答案1
这是一种可能性;由于问题中没有提供任何信息,我假设“I”和“J”是任意放置的:
代码:
\documentclass[border=3pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[>=latex]
\coordinate (O) at (0,0);
\coordinate (I) at ([shift={(30:2cm)}]O);
\coordinate (J) at ([shift={(120:5cm)}]O);
\coordinate (F) at ( $(I)+(J)$ );
% the circles (ellipses)
\draw
let
\p1=(I),
\p2=(J)
in
(0,\y1) ellipse [x radius=\x1,y radius=0.2cm]
(0,\y2) ellipse [x radius=\x2,y radius=0.2cm];
% the vector lines
\foreach \Valor in {I,J,F}
\draw[->] (O) -- (\Valor);
% the dashed lines
\draw[dashed]
(J) -- (F) -- (I);
% the labels
\foreach \Valor/\Pos in {O/below,I/right,J/left,F/above}
\node[\Pos] at (\Valor) {$\Valor$};
% axis lines (optional)
\draw[gray!20]
(O) -- ++(0pt,6cm);
\draw[gray!20]
([xshift=-3cm]O) -- ([xshift=3cm]O);
\end{tikzpicture}
\end{document}
如果圆(椭圆)的半径相同,则代码会更短:
\documentclass[border=3pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[>=latex]
\def\PosCirci{1}
\def\PosCircii{5}
\def\radius{2cm}
\coordinate (O) at (0,0);
\draw
(0,\PosCirci) ellipse [x radius=\radius,y radius=0.2cm]
(0,\PosCircii) ellipse [x radius=\radius,y radius=0.2cm];
\coordinate (I) at (\radius,\PosCirci);
\coordinate (J) at (-\radius,\PosCircii);
\coordinate (F) at ( $(I)+(J)$ );
% the vactor lines
\foreach \Valor in {I,J,F}
\draw[->] (O) -- (\Valor);
% the dashed lines
\draw[dashed]
(J) -- (F) -- (I);
% the labels
\foreach \Valor/\Pos in {O/below,I/right,J/left,F/above}
\node[\Pos] at (\Valor) {$\Valor$};
\end{tikzpicture}
\end{document}