TikZ 中的线到交叉点

Question

我想在 TikZ 中重现下图：

这两个向量是（-1/5, 2/5），（2/5, 1/5） 。有没有办法让 TikZ 绘制虚线直到与粗外框的交点，而不必实际计算交点并明确输入节点？到目前为止，我有：

\begin{tikzpicture}
    \node (a) at (-1, 2) {};
    \node (b) at (2, 1) {};

    \draw [very thick] (-5, 0) -- (5, 0);
    \draw [very thick] (-5, -5) -- (5, -5);
    \draw [very thick] (-5, 5) -- (5, 5);

    \draw [very thick] (-5, -5) -- (-5, 5);
    \draw [very thick] (0, -5) -- (0, 5);
    \draw [very thick] (5, -5) -- (5, 5);

    \draw [very thick, ->] (0, 0) -- (a);
    \draw [very thick, ->] (0, 0) -- (b);

    \node[label=right:{$1$}] at (5,0) {};
    \node[label=above:{$1$}] at (0,5) {};
\end{tikzpicture}

产生

Answer 1

可能使用极坐标会得到更好的结果（尤其是不用来node定义点 a 和 b，而是coordinate）。

\documentclass[tikz,border=3.14mm]{standalone}
\begin{document}
    \begin{tikzpicture}
        
        \begin{scope}
            \clip (-5,-5) rectangle (5,5);
            \draw[rotate=30,dotted,orange,cap=round] (-10,-10) grid[step=2cm] (10,10);
        \end{scope}
        \coordinate (a) at (30:2) {};
        \coordinate (b) at (120:2) {};
    
        \draw [very thick]  (-5, -5) rectangle (5, 5)
                            (-5,0) -- (5,0) node[right] {$1$}
                            (0,-5) -- (0,5) node [above] {$1$};
    
        \draw [very thick, ->] (0, 0) -- (a);
        \draw [very thick, ->] (0, 0) -- (b);
    
    \end{tikzpicture}
\end{document}

Answer 2

这是一个使用的解决方案grid：

\documentclass[tikz, border=1cm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\coordinate (a) at (-1, 2);
\coordinate (b) at (2, 1);    
\draw [very thick, ->] (0, 0) -- (a);
\draw [very thick, ->] (0, 0) -- (b);
\node[label=right:{$(1,0)$}] at (5,0) {};
\node[label=above:{$(0,1)$}] at (0,5) {};
\draw[very thick, step=5] (-5,-5) grid (5,5);
\clip (-5,-5) rectangle (5,5);
\draw[dashed, very thick, step=sqrt(5), rotate=atan(1/2)] (-10,-10) grid (10,10);
\end{tikzpicture}
\end{document}