The plane $lx+my+nz=0$ moves in such a way that its intersection with the planes $ax+by+cz+d=0$ and $a'x + b'y + c'z+d'=0$ are perpendicular. Show that the normal to the plane through the origin describes in general, a cone of the second degree and find its equation.
My analysis
Here the given plane $lx+my+nz=0$ passes through the origin, so considering a normal dropped from origin is an incorrect term
Where am I going wrong?
Soham
The intersection with first plane gives the following line (arrive at this just by linear algebra):
$$\frac x{cm-bn}=\frac y{an-cl}=\frac z{bl-am}$$
Similarly the intersection with the second line is:
$$\frac x{c'm-b'n}=\frac y{a'n-c'l}=\frac z{b'l-a'm}$$
For these lines to be perpendicular the direction ratios' inner product should be 0. i.e., $$(cm-bn)(c'm-b'n)+(an-cl)(a'n-c'l)+(bl-am)(b'l-a'm)=0$$ i.e., $$(bb'+cc')l^2+(cc'+aa')m^2+(aa'+bb')n^2-(ab'+a'b)lm-(bc'+b'c)mn-(ac'+a'c)ln=0$$
Try and show that this is the equation of a cone.