Coordinate geometry reflection of point

Question

I have point in $1st$ octant($ x, y, z$ all positive). Now I take the mirror image of that point about $xy$ plane. I guess that new point will be simple $ (x, y ,-z)$. Verify if I am right. Further assume that I am given a point S$(0, 0 ,ha)$ and other point $T(5, 6, 7)$. Now there will be a ray (just one as i have fixed two points) emanating from source S and after reflection from $xy$ plane will pass through T. I have to know the incident or reflection angle about normal drawn about the $xy$ plane from point of incidence. Thanks in advance.

Answer 1

You could use vector arithmetic, or you can just use coordinate geometry. The ray that reflects to $T=(5,6,7)$ also passes through $(5,6,-7).$ You can construct a right triangle, one of whose legs runs along the $z$-axis from $(0,0,-7)$ to $(0,0,ha),$ and the other leg goes from $(0,0,-7)$ to $(5,6,-7).$ One of the angles of this triangle is the angle you want. (It's the angle between your ray and the $z$-axis, which is also the angle of incidence of reflection since the $z$-axis is parallel to the normal vector at the point of incidence.)

Answer 2

To find the ray which passes from $S$ via one reflection to $T$, find the ray which would go from $S$ to the mirror image of $T$ in a straight line. I.e. the ray from $S$ to $T'(5,6,-7)$.

You can compute the angle that ray makes with the plane or its normal based on the dot product between the normalized direction vector of the ray and the normalized normal vector of the plane.