在tikz中绘制一个坐标值为sqrt(3)的矢量

Question

我正在尝试绘制一个坐标为 (sqrt(3), 1) 的向量。Latex 不接受平方根或 3^(1/2)。每次我有平方根坐标时，我是否需要手动写入十进制数？

\begin{tikzpicture}
\begin{axis}[my style, xtick={-1,0,...,1}, ytick={-3,-2,...,3},
xmin=-2, xmax=2, ymin=-3, ymax=3
]
\draw[->, thick, blue](0,0) -- (-\sqrt(3),1)node[right] {$\vec z_1$};
\draw[->, thick, blue](0,0) -- (\sqrt(3),0)node[above,sloped] {$\vec z_2$};
\draw[->, thick, blue](0,0) -- (0,1)node[above,sloped] {$\vec z_1+ \vec z_2$};
\end{axis}
\end{tikzpicture}

Answer 1

不，你不需要写小数。这样就可以了：

\documentclass[12pt]{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\draw[->, thick, blue](0,0) -- ({-sqrt(3)},1)node[right] {$\vec z_1$};
\draw[->, thick, blue](0,0) -- ({sqrt(3)},0)node[above,sloped] {$\vec z_2$};
\draw[->, thick, blue](0,0) -- (0,1)node[above,sloped] {$\vec z_1+ \vec z_2$};
\end{tikzpicture}
\end{document}