I'm trying to produce this simple 3d picture,
but, since it's the first time I draw a 3d pic, I'm not sure about how to get the plane passing through the line X. The following code shows a first attempt. I tried to use \filldraw
, with random points, but I'm sure this is not the best way to do that.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\coordinate (O) at (0, 0, 0);
\coordinate (A) at (2,3,1);
\draw[thick,->] (O) --++ (4.5,0,0) node[anchor=north east]{spot 0};
\draw[thick,->] (O) --++ (0,4.5,0) node[anchor=north east]{spot 1};
\draw[thick,->] (O) --++ (0,0,6) node[anchor=east]{spot 2};
\draw[->] (O)--(A) node[anchor=west]{$\Phi$};
\draw [thick] ($(O)!4cm!270:(A)$) -- ($(O)!3cm!90:(A)$) node[anchor=east]{$X$};
\filldraw[fill=blue!10, opacity=0.6] (2.5,-2.5,1) -- (2.5,1,1) -- (4,3,1) -- (4,-0.5,1) -- (2.5,-2.5,1);
\end{tikzpicture}
The plane H should intersect X and be perpendicular to p, which is why I defined first p, and then its orthogonal line X. Maybe should I define some coordinates on X , and then define somehow H? I'd like to get the "projection" of p also, as in the figure. That's not a projection in fact, it is there just to highlight that p is a vector of R3.
calc
does not support "real" 3d computations. Also the projection ofp
on the plane requires a prescription. Sincep
is the normal the naive projection is zero. What kind of projection do you have in mind? What is the relation between between your pointA
andp
?