我正在尝试使用 Ti 在 3D 中绘制的三角形中标记 45° 角 ACB钾Z:
围绕 C 的整个圆已在 3D 中围绕 Y 旋转,因此它处于正确的平面中。但是,我不知道如何对圆弧执行相同的操作:红色圆弧在 xy 平面中。我如何将此圆弧旋转到正确的平面?
\documentclass[10pt]{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[scale=6,z=.4cm]
\draw[thick,dotted] (.5,0,.4) node[anchor=north]{$B$} -- (.9,0,.7) node[anchor=south west]{$A$};
\draw[thick,dotted] (.5,0,.4) -- (.5,.5,.4) node[anchor=south]{$C$};
\draw[thick,dotted] (.5,.5,.4) -- (.9,0,.7);
\draw (.5,0.05,.4) -- (0.54, 0.05, 0.43) -- (0.54, 0, 0.43);
\draw[thick] (.5,.5,.4) circle[x={(.8,0,.6)},y={(0,1,0)},radius=.2];
\draw[thick,red] (.5,.3,.4) arc[x={(.8,0,.6)},y={(0,1,0)},start angle=-90,end angle=-45,radius=.2];
\end{tikzpicture}
\end{document}
答案1
也可以仅为圆弧定义坐标变化,使用{[options]arc...}
:
\documentclass[10pt]{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[scale=6,z=.4cm]
\draw[thick,dotted] (.5,0,.4) node[anchor=north]{$B$} -- (.9,0,.7) node[anchor=south west]{$A$};
\draw[thick,dotted] (.5,0,.4) -- (.5,.5,.4) node[anchor=south]{$C$};
\draw[thick,dotted] (.5,.5,.4) -- (.9,0,.7);
\draw (.5,0.05,.4) -- (0.54, 0.05, 0.43) -- (0.54, 0, 0.43);
\draw[thick] (.5,.5,.4) circle[x={(.8,0,.6)},y={(0,1,0)},radius=.2];
\draw[thick,red] (.5,.3,.4){[x={(.8,0,.6)},y={(0,1,0)}]arc[start angle=-90,end angle=-45,radius=.2]};
\end{tikzpicture}
\end{document}
答案2
经过一番尝试,我发现可以将自定义单位向量提供给命令\draw
而不是arc
命令,但必须事先定义弧的起始点,然后引用,否则新的 x 向量将用于计算起始坐标:
\documentclass[10pt]{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[scale=6,z=.4cm]
\draw[thick,dotted] (.5,0,.4) node[anchor=north]{$B$} -- (.9,0,.7) node[anchor=south west]{$A$};
\draw[thick,dotted] (.5,0,.4) -- (.5,.5,.4) node[anchor=south]{$C$};
\draw[thick,dotted] (.5,.5,.4) -- (.9,0,.7);
\draw (.5,0.05,.4) -- (0.54, 0.05, 0.43) -- (0.54, 0, 0.43);
\coordinate (arcstart) at (.5,.3,.4);
\draw[thick] (.5,.5,.4) circle[x={(.8,0,.6)},y={(0,1,0)},radius=.2];
\draw[x={(.8,0,.6)},y={(0,1,0)},thick,red] (arcstart) arc[start angle=-90,end angle=-45,radius=.2];
\end{tikzpicture}
\end{document}