次のコードは、直交平面上の 2 本の線のグラフを作成します。(線は、正の x 軸に対する傾斜角度で指定します。私が提供するコードが、これを行うのに効率的かどうかは疑問です。)TikZこれらの線を矢印付きで描画して、線上の点の座標が -3.75 から 3.75 の間になるようにします。視覚的には、これらの線は、軸に平行で軸から 3.75 単位離れた正方形で囲まれます。(軸の矢印は、軸に平行で軸から 4 単位離れた正方形と交差します。) パッケージを使用しようとしましたintersectionsがTikZ、最後の 6 つのコマンドはコンパイルされませんでした。
\documentclass{amsart}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,intersections,quotes,backgrounds}
\begin{document}
\begin{tikzpicture}[outer sep=0pt,p/.style={circle, fill,inner sep=1.5pt}]
\draw[draw=gray!30,latex-latex] (-3.75,0) +(-0.25cm,0) -- (3.75,0) -- +(0.25cm,0) node[below right] {$x$};
\clip (-3.75,-3.75) rectangle (3.75,3.75);
\draw[gray,dashed,line width=0.1pt] (-3.75,3.75) -- (3.75,3.75);
\draw[gray,dashed,line width=0.1pt] (-3.75,-3.75) -- (3.75,-3.75);
\draw[gray,dashed,line width=0.1pt] (-3.75,-3.75) -- (-3.75,3.75);
\draw[gray,dashed,line width=0.1pt] (3.75,-3.75) -- (3.75,3.75);
\draw[draw=blue!30,-latex] (0,0) -- (142:5);
\draw[draw=blue!30,-latex] (0,0) -- (-38:5);
\draw[draw=green!50,-latex] (0,0) -- (52:5);
\draw[draw=green!50,-latex] (0,0) -- (-128:5);
\coordinate[p,label={[fill=white]below right:$O$}] (O) at (0,0);
\coordinate (A) at (0:1);
\coordinate (B) at (52:1);
\path pic[draw, angle radius=5mm,"$\phi$",angle eccentricity=1.25] {angle = A--O--B};
\coordinate (a) at (180:1);
\coordinate (b) at (142:1);
\path pic[draw, angle radius=5mm,"$\theta$",angle eccentricity=1.25] {angle = b--O--a};
\coordinate (P) at (142:1);
\coordinate (Q) at (52:1);
\coordinate (R) at ($(O)!4mm! -45:(P)$);
\draw (R) -- ($(O)!(R)!(P)$);
\draw (R) -- ($(O)!(R)!(Q)$);
%The following code makes the right-angle mark and "colors" the inside of it white.
\begin{scope}[on background layer]
\draw[draw=gray!30,latex-latex] (0,3.75) +(0,0.25cm) node[above right] {$y$} -- (0,-3.75) -- +(0,-0.25cm);
\filldraw[fill=white] (O.center) -- ($(O)!(R)!(P)$) -- (R) -- ($(O)!(R)!(Q)$) -- cycle;
\end{scope}
%The following code is for placing arrowheads at the ends of the line segments.
%\path[name intersections={of=(-3.75,3.75) -- (3.75,3.75) and (0,0) -- (52:5), by=intersection-1}];
%\path[name intersections={of=(3.75,3.75) -- (3.75,-3.75) and (0,0) -- (-38:5), by=intersection-2}];
%\path[name intersections={of=(-3.75,-3.75) -- (3.75,-3.75) and (0,0) -- (-128:5), by=intersection-3}];
%\path[name intersections={of=(-3.75,3.75) -- (-3.75,-3.75) and (0,0) -- (142:5), by=intersection-4}];
%\draw[draw=green!50,latex-latex] (intersection-1) -- (intersection-3);
%\draw[draw=blue!30,latex-latex] (intersection-2) -- (intersection-4);
\end{tikzpicture}
\end{document}
答え1
(0,0)原点を通る角度度の線は、52角 と で定義される長方形(-3.75,-3.75)と(3.75,3.75)点で交差し、(52:{3.75/sin(52)})角度142は です(142:{3.75/cos(142)})。
\documentclass[10pt, border=5mm, tikz]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,quotes}
\begin{document}
\begin{tikzpicture}[outer sep=0pt,p/.style={circle, fill,inner sep=1.5pt}]
\draw[help lines] (-3.75,-3.75) rectangle (3.75,3.75);
\draw[draw=blue,latex-latex] (-4,0) coordinate (a) -- (4,0) coordinate (A) node[below right] {$x$};
\draw[draw=blue,latex-latex] (0,4) node[above right] {$y$} -- (0,-4);
\draw[red,latex-latex] (52:{3.75/sin(52)}) coordinate (B)--(52:{-3.75/sin(52)});
\draw[green,latex-latex] (142:{3.75/cos(142)}) --(142:{-3.75/cos(142)}) coordinate (b);
\coordinate[p,label={[fill=white]below right:$O$}] (O) at (0,0);
\path pic[draw, angle radius=5mm,"$\phi$",angle eccentricity=1.25] {angle = A--O--B};
\path pic[draw, angle radius=5mm,"$\theta$",angle eccentricity=1.25] {angle = b--O--a};
\coordinate (R) at ($(O)!4mm! -45:(b)$);
\draw (R) -- ($(O)!(R)!(b)$);
\draw (R) -- ($(O)!(R)!(B)$);
\end{tikzpicture}
\end{document}
答え2
あなたの言いたいことは分かると思います。次のコードを見てください。線は幅と高さが 2(3.75) = 7.5 センチメートルの正方形で囲まれています。ただし、矢印の先は切り取られています。
\documentclass[10pt]{amsart}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,quotes}
\begin{document}
\begin{tikzpicture}[outer sep=0pt,p/.style={circle, fill,inner sep=1.5pt}]
\draw[draw=gray!30,latex-latex] (-3.75,0) +(-0.25cm,0) -- (3.75,0) -- +(0.25cm,0) node[below right] {$x$};
\draw[draw=gray!30,latex-latex] (0,3.75) +(0,0.25cm) node[above right] {$y$} -- (0,-3.75) -- +(0,-0.25cm);
\clip (-3.75,-3.75) rectangle (3.75,3.75);
\draw[gray,dashed,line width=0.1pt] (-3.75,3.75) -- (3.75,3.75);
\draw[gray,dashed,line width=0.1pt] (-3.75,-3.75) -- (3.75,-3.75);
\draw[gray,dashed,line width=0.1pt] (-3.75,-3.75) -- (-3.75,3.75);
\draw[gray,dashed,line width=0.1pt] (3.75,-3.75) -- (3.75,3.75);
\draw[draw=blue!30,-latex] (0,0) -- (142:5);
\draw[draw=blue!30,-latex] (0,0) -- (-38:5);
\draw[draw=green!50,-latex] (0,0) -- (52:5);
\draw[draw=green!50,-latex] (0,0) -- (-128:5);
\coordinate[p,label={[fill=white]below right:$O$}] (O) at (0,0);
\coordinate (A) at (0:1);
\coordinate (B) at (52:1);
\path pic[draw, angle radius=5mm,"$\phi$",angle eccentricity=1.25] {angle = A--O--B};
\coordinate (a) at (180:1);
\coordinate (b) at (142:1);
\path pic[draw, angle radius=5mm,"$\theta$",angle eccentricity=1.25] {angle = b--O--a};
\coordinate (P) at (142:1);
\coordinate (Q) at (52:1);
\coordinate (R) at ($(O)!4mm! -45:(P)$);
\draw (R) -- ($(O)!(R)!(P)$);
\draw (R) -- ($(O)!(R)!(Q)$);
\end{tikzpicture}
\end{document}