Show that the plane $3x-2y-z=0$ cuts the cones $21x^2-4y^2-5z^2=0$ and $3yz-2zx+2xy=0$ in the same pair of perpendicular lines.

Question

Show that the plane $3x-2y-z=0$ cuts the cones $21x^2-4y^2-5z^2=0$ and $3yz-2zx+2xy=0$ in the same pair of perpendicular lines.

I am not understanding the problem. Don't what is the meaning of "same pair of perpendicular lines"? Please help me to solve the problem.

Answer 1

Substituting $z=3x-2y$ into the equations of the two cones, you get in both cases: $$ 2x^2+2y^2-5xy=0, \quad\hbox{that is:}\quad (2x-y)(x-2y)=0. $$ The last equation is equivalent to $2x-y=0$ or $x-2y=0$, which are the equations of two planes. Combining these with $z=3x-2y$ we then obtain two lines, whose direction vectors are $(1,2,-1)$ and $(2,1,4)$.