Cone in Latex with right angles and Label
I am new LaTex user and I have hard time plot the following shape in LaTex. I started the shape but I can't find way to finish it if you can help me please
Here is my code
documentclass{article}
usepackage{tikz}
usepackage{mathrsfs}
begin{document}
begin{tikzpicture}
draw (3,0) node {$mathcal{C}$};
draw (-2,2)--(0,0);
draw (3,1.5)--(0,0);
draw (-2,-2)--(0,0);
draw (3,-1.5)--(0,0);
end{tikzpicture}
end{document}
Here is the shape I am aiming to get if you can help me finish it please.

tikz-pgf
add a comment |
I am new LaTex user and I have hard time plot the following shape in LaTex. I started the shape but I can't find way to finish it if you can help me please
Here is my code
documentclass{article}
usepackage{tikz}
usepackage{mathrsfs}
begin{document}
begin{tikzpicture}
draw (3,0) node {$mathcal{C}$};
draw (-2,2)--(0,0);
draw (3,1.5)--(0,0);
draw (-2,-2)--(0,0);
draw (3,-1.5)--(0,0);
end{tikzpicture}
end{document}
Here is the shape I am aiming to get if you can help me finish it please.

tikz-pgf
2
Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.
– marmot
Jan 14 at 2:21
@marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C
– JesZ
Jan 14 at 2:35
add a comment |
I am new LaTex user and I have hard time plot the following shape in LaTex. I started the shape but I can't find way to finish it if you can help me please
Here is my code
documentclass{article}
usepackage{tikz}
usepackage{mathrsfs}
begin{document}
begin{tikzpicture}
draw (3,0) node {$mathcal{C}$};
draw (-2,2)--(0,0);
draw (3,1.5)--(0,0);
draw (-2,-2)--(0,0);
draw (3,-1.5)--(0,0);
end{tikzpicture}
end{document}
Here is the shape I am aiming to get if you can help me finish it please.

tikz-pgf
I am new LaTex user and I have hard time plot the following shape in LaTex. I started the shape but I can't find way to finish it if you can help me please
Here is my code
documentclass{article}
usepackage{tikz}
usepackage{mathrsfs}
begin{document}
begin{tikzpicture}
draw (3,0) node {$mathcal{C}$};
draw (-2,2)--(0,0);
draw (3,1.5)--(0,0);
draw (-2,-2)--(0,0);
draw (3,-1.5)--(0,0);
end{tikzpicture}
end{document}
Here is the shape I am aiming to get if you can help me finish it please.

tikz-pgf
tikz-pgf
asked Jan 14 at 1:37
JesZJesZ
82
82
2
Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.
– marmot
Jan 14 at 2:21
@marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C
– JesZ
Jan 14 at 2:35
add a comment |
2
Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.
– marmot
Jan 14 at 2:21
@marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C
– JesZ
Jan 14 at 2:35
2
2
Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.
– marmot
Jan 14 at 2:21
Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.
– marmot
Jan 14 at 2:21
@marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C
– JesZ
Jan 14 at 2:35
@marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C
– JesZ
Jan 14 at 2:35
add a comment |
1 Answer
1
active
oldest
votes
It is always a bit hard to undo the projection, i.e. to guess the 3d coordinates from their projections on the screen. So most likely I got some wrong, yet this may serve as a start. I recommend using tikz-3dplot because it allows you to do orthographic projections from 3d to the screen. That is, you can adjust the view angles to your needs.
documentclass[tikz,border=3.14mm]{standalone}
usepackage{tikz-3dplot}
begin{document}
tdplotsetmaincoords{70}{20}
begin{tikzpicture}[dot/.style={circle,fill,inner sep=1pt}]
tdplotsetrotatedcoords{30}{20}{0}
begin{scope}[tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw[dotted,thick] (0,0,3) node[dot,label=right:{$widetilde{theta}$}]{}
-- (0,5,3) node[dot,label=above:{$X$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw[dotted,thick] (1.5,0,0) node[dot,label=above:{$widetilde{theta}$}]{}
-- ++ (0,-3,-3) node[dot,label=right:{$X$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw[dotted,thick] (O) -- (0,3,-3) node[dot,label=left:{$X$}]{};
end{scope}
begin{scope}[xshift=7cm,tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw (0,0,3) node[dot,label=right:{$Q$}] (Q) {}
-- (0,5,3) node[dot,label=above:{$P$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw (1.5,0,0) node[dot,label=above right:{$R$}] (R) {}
-- ++ (0,-3,-3) node[dot,label=right:{$S$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw (R) to[out=80,in=-40] (Q);
end{scope}
end{tikzpicture}
end{document}

add a comment |
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It is always a bit hard to undo the projection, i.e. to guess the 3d coordinates from their projections on the screen. So most likely I got some wrong, yet this may serve as a start. I recommend using tikz-3dplot because it allows you to do orthographic projections from 3d to the screen. That is, you can adjust the view angles to your needs.
documentclass[tikz,border=3.14mm]{standalone}
usepackage{tikz-3dplot}
begin{document}
tdplotsetmaincoords{70}{20}
begin{tikzpicture}[dot/.style={circle,fill,inner sep=1pt}]
tdplotsetrotatedcoords{30}{20}{0}
begin{scope}[tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw[dotted,thick] (0,0,3) node[dot,label=right:{$widetilde{theta}$}]{}
-- (0,5,3) node[dot,label=above:{$X$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw[dotted,thick] (1.5,0,0) node[dot,label=above:{$widetilde{theta}$}]{}
-- ++ (0,-3,-3) node[dot,label=right:{$X$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw[dotted,thick] (O) -- (0,3,-3) node[dot,label=left:{$X$}]{};
end{scope}
begin{scope}[xshift=7cm,tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw (0,0,3) node[dot,label=right:{$Q$}] (Q) {}
-- (0,5,3) node[dot,label=above:{$P$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw (1.5,0,0) node[dot,label=above right:{$R$}] (R) {}
-- ++ (0,-3,-3) node[dot,label=right:{$S$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw (R) to[out=80,in=-40] (Q);
end{scope}
end{tikzpicture}
end{document}

add a comment |
It is always a bit hard to undo the projection, i.e. to guess the 3d coordinates from their projections on the screen. So most likely I got some wrong, yet this may serve as a start. I recommend using tikz-3dplot because it allows you to do orthographic projections from 3d to the screen. That is, you can adjust the view angles to your needs.
documentclass[tikz,border=3.14mm]{standalone}
usepackage{tikz-3dplot}
begin{document}
tdplotsetmaincoords{70}{20}
begin{tikzpicture}[dot/.style={circle,fill,inner sep=1pt}]
tdplotsetrotatedcoords{30}{20}{0}
begin{scope}[tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw[dotted,thick] (0,0,3) node[dot,label=right:{$widetilde{theta}$}]{}
-- (0,5,3) node[dot,label=above:{$X$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw[dotted,thick] (1.5,0,0) node[dot,label=above:{$widetilde{theta}$}]{}
-- ++ (0,-3,-3) node[dot,label=right:{$X$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw[dotted,thick] (O) -- (0,3,-3) node[dot,label=left:{$X$}]{};
end{scope}
begin{scope}[xshift=7cm,tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw (0,0,3) node[dot,label=right:{$Q$}] (Q) {}
-- (0,5,3) node[dot,label=above:{$P$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw (1.5,0,0) node[dot,label=above right:{$R$}] (R) {}
-- ++ (0,-3,-3) node[dot,label=right:{$S$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw (R) to[out=80,in=-40] (Q);
end{scope}
end{tikzpicture}
end{document}

add a comment |
It is always a bit hard to undo the projection, i.e. to guess the 3d coordinates from their projections on the screen. So most likely I got some wrong, yet this may serve as a start. I recommend using tikz-3dplot because it allows you to do orthographic projections from 3d to the screen. That is, you can adjust the view angles to your needs.
documentclass[tikz,border=3.14mm]{standalone}
usepackage{tikz-3dplot}
begin{document}
tdplotsetmaincoords{70}{20}
begin{tikzpicture}[dot/.style={circle,fill,inner sep=1pt}]
tdplotsetrotatedcoords{30}{20}{0}
begin{scope}[tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw[dotted,thick] (0,0,3) node[dot,label=right:{$widetilde{theta}$}]{}
-- (0,5,3) node[dot,label=above:{$X$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw[dotted,thick] (1.5,0,0) node[dot,label=above:{$widetilde{theta}$}]{}
-- ++ (0,-3,-3) node[dot,label=right:{$X$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw[dotted,thick] (O) -- (0,3,-3) node[dot,label=left:{$X$}]{};
end{scope}
begin{scope}[xshift=7cm,tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw (0,0,3) node[dot,label=right:{$Q$}] (Q) {}
-- (0,5,3) node[dot,label=above:{$P$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw (1.5,0,0) node[dot,label=above right:{$R$}] (R) {}
-- ++ (0,-3,-3) node[dot,label=right:{$S$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw (R) to[out=80,in=-40] (Q);
end{scope}
end{tikzpicture}
end{document}

It is always a bit hard to undo the projection, i.e. to guess the 3d coordinates from their projections on the screen. So most likely I got some wrong, yet this may serve as a start. I recommend using tikz-3dplot because it allows you to do orthographic projections from 3d to the screen. That is, you can adjust the view angles to your needs.
documentclass[tikz,border=3.14mm]{standalone}
usepackage{tikz-3dplot}
begin{document}
tdplotsetmaincoords{70}{20}
begin{tikzpicture}[dot/.style={circle,fill,inner sep=1pt}]
tdplotsetrotatedcoords{30}{20}{0}
begin{scope}[tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw[dotted,thick] (0,0,3) node[dot,label=right:{$widetilde{theta}$}]{}
-- (0,5,3) node[dot,label=above:{$X$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw[dotted,thick] (1.5,0,0) node[dot,label=above:{$widetilde{theta}$}]{}
-- ++ (0,-3,-3) node[dot,label=right:{$X$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw[dotted,thick] (O) -- (0,3,-3) node[dot,label=left:{$X$}]{};
end{scope}
begin{scope}[xshift=7cm,tdplot_rotated_coords]
draw (0,0,0) node[dot,label=left:{$O$}] (O) {}
-- (0,0,5) node[pos=1.1] (A) {$A$}; % A=z
draw (O) -- (4,0,0) node[pos=1.1] (B) {$B$}; % B=x
draw (O) -- (0,5,0) node[pos=1.1] (C) {$C$}; % C=y
draw (0,0,1) -- ++ (0,1,0) -- ++ (0,0,-1);
draw (0,0,3) node[dot,label=right:{$Q$}] (Q) {}
-- (0,5,3) node[dot,label=above:{$P$}]{};
draw (0,0,4)-- ++ (0,1,0) -- ++ (0,0,-1);
draw (O) -- (0,-3,-3) node[pos=1.1]{$D$};
draw (1,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
draw (1.5,0,0) node[dot,label=above right:{$R$}] (R) {}
-- ++ (0,-3,-3) node[dot,label=right:{$S$}]{};
draw (2.5,0,0) -- ++ (0,{-1/sqrt(2)},{-1/sqrt(2)}) -- ++ (-1,0,0);
path (B) -- (A) node[pos=0.3]{$mathcal{C}$};
draw (R) to[out=80,in=-40] (Q);
end{scope}
end{tikzpicture}
end{document}

answered Jan 14 at 2:38
marmotmarmot
93.3k4109204
93.3k4109204
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2
Welcome to TeX.SE! Please explain a bit more the context of the figure. Is this a 3d figure? If so, there is the possibility to draw this in 3d.
– marmot
Jan 14 at 2:21
@marmot Thank you. Yes it is 3d figure where the point X is projected to the Cone C
– JesZ
Jan 14 at 2:35