[Tex/LaTex] Drawing some simple orthogonal projections

drawtikz-pgf

I'm a beginner with TikZ and I'd like to show that a vector y can be decomposed as

How can I tell TikZ to draw a dashed line orthogonal to y_p onto X and a vector ε orthogonal to X, starting at the origin? This is the code, but the projections are missing.

\documentclass[12pt,a4paper]{article}

\usepackage{tikz}
\usetikzlibrary{calc} 
\usetikzlibrary{positioning,arrows.meta}
\begin{document}

\begin{tikzpicture}[dot/.style={circle,inner sep=1pt,fill,label={#1},name=#1},
  extended line/.style={shorten >=-#1,shorten <=-#1},
  extended line/.default=1cm]

\draw[thick,->] (-4.5,0) -- (4.5,0);
\draw[thick,->] (0,0) -- (0,4.5);

\coordinate (A) at (0,0);
\coordinate (B) at (-4,3);

\draw [extended line=0.5cm, <->] (A) -- (B) 
node[pos=0.8,left=1em, font=\small]{$X$};     
\draw [ ->] (0,0) -- (-2.6, 4.3) node[anchor=north east,font=\small] {$y$};

\end{tikzpicture}
\end{document}

Best Answer

One can project a coordinate, say, (yn) on the line between, say, (A) and (B) by using ($(A)!(yn)!(B)$), as explained in the post linked by Marian G.. To get the picture, you may thus want to simply (meaning no intersections, explicit angles and so on) do

\documentclass[12pt,a4paper]{article}
\usepackage{tikz}
\usetikzlibrary{calc,positioning}
\begin{document}
\begin{tikzpicture}[dot/.style={circle,inner sep=1pt,fill,label={#1},name=#1},
  extended line/.style={shorten >=-#1,shorten <=-#1},
  extended line/.default=1cm]
 \draw[thick,-stealth] (-4.5,0) -- (4.5,0);
 \draw[thick,-stealth] (0,0) -- (0,4.5);
 \coordinate (A) at (0,0);
 \coordinate (B) at (-4,3);
 \draw [extended line=0.5cm, stealth-stealth] (A) -- (B) node[pos=1.15,font=\small]{$X$};     
 \draw [ -stealth] (0,0) -- (-2.6, 4.3) coordinate (yn) node[right]{$y$}; 
 \draw[dashed] (yn) --  node[midway,above left]{$\varepsilon$} ($(A)!(yn)!(B)$) node[below left]{$y_p$};
\end{tikzpicture}
\end{document}

Related Solutions

[Tex/LaTex] Rotate a node but not its content: the case of the ellipse decoration

The source of the difficulty is that ellipses are constructed in a particular way in TikZ. They are paths that start from the x-axis and proceed counter-clockwise around their centre. The vast majority of the time, the exact parametrisation doesn't matter. You appear to have found the one situation where it does!

In the actual question, you only want to be able to mirror the ellipse, and so draw it starting from the negative x-axis (the title of the question suggests a more flexible approach). That's actually not too hard since we can exploit the symmetry of the ellipse. The key is to provide it with a negative x-radius, since then it will start from the negative x-axis (and proceed clockwise, but we could correct for that by negating the y-radius as well). To do this, we interrupt the call from the node shape to the drawing command and flip the sign of the x-radius. The simplest way to do this is to redefine the \pgfpathellipse macro to do the negation and then call the original macro. The following code does this.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations,shapes,decorations.markings}
\makeatletter
\let\origpgfpathellipse=\pgfpathellipse
\def\revpgfpathellipse#1#2#3{%
  #2%
  \pgf@xa=-\pgf@x
  \origpgfpathellipse{#1}{\pgfqpoint{\pgf@xa}{0pt}}{#3}}
\makeatother

\tikzset{
  reversed ellipse/.style={
    ellipse,
    reverse the ellipse%
  },
  reverse the ellipse/.code={
    \let\pgfpathellipse=\revpgfpathellipse
  }
}
\begin{document}
\begin{tikzpicture}
\node[ellipse,
  draw,
  postaction={
    decorate,
    decoration={
      markings,
      mark=at position 1 with {
        \arrow[line width=5pt,blue]{>}
      }
    }
  }
] at (0,0) {hello world};

\node[reversed ellipse,
  draw,
  postaction={
    decorate,
    decoration={
      markings,
      mark=at position 1 with {
        \arrow[line width=5pt,blue]{>}
      }
    }
  }
] at (0,-2) {hello world};
\end{tikzpicture}

\end{document}

Here's the result:

rotated ellipse

(the arrow got clipped, but you can see where it lies)

[Tex/LaTex] How to define the default vertical distance between nodes

I don't really get the question so I hope this is what you wanted. If you include a full document (such that we copy paste and see the problem on our systems) things are much more easier.

Here, you can change the default setting within a scope but your block style had a node distance which was resetting every time it is issued. I've made it 2mm such that we can see the difference easier.

\documentclass[tikz]{standalone}
\usetikzlibrary{arrows,shapes.geometric,positioning}

\begin{document}
\begin{tikzpicture}[decision/.style={diamond, draw, text width=4.5em, text badly centered, node distance=3.5cm, inner sep=0pt},
                    block/.style   ={rectangle, draw, text width=6em, text centered, rounded corners, minimum height=4em, minimum height=2em},
                    cloud/.style   ={draw, ellipse, minimum height=2em},
                    line/.style    ={draw,-latex'},
                    node distance = 1cm, 
                    auto]
    \node [block] (1st) {1st};
    \node [block, right= of 1st]  (2nd1) {2nd1};
\begin{scope}[node distance=2mm and 10mm]%Here we change it for everything inside this scope
    \node [block, above= of 2nd1] (2nd2) {2nd2};
    \node [block, below= of 2nd1] (2nd3) {2nd3};
    \node [block, right= of 2nd1] (3rd1) {3rd1};
    \node [block, above= of 3rd1] (3rd2) {3rd2};
    \node [block, above= of 3rd2] (3rd3) {3rd3};
\end{scope}
    \node [block, below= of 3rd1] (3rd4) {3rd4};
    \node [block, below= of 3rd4] (3rd5) {3rd5};

    \path [line] (1st) -- (2nd1);
    \path [line] (2nd1) -- (2nd2);
    \path [line] (2nd1) -- (2nd3);
    \path [line] (2nd2) -- (3rd3);
    \path [line] (2nd1) -- (3rd1);
    \path [line] (1st) -- (2nd1);
\end{tikzpicture}
\end{document}

enter image description here

Best Answer

Related Solutions

[Tex/LaTex] Rotate a node but not its content: the case of the ellipse decoration

[Tex/LaTex] How to define the default vertical distance between nodes

Related Question