How to find the intersection of plane with column and then reflection of a general point in that intersection about a line?

Question

Let there be a column of square cross section having side length 2 units and of infinite length centred on z axis. Let P be a random point lying in the section cut off by the plane $x+2y+2z=30$ in the column. Q is a reflection of P in the plane z=15 and R is locus of point Q.
We Need to find the area of R and plane in which R lies.
I don't know how to proceed even one step further because I don't see a way of finding intersection between the plane and the column. Thank you

Answer 1

The area of a reflection of a planar area is the same as the original area. So the area of the region spanned by $Q$ is the same as the area of the region spanned by $P$, and the latter is just the area of the cut of the column by the given plane.

This area is given by

$ A_{Cut} = \dfrac{A_0 }{ \cos \theta } $

where $A_{Cut}$ is the area of the cut, and $A_0$ is the cross sectional area of the column, and $\theta$ is the angle between the normal to the cross section (which is the vector $k =[0,0,1]^T $) and the normal to the cutting plane, which is $[1,2,2]^T$. Hence

$ \cos \theta = \dfrac{2}{\sqrt{1^2 + 2^2 + 2^2} } = \dfrac{2}{3} $

Therefore, the required area is

$ \text{Area} = \dfrac{3}{2} (2)^2 = 6 $

As for the plane of $R$, it is the reflection of the plane $x + 2 y + 2 z= 30 $ about the plane $z = 15$. Thus if $(x,y,z)$ is on the plane $x + 2y + 2z = 30 $, then when reflecting about the plane $z=15$, on the $z$ coordinate changes, while the $x$ and $y$ coordinates remain the same. The image of $z$ is $z' = 15 - (z - 15) = 30 - z$, Hence $z = 30 - z'$. Substituting this into the equation of the original plane we get

$ x + 2 y + 2 (30 - z') = 30 $

Now we replace $z'$ with $z$, and we get the equation of the reflection of the original plane as

$ x + 2 y - 2 z = - 30 $