El siguiente código muestra la gráfica de dos rectas en el plano cartesiano. (Especifico las líneas por los ángulos en los que están inclinadas con respecto al eje x positivo. Dudo que el código que proporciono sea eficiente para hacer esto). Estoy tratando de dibujar TikZestas líneas, con puntas de flecha, por lo que que las coordenadas de los puntos de estas rectas están entre -3,75 y 3,75. Visualmente, estas líneas estarán delimitadas por un cuadrado con lados paralelos a los ejes y a 3,75 unidades de ellos. (Las puntas de flecha de los ejes intersecan un cuadrado con lados paralelos a los ejes y a 4 unidades de ellos). Intenté usar el intersectionspaquete. TikZSin embargo, no compilaría los últimos seis comandos.
\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}
Respuesta1
Una línea que pasa por el origen (0,0)con un ángulo 52de grados intersecta un rectángulo definido por las esquinas (-3.75,-3.75)y (3.75,3.75)en un punto (52:{3.75/sin(52)})mientras que en un ángulo 142es (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}
Respuesta2
Creo que sé lo que quieres decir. Mire el siguiente código. Tengo las líneas delimitadas por un cuadrado de 2(3,75) = 7,5 centímetros de ancho y alto. Sin embargo, las puntas de las flechas quedaron cortadas.
\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}