在乳胶中可以做交叉点吗？
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{tikz}
\usetikzlibrary{shapes,arrows,shadows,shadows.blur,positioning,calc,arrows.meta,automata}
\usetikzlibrary{matrix,calc}
\usepackage{amsmath}
\begin{document}
\begin{tikzpicture}[every node/.style={anchor=north east,fill=white,minimum width=1.4cm,minimum height=7mm}]
\matrix (mA) [draw,matrix of math nodes]
{
(w_{3,1,1}) & (w_{3,2,1}) & (\hdots) & (w_{3,J,1}) \\
(w_{3,1,2}) & (w_{3,2,2}) & (\hdots) & (w_{3,J,2}) \\
(\vdots) & (\vdots) & (\hdots) & (\vdots) \\
(w_{3,1,F}) & (w_{3,2,F}) & (\hdots) & (w_{3,J,F}) \\
};
\matrix (mB) [draw,matrix of math nodes] at ($(mA.south west)+(1.5,0.7)$)
{
(w_{2,1,1}) & (w_{2,2,1}) & (\hdots) & (w_{2,J,1}) \\
(w_{2,1,2}) & (w_{2,2,2}) & (\hdots) & (w_{2,J,2}) \\
(\vdots) & (\vdots) & (\hdots) & (\vdots) \\
(w_{2,1,F}) & (w_{2,2,F}) & (\hdots) & (w_{2,J,F}) \\
};
\matrix (mC) [draw,matrix of math nodes] at ($(mB.south west)+(1.5,0.7)$)
{
(w_{1,1,1}) & (w_{1,2,1}) & (\hdots) & (w_{1,J,1}) \\
(w_{1,1,2}) & (w_{1,2,2}) & (\hdots) & (w_{1,J,2}) \\
(\vdots) & (\vdots) & (\hdots) & (\vdots) \\
(w_{1,1,F}) & (w_{1,2,F}) & (\hdots) & (w_{1,J,F}) \\
};
\draw[dashed](mA.north east)--(mC.north east);
\draw[dashed](mA.north west)--(mC.north west);
\draw[dashed](mA.south east)--(mC.south east);
\end{tikzpicture}
\end{document}
我在乳胶中制作了这个 3d 矩阵，并希望其中一个列实体同时具有垂直点和水平点，以表明它在两个方向上延伸。
看起来点命令没有这样的选项？
答案1
不清楚你想要什么样的对齐方式，也不清楚你想要多少个点，所以我给出了几种可能性。在所有情况下，它们都是\dots和的叠加\vdots。
\documentclass{article}
\usepackage{amsmath,stackengine}
\stackMath
\begin{document}
$a\stackengine{4pt}{\vdots}{\dots}{O}{c}{F}{F}{L}$
$a\stackengine{-4pt}{\dots}{\vdots}{O}{c}{F}{F}{L}$
$a\stackengine{0pt}{\stackengine{0pt}{\kern9pt\dots}{\dots\kern9pt}{O}{c}{F}
{F}{L}}{\stackengine{-8pt}{\vdots}{\vdots}{O}{c}{F}{F}{L}}{O}{c}{F}{F}{L}$
\end{document}
