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.


1 Answer 1



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 $z_C$.

  • $\begingroup$ I can' not understand how to write these two equations. Can you be more specific please ? IBI=SQRT(4^2+0^2+zb^2) ICI=SQRT(2^2+2SQRT(3)^2+zc^2) BxC=(0*zc-zb*3.46;-(4*zc-zb*2);4*3.46-0*2) ??? so vector n=(nx,ny,nz) So I can not understand how these 2 equations looks like, can you help? $\endgroup$ Jul 19, 2019 at 10:41
  • $\begingroup$ Sorry, there was a trivial error in my formula for $\vec n$. I corrected it and added the equations for case (a). $\endgroup$ Jul 19, 2019 at 16:40
  • $\begingroup$ As a matter of fact, in both cases I found no solution, even if in case (a) there is a "limiting" solution for $z_B=-z_C \to \pm\infty$. $\endgroup$ Jul 19, 2019 at 16:55

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