Find the third vertex of a triangle in $3D$ I'm trying to get the third vertex of triangle in the $3D$ space $xyz$ and I have:


*

*$2$ points

*Normal of the plane where the triangle is

*I know the triangle has an angle of $90^{\circ}$ that is on the third unknown point
Does anyone have any idea on how to solve this? I've been searching for a solution to this, but I didn't have any success...
Any help would be greatly appreciated!

Use the fact that for a plane passing through a point $\mathbf{x_1}$ with normal $\mathbf{n}$, all points $x$ on it satisfy the equation:
$$\mathbf{n}\cdot(\mathbf{x} - \mathbf{x_0}) =0 $$
You have $\mathbf{n},\mathbf{x_1},\mathbf{x_2}$ and want to find $\mathbf{x_3}$. 
Since the angle at $\mathbf{x_3}$ is $90^\circ$, we have two equations:
$$(\mathbf{x_3} - \mathbf{x_1})\cdot(\mathbf{x_3}-\mathbf{x_2})=0$$
$$ \mathbf{n}\cdot(\mathbf{x} - \mathbf{x_3}) =0 $$
But we have three unknowns! this means the system is underdetermined, and that there are an infinite number of solutions, which indeed there are:
$$ $$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
