# Angle of inclination of the plane to the $x$ and $y$ axis

I have three points that support the plane $$z=0$$ (create an equilateral triangle at the beginning) :

$$A=(0,0,0) \qquad B=(4,0,0) \qquad C=(2,3.46,0)$$

Points $$B$$ and $$C$$ can change their position by changing the $$z$$-coordinate, up and down (independently of each other).

How to change the coordinates of points $$B$$ and $$C$$ so that the plane forms

$$(\textrm{a})$$ An angle of $$30$$ degrees with the $$x$$-axis and an angle of $$60$$ degrees with the $$y$$-axis.

$$(\textrm{b})$$ An angle of $$60$$ degrees with $$x$$-axis and an angle of $$45$$ degrees with $$y$$-axis.

Given $$B=(4,0,z_B)$$, $$C=(2,2\sqrt3,z_C)$$, then the unit vector normal to plane $$ABC$$ is $$\vec n={B\times C\over |B\times C|}$$ and angles $$\theta_x$$, $$\theta_y$$ formed by the plane with $$x$$, $$y$$ axes satisfy: $$\sin\theta_x=|n_x|,\quad \sin\theta_y=|n_y|.$$ In case (a), for instance, these become: $${2\sqrt3 |z_B|\over\sqrt{12z_B^2+192+(2z_B-4z_C)^2}}={1\over2},\quad {|2z_B-4z_C|\over\sqrt{12z_B^2+192+(2z_B-4z_C)^2}}={\sqrt3\over2}.$$ From these two equations you can get $$z_B$$ and $$z_C$$.
• Sorry, there was a trivial error in my formula for $\vec n$. I corrected it and added the equations for case (a). Jul 19, 2019 at 16:40
• As a matter of fact, in both cases I found no solution, even if in case (a) there is a "limiting" solution for $z_B=-z_C \to \pm\infty$. Jul 19, 2019 at 16:55