I am trying to figure this out for implementation into a Graphics manipulator I've been trying to work out. I need to find a plane (a normal vector to the plane will suffice) and I know some of its properties:
- A point on the plane $p = (x, y, z)$
- A vector that lies on the plane $\vec{v} = (v1, v2, v3)$
- Another vector that lies on the plane, perpendicular to $v$, originating from $p$, and parallel to another plane with a normal vector $\vec{n} = (n1, n2, n3)$.
How would I go about finding this plane? I know that it is unique because these three constraints define a unique plane in $R^3$ but I am lost on where to start. I believe the first step is finding the vector defined by 3 and then it should be pretty straightforward from there. Thanks in advance.
There's something that I'd like to get clarified, are $\vec{v}$ and $\vec{n}$ vectors originating form $p$ (1) or are they originating from the origin and giving a point in the plane? (2)
Try using the cross product. The cross product of two vectors, $\vec{a}\times\vec{b}$ gives you exactly another vector which is orthogonal to both $\vec{a}$ and $\vec{b}$.
So if the answer is (1) you just need to take the cross product $\vec{v}\times \vec{n}$.
If the answer is (2) you must take the cross product of $(\vec{v}-\vec{p})\times (\vec{n}-\vec{p})$ ($\vec{v}-\vec{p}$ and $\vec{n}-\vec{p}$ are actually the vectors with origin in $p$ and endpoints $v$ and $n$ respectively).
Edit: I see you already know you must use the cross product. Once you calculate it the equation of the plane you need will be $\vec{n}\cdot (\vec{x}-\vec{p})=0$. Where $n$ is the normal vector, $x$ your coordinates and $p$ the point the plane must pass through.