex,line width=0.5pt,Gray},
    helix/.style = {-{Latex[scale=0.75,sep=-10pt]},line width=2pt,black},
  ]
  % \path (tpp cs:x=4, y=5, z=0) node [font=\Large, below left=-3mm and 5mm] {$\mathcal{S}$};
  \draw[axes] (0,0,0) -- (4,0,0) node [pos=1.1,black] {$\bar{a}$};
  \draw[axes] (0,0,0) -- (0,3,0) node [pos=1.1,black] {$\bar{b}$};
  \draw[axes] (0,0,0) -- (0,0,5) node [pos=1.1,black] {$\bar{a} \times \bar{b}$};
  \draw [helix] (0.35,0,0) \foreach \t [
    evaluate=\t as \x using 0.35*cos(\t),
    evaluate=\t as \y using 0.35*sin(\t),
    evaluate=\t as \z using 0.0015*\t,
  ] in {10,20,...,3000} {-- (\x,\y,\z)};
  \draw [-Latex] (tpp cs:x=2, y=0, z=0) arc (0:90:2) node [pos=0.5,above] {$\phi$};
\end{tikzpicture}
\end{document}

Answer 2

我的代码很长，因为至少在我看来，在空间中生成一个带有向量的弹簧并不容易。有许多变量和常量（对于向量 a 和 b，以及它们的叉积，对于弹簧的半径等等）。您也可以尝试改变视角...

代码

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{math, calc, arrows.meta}

\begin{document}
\tikzset{%
  view/.style 2 args={%  observer longitude and latitude (y upwards)
                      %  Remark. lomg=0 means x=0
    z={({-sin(#1)}, {-cos(#1)*sin(#2)})},
    x={({cos(#1)}, {-sin(#1)*sin(#2)})},
    y={(0, {cos(#2)})},
    evaluate={%
      \tox={sin(#1)*cos(#2)};
      \toy={sin(#2)};
      \toz={cos(#1)*cos(#2)};
    },
    longitude = #1,
    latitude = #2
  }
}
\pgfkeys{/tikz/.cd,
  latitude/.store in=\aLatit,  % observer's latitude
  latitude=0
}
\pgfkeys{/tikz/.cd,
  longitude/.store in=\aLongit,  % observer's longitude
  longitude=0  % corresponds to x=0
}

\tikzset{
  arc from/.style args={#1 towards #2}{%
    insert path={coordinate (tmp) let
      \p1 = (tmp),
      \p2 = (#1),
      \p3 = (#2),
      \n1 = {veclen(\x1-\x2, \y1-\y2)},
      \n2 = {atan2(\y2-\y1, \x2-\x1)},
      \n3 = {atan2(\y3-\y1, \x3-\x1)}
      in (\p2) arc (\n2: \n3 : \n1)
    }
  }
}

\tikzmath{
  real \ax, \ay, \az, \bx, \by, \bz, \cx, \cy, \cz;
  \ax = 2; \ay = -1.5; \az = 1;
  \bx = 1.5; \by = .5; \bz = -.75;
  \cx = \ay*\bz -\az*\by;
  \cy = -\ax*\bz +\az*\bx;
  \cz = \ax*\by -\ay*\bx;
  real \an, \bn, \ux, \uy, \uz, \vx, \vy, \vz, \vn, \uv, \wx, \wy, \wz;
  \an = {sqrt(\ax*\ax +\ay*\ay + \az*\az)};
  \ux = \ax/\an; \uy = \ay/\an; \uz = \az/\an; 
  \bn = {sqrt(\bx*\bx +\by*\by + \bz*\bz)};
  \vx = \bx/\bn; \vy = \by/\bn; \vz = \bz/\bn; 
  \uv = \ux*\vx +\uy*\vy +\uz*\vz;
  \vx = \vx -\uv*\ux;
  \vy = \vy -\uv*\uy;
  \vz = \vz -\uv*\uz;
  \vn = {sqrt(\vx*\vx +\vy*\vy + \vz*\vz)};
  \vx = \vx/\vn; \vy = \vy/\vn; \vz = \vz/\vn;
  \wx = \uy*\vz -\uz*\vy;
  \wy = -\ux*\vz +\uz*\vx;
  \wz = \ux*\vy -\uy*\vx;
  real \r, \wl;
  \r = .2;
  \wl = .5;
  integer \N, \nbPoints;
  \N = 4;
  \nbPoints = 24;
}
\begin{tikzpicture}[view={45}{28}, line width=.8pt, every node/.style={scale=.7}]
  % canonical coordinate system
  \begin{scope}[color=gray!80, line width=.2pt]
    \draw[->] (0, 0, 0) -- (1, 0, 0) node[shift={(.2, .1, 0)}] {$x$};
    \draw[->] (0, 0, 0) -- (0, 1, 0) node[shift={(.2, .2, 0)}] {$y$};
    \draw[->] (0, 0, 0) -- (0, 0, 1) node[shift={(0, .2, .2)}] {$z$};
  \end{scope}

  % the vectors
  \draw[arrows={-Latex}] (0, 0, 0) -- (\ax, \ay, \az)
  node[pos=.7, left] {$\vec{a}$};
  \draw[arrows={-Latex}] (0, 0, 0) -- (\bx, \by, \bz)
  node[pos=.7, above] {$\vec{b}$};
  \draw (0, 0, 0) -- (\cx, \cy, \cz)
  node[pos=.7, above right] {$\vec{a}\times\vec{b}$};

  % Gram Schmidt on the vectors
  \begin{scope}[color=orange!80, line width=.4pt]
    \draw[->] (0, 0, 0) -- (\ux, \uy, \uz) coordinate (U);
    \path (\bx/\bn, \by/\bn, \bz/\bn) coordinate (V);
    \draw[->] (0, 0, 0) -- (\vx, \vy, \vz);
    % \draw[->] (0, 0, 0) -- (\wx, \wy, \wz);
  \end{scope}

  % the arc
  \draw[->, very thin] (0, 0, 0) [arc from=U towards V]; 

  % cross product inside the spring
  \draw[white, opacity=.7, line width=3pt]
  (0, 0, 0) -- (${(\N -1)*\wl}*(\wx, \wy, \wz)$) coordinate (seen);

  % the spring
  \foreach \j [parse=true, evaluate=\j as \s using {(\j -1)/\nbPoints*360},
  evaluate=\j as \t using {\j/\nbPoints*360}]
  in {1, ..., \N*\nbPoints}{%
    \tikzmath{%
      \x1 = \r*cos(\s)*\ux +\r*sin(\s)*\vx +(\j -1)/\nbPoints*\wl*\wx;
      \y1 = \r*cos(\s)*\uy +\r*sin(\s)*\vy +(\j -1)/\nbPoints*\wl*\wy;
      \z1 = \r*cos(\s)*\uz +\r*sin(\s)*\vz +(\j -1)/\nbPoints*\wl*\wz;
      \x2 = \r*cos(\t)*\ux +\r*sin(\t)*\vx +\j/\nbPoints*\wl*\wx;
      \y2 = \r*cos(\t)*\uy +\r*sin(\t)*\vy +\j/\nbPoints*\wl*\wy;
      \z2 = \r*cos(\t)*\uz +\r*sin(\t)*\vz +\j/\nbPoints*\wl*\wz;
    }
    \draw[blue, preaction={draw=white, opacity=.9, line width=2pt}]
    (\x1, \y1, \z1) -- (\x2, \y2, \z2);
  }

  % the cross product over the spring
  \draw[arrows={-Latex}, preaction={draw=white, opacity=.9, line width=2pt}]
  (seen) -- (\cx, \cy, \cz);
\end{tikzpicture}
\end{document}