Plane rotated about line of intersection to another plane?

Question

The plane ax+by=0 is rotated about its line of intersection with the plane z=0 through an angle n. What is the equation of the plane in its new position? I saw a question in stack exchange relating to 3d plane rotation but its answers were of a more advanced level. Would someone please help me with an easy visualisation and solving of this problem?

Answer 1

A general plane has the equation

$$px+qy+rz=s.$$

If you plug the condition $z=0$, the equation of all planes by the line $z=0$ are

$$px+qy=s.$$

If you observe that the origin $(0,0,0)$ belongs to the line of intersection, $s=0$ and the equation is

$$px+qy=0.$$

Now we want to express that the angle between the given plane and the unknown one is $\theta$, or that the angle between the unit normal vectors is $\theta$, we have that their dot product is the cosine of $\theta$, and

$$\frac{ap+bq}{\sqrt{(a^2+b^2)(p^2+q^2)}}=\cos\theta.$$

To solve this equation for $p,q$, it is easier to express $(a,b)$ and $(p,q)$ in polar coordinates. After simplification,

$$\cos(\phi_{ab}-\phi_{pq})=\cos\theta.$$

Answer 2

Take a unit vector $\bf t$ parallel to the intersection line, e.g. take the difference of two points that satisfy the equations of both planes, and then make it unitary.
Otherwise take the cross product of the unit normals to each plane $ \bf n_1 \times \bf n_2$.

After that $\bf m = \bf n_1 \times \bf t$ is a unit vector that lies on the first plane and is normal to the intersection line.
The set $(\bf t, \bf m, \bf n_1)$ has the same chirality as $(x,y,z)$.
Rotating the plane by an angle $\alpha$ around $t$ according to the right-hand rule, gives a new normal to the plane which is $\bf n_1'= - \sin \alpha\, \bf m + \cos \alpha \, \bf n_1$.

Having the new normal $\bf n_1'$ and a point on the intersection line, you have the equation of the rotated plane.