Question

The plane $lx+my=0$ is rotated about its line of intersection with the plane $z=0$ through angle measure $\alpha$. Prove that the equation of the plane in new position is $lx+my\pm z\sqrt{l^2+m^2}\tan \alpha =0$.

Answer 1

Plane:$lx+my=0$ rotated about $z=0$ through $\alpha$.

Plane through $lx+my=0$ and $z=0$ line of intersection is

$lx+my+\lambda z=0$

Now the rotated plane also passes through line of intersection hence its equation will also be $lx+my+\lambda z=0$

If the plane is rotated by angle $\alpha$ then the angle also gets rotated by $\alpha$.Hence ,angle between the normal of earlier plane and rotated must be $\alpha$

$\therefore \cos \alpha =\frac{l^2+m^2}{\sqrt{l^2+m^2}\sqrt{l^2+m^2+{\lambda }^2}}\cos {}^2\alpha =\frac{l^2+m^2}{l^2+m^2+{\lambda }^2}(l^2+m^2+{\lambda }^2)\cos {}^2\alpha =l^2+m^2{\lambda }^2\cos {}^2\alpha =(l^2+m^2)(1-\cos {}^2\alpha ){\lambda }^2=(l^2+m^2)\frac{\sin {}^2\alpha }{\cos {}^2\alpha }\lambda =\pm \sqrt{l^2+m^2}\tan \alpha$

$\therefore$ equation of rotated plane will be,

$lx+my\pm \sqrt{l^2+m^2}\tan \alpha =0$