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I came across a problem which requires to prove that plane $ax + by + cz=0$ cuts cone $xy+yz+xz = 0$ in perpendicular lines if $1/a + 1/b + 1/c = 0$

Solution to the problem says that since given cone is generated by three mutually perpendicular planes, hence plane $ax + by + cz=0$ will cut it in perpendicular lines if normal to plane through vertex (0,0,0) lies on cone itself.

I am unable to visualise graphically how such a plane can cut cone in perpendicular lines. Why is it necessary for normal to plane through $(0,0,0)$ to lie on cone?

I am assuming that lines being referred in question are the boundaries of the cone which plane would touch when cutting across cone. Any graph/picture would be thankful.

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  • $\begingroup$ This cone is the surface of rotation of any of the coordinate axes about the line $x=y=z$. So in particular, each of the coordinate planes intersects this cone in a pair of perpendicular lines: two of the coordinate axes. $\endgroup$
    – amd
    Commented Sep 15, 2017 at 21:09
  • $\begingroup$ Is it possible to visualise this somewhere? $\endgroup$ Commented Sep 18, 2017 at 8:50
  • $\begingroup$ You could plot the cone and play around with various planes through the origin in Wolfram|Alpha or a tool like GeoGebra. $\endgroup$
    – amd
    Commented Sep 18, 2017 at 9:19

1 Answer 1

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Plane cutting cone in perpendicular lines

You can see here for interactive graph

https://www.geogebra.org/3d/ag7rjf5c
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