Without using intersection method, can we find the following vector component?

Question

Given that two vectors, i.e., green and red ones. The objective is to find the blue one such that the resultant of the green and blue ones is parallel to the red one and with the same direction as the direction of the red vector.

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

\begin{document}

\begin{pspicture}[showgrid=true](4,3)
\pnode(0,0){O}
\pnode(4,3){A}
\pnode(0,2){B}
\pnode(A|B){C}
\psset{arrows=->}
\psline[linecolor=red](A)
\psline[linecolor=green](B)
\pstInterLL[PointName=none,PointSymbol=none]{B}{C}{O}{A}{D}
\psline[linecolor=blue](D|O)
\psline[linestyle=dashed,linecolor=gray,arrows=-](B)(D)(D|O)
\end{pspicture}

\end{document}

How to get the blue one in the simplest way without using intersection?

I want to apply the method in question for drawing the tension vector and its horizontal component as shown in the following figure (but such vectors have not been implemented yet):

Answer 1

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

\begin{pspicture}[showgrid=bottom](4,3)
\pnode(4,3){A}\pnode(0,2){B}
\psline[linecolor=red]{->}(A)
\psline[linecolor=green]{->}(B)
\pnode(!\psGetNodeCenter{A} 
        \psGetNodeCenter{B} 
        B.y A.y div A.x mul 0 ){D}
\psline[linestyle=dashed,linecolor=gray](B)(D|B)(D)
\psline[linecolor=blue]{->}(O)(D)
\end{pspicture}

\end{document}

the following works with the Beta version of pst-node (http://texnik.dante.de/tex/generic/pst-node/pst-node.tex) the coordinates of Nodes are available as N-<name>.x and N-<name>.y. Maybe that I'll change it to <name>.x and <name>.y

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

\begin{pspicture}[showgrid=bottom](4,3)
\pnode(4,3){A}
\pnode(0,2){B}
\psline[linecolor=red]{->}(A)
\psline[linecolor=green]{->}(B)
\pnode(!N-B.y N-A.y div N-A.x mul 0 ){D}
\psline[linestyle=dashed,linecolor=gray](B)(D|B)(D)
\psline[linecolor=blue]{->}(O)(D)
\end{pspicture}

\end{document}

Answer 2

Here is a TikZ solution:

\documentclass[margin=2mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
  \draw[green] (0,0) -- (0,2) coordinate (b);
  \draw[red] (0,0) -- (4,3) coordinate (a);
  \draw[blue] let \p1=(a), \p2=(b) in (0,0) -- (\y2/\y1*\x1,0);
\end{tikzpicture}
\end{document}

Answer 3

I complement what has been answered by Herbert Voss. If the vector tails are not on the origin then the calculation must include the tail vectors as follows.

With the latest PSTricks core,

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

\begin{pspicture}[showgrid=bottom](-1,-2)(4,3)
\pnode(-1,-2){O}
\pnode(4,3){A}
\pnode(0,2){B}
\psline[linecolor=red]{->}(O)(A)
\psline[linecolor=green]{->}(O)(B)
\pnode(!N-B.y N-O.y sub N-A.y N-O.y sub div N-A.x N-O.x sub mul N-O.x add 0){C}
\psline[linestyle=dashed,linecolor=gray](B)(C|B)(C)
\psline[linecolor=blue]{->}(O)(C)
\end{pspicture}

\end{document}