Computing the normal of a plane given 3 points

Question

I have a plane in 3D space and I am trying to compute its normal vector. This is the strategy I'm using right now.

1) Choose any 3 points on the plane. $A, B, C$

2) Find line $AB = B - A$ and line $AC = C - A$

3) $\mathbf{n} = AB\times AC$

4) $\mathbf{\hat{n}} = \frac{1}{|\mathbf{n}|}\mathbf{n}$

I'm pretty sure I'm going wrong with step 1). because I know that the points on the plane must not be collinear. However, I'm not really sure what collinear means in the the context of a 3D plane.

EDIT: these are the 3 points I'm using

However, I sample the corresponding points after I hit the plane with a 4x4 matrix which represents an affine transform. Will these points no longer be collinear?

Answer 1

It's almost perfectly correct but Step (1) should be replaced with

(1) Choose any 3 non-collinear points on the plane, A, B, C.

Collinear means on the same straight line. You would detect collinearity in Step (3) where the cross product would then yield the zero vector.

(However, I think you'd be unlucky if you picked 3 collinear points at random.)

Step (4) is only necessary if you require a unit vector.

After question Edit:

The points in the circles are not collinear. There is no single straight line that passes through all 3 points. If you want to know the effect of affine transformations on points then it sounds like it could be asked as a new question.

Answer 2

It is absolutely right for A,B,C not on the same line ( that is not collinear) since cross product $u\times v$ gives a vector orthogonal to the u-v plane.

Note also that you don’t need step 4 since $\vec n$ is just fine as normal vector (normalization is not necessary).

Finally observe that when we know the plane equation

$$ax+by+cz=d$$

a normal vector is $$\vec n=(a,b,c)$$