Is it possible to get a normal vector to the plane if i only know 3 points which exist on this plane?

Question

I have 3 points: $A(0;0;0), B(0;0;1), C(2;2;1) $. They exist on the plane. I assumed that scalar product of the normal vector and a line which exists on the same plane will be equal to 0. Scalar product equals to $x*2+y*2+z*1=0$ where $x,y,z$ are coordinates of the normal vector. Finally i can get $x,y,z$ using a selection method. For example $x,y,z$ can be $0,1,-1$. However i think it's a wrong assumption.

Answer 1

Let us say that we have any three arbitrary points $A$, $B$, $C$ that are unique. Define the following vectors:

$$\overrightarrow{AB} = B - A$$

$$\overrightarrow{AC} = C - A$$

which both lie along the plane that contains the three points. Now we form the normal vector by using the cross product:

$$\overrightarrow{AB} \times \overrightarrow{AC}$$

which is guaranteed to be perpendicular to the two above.

Answer 2

As long as there are enough points to define a plane, there will exist a normal vector to the plane; as a matter of fact, a plane is defined by $$ \vec{N}\cdot\left( \vec{a} + \vec{b} \right) = 0 $$ Where $\vec{a}$ and $\vec{b}$ are vectors that go from one point in space to another. We call the vector $\vec{N}$ that satisfies the above a normal vector.