For my project, I want to find a symmetrical plane between two planes. I have two equation of planes in form ax + by + cz + d = 0 which intersect each other(at approx. 90 degree). I created these two planes using 3 points to form each plane. Now I want to find a symmetrical plane (i.e approx 45 to each of these planes) which passes through the intersection line of two planes.
How can I do it?
Let's indicate with $n_1=(a_1,b_1,c_1)$ and $n_2=(a_2,b_2,c_2)$ the normalized normal vectors to the given planes $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$, thus for the third plane you can choose:
$$n_3=(a_3,b_3,c_3)=(a_1\pm a_2,b_1\pm b_2,c_1\pm c_2)$$
note: $\pm$ sign need to be choosen depending on the direction of $n_1$ and $n_2$
and find $d_3$ by the condition that it passes through a point in common with the 2 given planes.