A, B and C are three points in 3D space. The points A and B are fixed. C is below A and B. $H_{p}$ is any given plane.
Point C moves along the plane $H_{p}$.
How to find the location of point C to minimise the distance $d_{a}+d_{b}$?
$d_{a}$- The distance between A to C
$d_{b}$- The distance between B to C
Let $A^\prime$ and $B^\prime$ be the projections of $A$ and $B$ onto the plane $H_p$. Furthermore, let $h_a$ and $h_b$ be the distances from $A$ to $A^\prime$ and from $B$ to $B^\prime$ respectively and let $w$ be the distance between $A^\prime$ and $B^\prime$.
Clearly, the point $C$ must lie on the line between $A^\prime$ and $B^\prime$, say at a distance $wt$ from $A^\prime$ and distance $w(1-t)$ from $B^\prime$.
Then $$d_a+d_b = \sqrt{h_a^2+w^2t^2}+\sqrt{h_b^2+w^2(1-t)^2}.$$ To find the optimal $t$ we set $\frac{\mathrm{d}(d_a+d_b)}{\mathrm{d}t}=0$ or $$0 = \frac{w^2t}{\sqrt{h_a^2+w^2t^2}}+\frac{-w^2(1-t)}{\sqrt{h_b^2+w^2(1-t)^2}}.$$ Thus $$\frac{w^2t}{\sqrt{h_a^2+w^2t^2}} = \frac{w^2(1-t)}{\sqrt{h_b^2+w^2(1-t)^2}}$$ and squaring both sides and rewriting a bit gives $$w^4t^2(h_b^2+w^2(1-t)^2) = w^4(1-t)^2(h_a^2+w^2t^2).$$ Cancel some terms and we obtain $t^2h_b^2 = (1-t)^2h_a^2$, thus $th_b = (1-t)h_a$ and $t = \frac{h_a}{h_a+h_b}$.