我需要做一些看起来像这样的东西(抱歉图片质量太差)
所以我输入了这个但它不能正常工作:
\begin{tikzpicture}
\node (A) at (0,0) {$ \vec{\bf r} = \vec{\bf R}(\vec{\bf r},\bf t)$};
\node (B) at (0, 20)
{$
\[
\begin{cases}
{$ X_1 = X_1(X_1,X_2,X_3,t) $};
\dots
{$ x_3 = x_3(X_1,X_2,X_3,t) $};
\end{cases};
\]
$}
\draw[->] (A) -| node[near start,below] (B);
\end{tikzpicture}
以下是一些错误(与该代码部分相关).log
:
line 0: Argument of \tikz@scan@no@calculator has an extra }
line 0: Package tikz Error: Giving up on this path. Did you forget a semicolon?
line 277: Bad math environment delimiter. {$
line 277: Missing $ inserted {$ X_
line 277: Missing \cr inserted {$ X_1 = X_1(X_1,X_2,X_3,t) $}
line 0: Misplaced \crcr
line 0: Missing \endgroup inserted
line 283: Missing $ inserted \]
line 283: Missing } inserted \]
line 284: Package tikz Error: Giving up on this path. Did you forget a semicolon? $}
line 285: Missing $ inserted I've inserted a begin-math/end-math symbol since I think
答案1
一种可能性是使用matrix of math nodes
:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{matrix,positioning}
\begin{document}
\begin{tikzpicture}[every left delimiter/.style={xshift=1ex}]
\node (r) {$\vec{r}=\vec{r}(\vec{R},t)$};
\matrix[matrix of math nodes,left delimiter=\lbrace,below = 10pt of r] (mat)
{
x_1 = F(x) \\
\cdots \\
x_1 = F(x) \\
};
\draw[->,shorten >= 6pt] (r.west) -- +(-15pt,0) |- (mat);
\end{tikzpicture}
\end{document}