Calculation of third vertices of a triangle given a vector that should be perpendicular to the triangle plane

Question

I have an isosceles triangle (2 sides same length) with vertices O, A and B. OA and OB are the same length.

Vertices O and A are known, with O at origin (0,0,0).

A known vector V, should be perpendicular to the plane formed by the triangle.

Calculate vertices B.

This is the opposite of the fairly easy calculation of determining the perpendicular vector to the plane of a triangle, which is calculated by the dot product of the vectors OA and OB.

Answer 1

With the given data, there isn't that much to say about the vertex $B$. What we can say, is where $B$ might be. When given a vector $v$, we can find the perpendicular plane through the origin and know that all of $O$, $A$ and $B$ lie in this plane. Since $OA=OB$, all possibilities for $B$ lie on a circle in this plane with centre $O$ and radius $OA$. (In case one does not want to consider degenerate triangles, the points $A$ and $-A$ should be removed.)

It might be a somewhat different story, though, if the length of the perpendicular somehow depends on the angle $\angle AOB$, but I don't see any mention of this being the case in the question.