Given R find P

We have that R = (c,s,h) where $c = cos(\theta)$ and $s = sin(\theta)$ for some $\theta$. Introduce a new point Q which is the intersection of the cylinder x² + y² = r² with the line segment $\overline{RP}$. Then, refering to a top view, the x and y coordinates of Q are $rc_2 = r cos(2 \theta) = r(2 c^2 - 1)$ and $r s_2 = r sin(2 \theta) = 2 r c s$ respectively. Also, in the Z coordinate, Q is as far below R as E is above R. Thus, the Z coordinate of Q is h-(e-h) = 2 h - e. Thus,

Q = ( r ( 2 c² - 1 ), 2 r c s, 2 h - e) (6)

P is on the intersection of the z=0 plane and the line through R and Q. Thus,

P = R + t(Q-R) (7)

has z coordinate 0. Thus, h + t(2h-e-h) = 0 gives t = h/(e-h) which, when substituted into (7) gives

$$P = (1-\frac{h}{e-h})(c,s,0) + \frac{h r}{e - h} (2 c^2-1, 2 c s, 0).$$ (8)