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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?

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  • $\begingroup$ 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 $\endgroup$ – Tom Balmas Feb 23 '20 at 20:13
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    $\begingroup$ There is an infinite number of such planes. $\endgroup$ – amd Feb 23 '20 at 20:24
  • $\begingroup$ So to draw this in geogebra do we take specific values for $A$ and $B$ ? @amd $\endgroup$ – Mary Star Feb 23 '20 at 20:48
  • $\begingroup$ So to draw this in geogebra do we take specific values for $A$ and $B$ ? $\endgroup$ – Mary Star Feb 23 '20 at 20:48
  • $\begingroup$ 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. $\endgroup$ – amd Feb 23 '20 at 20:53
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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 $\alpha$.

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