# Right triangle annotations on vectors

I am drawing some force diagrams, and would really like the ability to draw a small right triangle on a line, like in the snapshot below.

Right now I'm drawing my force vectors like so:

\draw[force,->] (O) -- (C)
node[below] {$F_1 = \SI{850}{\newton}$};


I tried to create another node node[midway] (midPoint) {}, then \draw (midPoint) -- ++(0,-3mm) -- ++(4mm,0) however, there is some ugly gap between the vector and this tiny triangle.

Is there a library that I don't know about to help create annotations like these? Or perhaps more idiomatic tikz way of doing this? Thanks for your time!

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

EDIT: So I repurposed a tangentially related answer (Triangle as constraints in structural engineering) to some extent:

RightAngle/.style=
{
decoration={
markings,
mark=at position 0.5 with {
\draw[-, thin] (0,0) -- (0,-3mm)
node[midway, left] {\tiny 3};
\draw[-,thin] (0,-3mm) -- (4mm, 0)
node[midway, below] {\tiny 4};
}
},
postaction=decorate
}

\draw[force,->, RightAngle] (O) -- (C)
node[below right] {$F_1 = \SI{850}{\newton}$};


Which gets me almost there! However I'm still missing something to do with the new local coordinate system. How do I make that 3 leg come straight down (relative to the standard coordinate system)? • Welcome to TeX.SE. While code snippets are useful in explanations, it is always best to compose a fully compilable MWE that illustrates the problem including the \documentclass and the appropriate packages so that those trying to help don't have to recreate it. – Peter Grill May 20 '14 at 6:10

WIth tikz here is one way to do it: select the start point and end point on the vector itself:

\coordinate (Start Point) at ($(O)!0.3!(C)$);
\coordinate (End Point)   at ($(O)!0.5!(C)$);


In the above the (Start Point) is 0.3 along the length, and the (End Point) is at 0.5 along the path, and use |- to connect them which draws a vertical line followed by a horizontal line: ## Code:

\documentclass{article}
\usepackage{siunitx}
\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}
\begin{tikzpicture}
\coordinate (O) at (0,0);
\coordinate (B) at (3,4);
\coordinate (C) at (3,-2);

\draw[ultra thick, red, -latex] (O) -- (B) node[above] {$F_1 = \SI{950}{\newton}$};
\coordinate (Start Point) at ($(O)!0.3!(B)$);
\coordinate (End Point)   at ($(O)!0.5!(B)$);
\draw [red] (Start Point) |- (End Point);

\draw[ultra thick, blue, -latex] (O) -- (C) node[below] {$F_1 = \SI{850}{\newton}$};
\coordinate (Start Point) at ($(O)!0.3!(C)$);
\coordinate (End Point)   at ($(O)!0.5!(C)$);
\draw [blue] (Start Point) |- (End Point);
\end{tikzpicture}
\end{document}

• The point C should be (4,-3) to make it have a consistent slope as described by the slope triangle. In addition, End Point must be $(O)!0.8!(C)$. – kiss my armpit May 20 '14 at 6:28
• @MoneyOrientedProgrammer: Thanks, but it is not clear to me that that is the requirement. The 3-4-5 triangle won't fit many vectors so the ability to draw that is rather limited. I went with a general purpose method which will draw a right triangle on any vector.. – Peter Grill May 20 '14 at 6:37

Just for typing exercise with PSTricks.

## Option 1

\documentclass[pstricks,border=12pt,12pt]{standalone}
\usepackage{pst-eucl}
\psset
{
PointName=none,
PointSymbol=none,
shortput=nab,
linewidth=2pt,
linecap=1,
}
\usepackage{siunitx}
\begin{document}
\begin{pspicture}(5,-4)
\pstGeonode
{A}
(4,-3){B}
([nodesep=1]{B}A){A'}
([offset=-30pt]A'){C}
([nodesep=40pt]C){B'}
\ncline[linestyle=none]{A'}{B'}^{5}
\ncline[linecolor=gray]{A'}{C}_{3}
\ncline[linecolor=gray]{C}{B'}_{4}
\ncline{->}{A}{B}
\uput[-90](B){$F_1=\SI{850}{\newton}$}
\end{pspicture}
\end{document}


## Option 2

\documentclass[pstricks,border=12pt,12pt]{standalone}
\usepackage{pst-eucl}
\psset
{
PointName=none,
PointSymbol=none,
shortput=nab,
linewidth=2pt,
linecap=1,
}
\usepackage{siunitx}
\begin{document}
\begin{pspicture}(5,-4)
\pstGeonode{A}(4,-3){B}([nodesep=1]{B}A){A'}([nodesep=50pt]{B}A'){B'}(A'|B'){C}
\ncline[linestyle=none]{A'}{B'}^{5}
\ncline[linecolor=gray]{A'}{C}_{3}
\ncline[linecolor=gray]{C}{B'}_{4}
\ncline{->}{A}{B}
\uput[-90](B){$F_1=\SI{850}{\newton}$}
\end{pspicture}
\end{document} Another PSTricks solution. Run with xelatex:

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-node}
\begin{document}

\begin{pspicture}[shortput=nab,arrowscale=1.5](5,-4)
\pnodes(0,0){A}(4,-3){B}
\pcline[linewidth=1.5pt,arrows=->](A)(B)
\psLNode(A)(B){0.15}{C} \psLNode(A)(B){0.4}{D}
\pcline[linewidth=0.5pt](C)(C|D)_3
\pcline[linewidth=0.5pt](C|D)(D)_4
\pcline[linestyle=none](C)(D)^5
\uput[-90](B){$F_1=850\,\textrm{N}$}
\end{pspicture}

\end{document} 