以下程式碼給出了笛卡爾平面上兩條直線的圖形。 （我透過相對於正 x 軸傾斜的角度來指定線條。我懷疑我提供的程式碼是否有效執行此操作。）我試圖TikZ用箭頭繪製這些線條，所以這些線上的點的座標在-3.75 和3.75 之間。從視覺上看，這些線將由一個正方形界定，該正方形的邊與軸平行，距軸 3.75 個單位。 （軸的箭頭與一個正方形相交，該正方形的邊平行於軸並距軸 4 個單位。）我嘗試使用該intersections包。TikZ但不會編譯最後六個命令。
\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}
答案1
穿過原點(0,0)且角度為 的直線與由角和52定義的矩形相交於點，角度為。(-3.75,-3.75)(3.75,3.75)(52:{3.75/sin(52)})142(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}
答案2
我想我知道你的意思。看下面的程式碼。我的線條以 2(3.75) = 7.5 公分寬和高的正方形為界。不過，箭頭被剪掉了。
\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}