I have a plane $P_1$ given by the equation: $$ ax+by+cz+d=0$$ I want to find an orthogonal plane to that. I know that their normal vectors should be orthogonal so the normal vector $v_1$ of $P_1$ is $(a,b,c)$ and to find the normal vector $v_2=(a',b',c')$ you could fix random $a'$ and $b'$ and then you solve the equation of their dot product to be equal $0$ since they are orthogonal to find $c'$. But then from $v_2$ how do i proceed to define the plane $P_2$?
[Math] How do i define a plane orthogonal to a given one
algebraic-geometrygeometrylinear algebra
Best Answer
There are infinitely many planes orthogonal to your given plane (so you can't ask for "the" plane $P_2$). That said...
One (fairly simple) way to find a plane orthogonal to $ax+by+cz+d=0$ is to pick to points on your plane, say, $(x_1,y_1,z_1)$ (where $ax_1+by_1+cz_1+d=0$) and $(x_2,y_2,z_2)$ (where $ax_2+by_2+cz_2+d=0$) -- this can be done by randomly picking a couple of (x,y) coordinate pairs and then solving for the corresponding $z$-coordinate.
Once you have these two points, ${\bf n} = (x_2-x_1,y_2-y_1,z_2-z_1)$ is a vector parallel to your original plane and so it is normal to your desired plane. Thus $$(x_2-x_1)(x-x_1) + (y_2-y_1)(y-y_1) + (z_2-z_1)(z-z_1) = 0$$
is orthogonal to your original plane.