Equation of cone

Question

The plane $lx+my+nz=0$ moves in such a way that its intersection with the planes $$ax+by+cz+d=0$$ $$a'x+b'y+c'z+d'=0$$ are perpendicular. Show that the normal to the plane through the origin describes in general a cone of second degree and find its equation?

Answer 1

The fact that the plane $\pi_0 \;:\, lx+my+nz=0$ "moves" means of course that the parameters $l,m,n$ vary.

The intersection between two planes is a line normal to both the normals to the planes.

Hence $$ \eqalign{ & {\bf t}_{\,1} = \left( {l,m,n} \right) \times \left( {a,b,c} \right) = \left( {cm - bn, - cl + an,bl - am} \right) \cr & {\bf t}_{\,2} = \left( {l,m,n} \right) \times \left( {a',b',c'} \right) = \left( {c'm - b'n, - c'l + a'n,b'l - a'm} \right) \cr} $$ and we shall impose that the two lines remain normal to each other, i.e. $$ {\bf t}_{\,1} \cdot {\bf t}_{\,2} = 0 = \left( {cm - bn} \right)\left( {c'm - b'n} \right) + \left( { - cl + an} \right)\left( { - c'l + a'n} \right) + \left( {bl - am} \right)\left( {b'l - a'm} \right) $$

Expand that to get a quadratic equation in $l,m,n$. I think you can proceed by yourself now.