两个向量之间的角度 tikz LaTeX

Question

我需要在蓝色的两个向量之间建立一个角度，我无法完全理解我看到的代码，如果有人可以帮助我。

    \begin{tikzpicture}[scale=1]
    \draw[->] (-1,0)--(5,0) node[right]{$x$};
    \draw[->] (0,-1)--(0,5) node[above]{$y$};
    
    \draw[name path=fr1, blue,-stealth,line width = 0.5mm] (0,0) -- (5,2) node[pos=0.5, anchor=south]{$F_1 $};
    
    \draw[name path=fr2, blue,-stealth,line width = 0.5mm] (0.7,4) -- (0,0) node[pos=0.5, anchor=west]{$F_2 $};
    
    \draw[name path=fr3, red,-stealth,line width = 0.5mm] (0.7,4) -- (5,2) node[pos=0.5, anchor=south]{$F_3 $}; 
    
    \path[name path=circ] (0,0) circle (1);
    \draw[red, -, intersection segments = {of= fr1 and fr2, sequence=R1}];
    \end{tikzpicture}

Answer 1

\documentclass{standalone} 
\usepackage{tkz-base,tkz-euclide}
\begin{document} 
\begin{tikzpicture}
  \tkzInit[xmax=5,ymax=5]
 \tkzDrawXY[noticks]
 \tkzDefPoints{0/0/A,5/2/B,0.7/4/C}
 \tkzDrawSegments[vector style,blue](A,B C,A)
 \tkzDrawSegment[vector style,red](C,B)
 \tkzMarkAngle[<->](B,A,C)
 \tkzLabelAngle[pos=1.2](B,A,C){$\alpha$}
 \tkzLabelSegment[blue](A,B){$F_1$}
 \tkzLabelSegment[blue,swap](A,C){$F_2$}
 \tkzLabelSegment[red](C,B){$F_3$}
\end{tikzpicture}
\end{document}

Answer 2

\documentclass[border=5pt,tikz]{standalone}
\usepackage{tikz}
\usetikzlibrary{angles, quotes, intersections}
\begin{document}

 \begin{tikzpicture}[scale=1]
        \draw[->] (-1,0)--(5,0) node[right]{$x$};
        \draw[->] (0,-1)--(0,5) node[above]{$y$};
        
        \draw[name path=fr1, blue,-stealth,line width = 0.5mm] (0,0)coordinate(aux) -- (5,2) node[pos=0.5, anchor=south]{$F_1 $}coordinate(aux1);
        
        \draw[name path=fr2, blue,-stealth,line width = 0.5mm] (0.7,4)coordinate(aux2) -- (0,0) node[pos=0.5, anchor=west]{$F_2 $};
        
        \draw[name path=fr3, red,-stealth,line width = 0.5mm] (0.7,4) -- (5,2) node[pos=0.5, anchor=south]{$F_3 $}; 
        
        \path[name path=circ] (0,0) circle (1);

\pic[ draw,<->,>=stealth,red!60!black, "$\theta_1$"{fill=white},inner sep=1pt, circle, angle eccentricity=1.1, angle radius = 10mm] {angle = aux1--aux--aux2};   

    \end{tikzpicture}
\end{document}

Answer 3

与 @js bibra 答案（+1）类似，但使用quotes所有边标签的库且不使用intersections库：

\documentclass[border=3mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{angles, arrows.meta,
                quotes}

\begin{document}
    \begin{tikzpicture}[
         > = Straight Barb,
arr/.style = {color=#1, very thick, line cap=round, -Stealth},
every edge quotes/.append style = {auto, sloped}
                        ]
% axis
\draw[->] (-0.5,0)--(5,0) node [right] {$x$};
\draw[->] (0,-0.5)--(0,5) node [above] {$y$};
% vectors
\draw[arr=blue] (0.0,0) coordinate (O)  to ["$F_1$"]   (5,2);
\draw[arr=blue] (0.7,4) coordinate (f2) to ["$F_2$" '] (0,0);
\draw[arr=red]  (0.7,4) to ["$F_3$"] (5,2) coordinate  (f3);
% angle
\pic [draw, red, <->,
      angle radius = 7mm,
      angle eccentricity=1.3, 
      "$\alpha$"] {angle = f3--O--f2};
    \end{tikzpicture}
\end{document}

Answer 4

使用tzplot包裹：

\documentclass[tikz]{standalone}
    
\usepackage{tzplot}[scale=.6]

\begin{document}

\begin{tikzpicture}[font=\scriptsize]
\tzaxes*(-1,-1)(5,5){$x$}{$y$}
\tzcoors(0,0)(aux)(0.7,4)(aux2)(5,2)(aux1);
\tzline[blue,->,line width=0.5mm](aux){$F_1$}[a](aux1)
\tzline[blue,->,line width=0.5mm](aux2){$F_2$}[r](aux)
\tzline[red,->,line width=0.5mm](aux2){$F_3$}[a](aux1)
\tzanglemark[<->](aux1)(aux)(aux2){$\theta_1$}[pos=1,fill=white](1cm)
\end{tikzpicture}

\end{document}