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

Question

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.

Answer 1

HINT.

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 $z_C$.