# Planes through $OX$ and $OY$ include an angle $\alpha,$ show that their line of intersection lies on the cone $z^2(x^2+y^2+z^2)=x^2y^2\tan^2\alpha$

Planes through $OX$ and $OY$ include an angle $\alpha,$ show that their line of intersection lies on the cone $z^2(x^2+y^2+z^2)=x^2y^2\tan^2\alpha$

The lines of intersection of the planes through $OX$ and $OY$ will lie on the $xy-$plane but i do not know how to prove this question.

The line of intersection of the planes through $OX$ and $OY$ does not lie on the $xy-$plane. Let $n_1$ and $n_2$ be the unit vectors perpendicular to the planes. As the first plane passes through $OX$ then $n_1$ will be parallel to the $yz-$plane, and similarly $n_2$ will be parallel to the $xz-$plane. Thus: $$n_1=(0,a,b);\quad n_2=(c,0,d);$$ where: $$a^2+b^2=1,\quad c^2+d^2=1\quad\hbox{and}\quad bd=\cos\alpha.$$ The last equation comes from the condition on the angle formed by the planes, which is equivalent to $n_1\cdot n_2=\cos\alpha$. We then obtain: $$d=(\cos\alpha)/b,\quad a=\sqrt{1-b^2},\quad c=\sqrt{1-(\cos^2\alpha)/b^2}.$$ The direction of the intersection line is given by $n_3=n_1\times n_2=(ad,bc,-ac)$ and the intersection line itself has a parametric equation given by $(x,y,z)=n_3t=(ad,bc,-ac)t$. You can now easily check that this line lies on the surface having equation $$z^2(x^2+y^2+z^2)=x^2y^2\tan^2\alpha.$$ Just plug there: \begin{align} &x=ad={\cos\alpha\over b}\sqrt{1-b^2},\\ &y=bc=\sqrt{b^2-\cos^2\alpha},\\ &z=-ac=-\sqrt{(1-b^2)\left(1-{\cos^2\alpha\over b^2}\right)}\\ \end{align} and verify that both sides of the equation give the same result.