An equilateral triangle $ABC$ is projected orthogonally from a given plane P to another plane P' intersecting P.
- Show that if two lines are perpendicular in P, then then their orthogonal projections into P' are also perpendicular.
- Show that if $L$ is the line of intersection of the two planes and if $AB$ makes an angle of $\phi$ with $L$, $BC$ and $CA$ make angles of $f(\phi+\pi/3)$ and $f(\phi+2/3 \pi)$ with $L$ or vice versa, where $f(x) = \min\{x, \pi - x\}$.
Here's my attempt to prove 1:
Suppose we have a plane $P$ given by the equation $a_1 x + b_1y+c_1 z = d_1, a_1^2+b_1^2+c_1^2\neq 0$ and another (distinct) plane $P'$ intersecting $P$ given by $a_2x+b_2 y + c_2 z = d_2, a_2^2+b_2^2+c_2^2\neq 0.$ Note that $P'$ must have a different normal than $P$ (one can see this by considering the equations of the planes). From here on, all vectors should be treated as column vectors but will be written in the form $(a,b,c)$ for real numbers $a,b,c$ for convenience. Now, consider the following derivation of the projection of a point $(x_1,x_2,x_3)$ onto P, which can be arbitrary: The projection is the intersection $k$ of the line with equation coordinates $L(t) = (x_1,y_1,z_1) + t(a_1,b_1,c_1)$ with $P$, since $(x_1,x_2,x_3) - k$ is orthogonal to P (being parallel to the normal of P). We get $t(a_1^2 + b_1^2 + c_1^2) + a_1x_1+b_1y_1 + c_1z_1 = d_1\Rightarrow t = \dfrac{d_1-a_1x_1-b_1y_1-c_1z_1}{a_1^2 + b_1^2 + c_1^2}$ and hence $k= (\dfrac{a_1d_1+(b_1^2+c_1^2)x_1-a_1b_1y_1-a_1c_1z_1}{a_1^2 + b_1^2 + c_1^2},\\ \dfrac{b_1d_1-b_1a_1x_1+(a_1^2+b_1^2)y_1-b_1c_1z_1}{a_1^2 + b_1^2 + c_1^2} , \\\dfrac{c_1d_1-c_1a_1x_1-c_1b_1 y_1+(a_1^2+b_1^2)z_1}{a_1^2 + b_1^2 + c_1^2}).$ Let $(x_i,y_i,z_i)$ be in $P$ for $1\leq i\leq 3$. Let $L_i$ be the line through $(x_i,y_i,z_i)$ and $(x_{i+1},y_{i+1},z_{i+1})$ for $i=1,2$. If lines $L_1$ and $L_2$ in plane $P$ are orthogonal (and thus their dot product is zero), then they are also orthogonal in $P'.$ To see why, observe that it suffices to show that if $(x_i',y_i',z_i')$ is the projection of $(x_i,y_i,z_i)$ into $P', $ $(x_3'-x_2',y_3'-y_2',z_3'-z_2')$ is perpendicular to $(x_2'-x_1',y_2'-y_1',z_2'-z_1').$ It seems very tedious to perform the required calculations for this however.
To prove 2, I was thinking of something similarly tedious that involves considering separately the cases where $\phi \leq 60$ and when $\phi > 60$. I think it might be useful to observe that in the diagram below, $BC$ is a rotation of $AB$ about B that's $\pi/3$ radians counterclockwise. I know how to get a formula for the coordinates of a point when it's rotated $\theta$ radians about another point that involves complex numbers, though I think there's a simpler approach to this problem. Diagram (fairly rough):
