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I have a simple tikzsketch that draws a vector relative to the center of a triangle and perpendicular to one of it's edges.

enter image description here

I would like to draw another path (vector) at the tip of the blue arrow in my sketch. I figure what I want to do is name the coordinate at the end of the path, but am not sure about the syntax of how to do that.

How do I capture / name a coordinate in a calculated path?

\documentclass[border=20pt]{standalone}

\usepackage{tikz}
\usetikzlibrary{intersections, calc,through,backgrounds}

\begin{document}

    \begin{tikzpicture}[scale=2]
    
        % Coordinates for a triangle
        \coordinate (P) at ($(0,0) + (rand,rand)$);
        \coordinate (Q) at ($(2,-2) + .5*(rand, rand)$);
        \coordinate (R) at ($(-2, -2) + .5*(rand, rand)$);
        
        \coordinate (O) at (barycentric cs:P=1,Q=1,R=1) ;
        
        %projection of O on PQ 
        %use this for the intersection point for the vector below
        \coordinate (E) at ($(P)!(O)!(Q)$);
       
        % Vector perpendicular to edge PQ and located at O
        % I would like to get the coordinate at the tip of this vector
        \draw[->, blue] let 
                \p1 = ($(P)-(Q)$)
            in
                (O)--node[midway, sloped, above]{$\vec{A}$}node[above,at end]{${\color{black}S}$}++($(O)!sqrt(\x1*\x1+\y1*\y1)!(E)-(O)$);
    
        
        %\fill[red](S)circle(2pt)
        
        \draw[thin] (P) -- (Q) -- (R) --cycle;
    \end{tikzpicture}
\end{document}
1
  • the answer below may point you in the right direction
    – js bibra
    Commented Feb 14, 2021 at 8:45

2 Answers 2

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I think I already answered in your first question about that matter.

In the following line:
(O)--node[midway, sloped, above]{$\vec{A}$}node[above,at end]{${\color{black}S}$}++($(O)!sqrt(\x1*\x1+\y1*\y1)!(E)-(O)$);
you can add coordinate[at end](S) right before drawing your point S.

Then you get :
(O)--node[midway, sloped, above]{$\vec{A}$} coordinate[at end](S) node[above,at end]{${\color{black}S}$}++($(O)!sqrt(\x1*\x1+\y1*\y1)!(E)-(O)$);

1

enter image description here

\documentclass[border=20pt]{standalone}

\usepackage{tikz}
\usetikzlibrary{intersections, calc,through,backgrounds}

\begin{document}

    \begin{tikzpicture}[scale=2]
    
        % Coordinates for a triangle
        \coordinate (P) at ($(0,0) + (rand,rand)$);
        \coordinate (Q) at ($(2,-2) + .5*(rand, rand)$);
        \coordinate (R) at ($(-2, -2) + .5*(rand, rand)$);
        
        \coordinate (O) at (barycentric cs:P=1,Q=1,R=1) ;
        
        %projection of O on PQ 
        %use this for the intersection point for the vector below
        \coordinate (E) at ($(P)!(O)!(Q)$);
       
        % Vector perpendicular to edge PQ and located at O
        % I would like to get the coordinate at the tip of this vector
        \draw[->, blue] let 
                \p1 = ($(P)-(Q)$)
            in
                (O)--
                node[midway, sloped, above]{$\vec{A}$}
                coordinate[at end, label=aux, fill=red, circle, inner sep=0.5pt ]
                ++(
                    $(O)!sqrt(\x1*\x1+\y1*\y1)!(E)-(O)$
                    );
    
        
%        \fill[red](aux)circle(2pt)
        
        \draw[thin] (P) -- (Q) -- (R) --cycle;
    \end{tikzpicture}
\end{document}

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