# How can I draw the plane in geogebra?

I want to find a plane that goes through the origin and intersects perpendicularily the plane $$2x+3y+z=12$$.

I have done the following:

The equation of the plane is $$Ax+By+Cz=D$$.

Since it goes through the origin, we get that $$D=0$$.

The plane is perpendicular to $$2x+3y+z=12$$. Therefore their normal vectors are perpendicular, i.e. their dot product has to be equal to $$0$$. $$\begin{equation*}(A,B,C)\cdot (2,3,1)=0 \Rightarrow 2A+3B+C=0 \Rightarrow C=-2A-3B\end{equation*}$$

So a desired plane is of the form $$\begin{equation*}Ax+By+(-2A-3B)z=0\end{equation*}$$ Is that correct? Or can we calculate more coefficients?

Then I want to draw the two planes in geogebra. How can I draw the plane where we don't know all the coefficients?

• geogebra.org/3d?lang=en Here you can draw 3d planes and objects . I think that CalcPlot3D is better at 3d: monroecc.edu/faculty/paulseeburger/calcnsf/CalcPlot3D – Tom Balmas Feb 23 '20 at 20:13
• There is an infinite number of such planes. – amd Feb 23 '20 at 20:24
• So to draw this in geogebra do we take specific values for $A$ and $B$ ? @amd – Mary Star Feb 23 '20 at 20:48
• So to draw this in geogebra do we take specific values for $A$ and $B$ ? – Mary Star Feb 23 '20 at 20:48
• I might pick a specific perpendicular plane—use some specific values for $A$ and $B$—and then add a parameterized rotation about the original plane’s normal to be able to view all of the perpendicular planes. – amd Feb 23 '20 at 20:53

First of all draw with GeoGebra the given plane $$\alpha$$, by writing its equation $$2x+3y+z=12$$ into the input bar.
Select then the tool "Perpendicular line" and use it to draw the line $$r$$ passing through the origin and perpendicular to plane $$\alpha$$. Create then any other point $$P$$ in space not on line $$r$$ (for instance on plane $$\alpha$$) and finally draw the plane passing through line $$r$$ and point $$P$$: that plane is by definition perpendicular to $$\alpha$$.