- Thread starter
- #1
I don't need a x,y,z axis. A cube in free space only with vectors. I still need help.Do you still need help with this? There are many examples on the web. Do you need the origin to be in the center of the cube or in one of its vertices?
\usetikzlibrary{arrows}
\begin{tikzpicture}[>=stealth',x=2cm,y=2cm,z=0.77cm]
\fill (0,0) circle (1.5pt);
\draw (0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle;
\begin{scope}[shift={(0,0,1)}]
\draw (0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle;
\end{scope}
\draw (0,0) -- (0,0,1) (0,1) -- (0,1,1) (1,0) -- (1,0,1) (1,1) -- (1,1,1);
\end{tikzpicture}
This is one way.
View attachment 362Code:\usetikzlibrary{arrows} \begin{tikzpicture}[>=stealth',x=2cm,y=2cm,z=0.77cm] \fill (0,0) circle (1.5pt); \draw (0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle; \begin{scope}[shift={(0,0,1)}] \draw (0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle; \end{scope} \draw (0,0) -- (0,0,1) (0,1) -- (0,1,1) (1,0) -- (1,0,1) (1,1) -- (1,1,1); \end{tikzpicture}
For the description of the XYZ coordinate system, see section 22.2 on p. 249 in the v. 2.10 TikZ manual.
\begin{tikzpicture}[>=stealth',x = 2cm,y = 2cm,z = 0.77cm]
\draw[->] (.9,159/110,0)-- (-.5,-107/110,0) node[anchor = south east]{$\frac{\partial }{\partial y}$};
\draw[->] (-1.3,.15) -- (1.5,.15) node[anchor = north east]{$\frac{\partial }{\partial x}$};
\draw[->] (.15,-1.3) -- (.15,1.5) node[anchor = north east]{$\frac{\partial }{\partial z}$};
\draw (-.5,-.5) -- (.5,-.5) -- (.5,.5) -- (-.5,.5) -- cycle;
\begin{scope}[shift = {(0,0,1)}]
\draw (-.5,-.5) -- (.5,-.5) -- (.5,.5) -- (-.5,.5) -- cycle;
\end{scope}
\draw (-.5,-.5) -- (-.5,-.5,1) (.5,-.5) -- (.5,-.5,1) (.5,.5) -- (.5,.5,1) (-.5,.5) -- (-.5,.5,1);
\end{tikzpicture}