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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.

messi
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  • All three coordinate axes lie on this cone, so, for example, every coordinate plane cuts the cone in perpendicular lines. Can you visualize that? Now rotate this plane about the cone’s axis. – amd Sep 17 '19 at 01:31

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