I am dealing with normal vector equations at work and re-reviewing the subject and something bothers me,
From this document, like many others available online we can see that normal is modeled as N = (A,B,C) , without any position vector as N = P0 + (A,B,C) but is not drawn from the origin but simply from a random point in which it will be perpendicular to the plane,
http://jongarvin.com/up/MCV4U/slides/scalar_plane_handout.pdf
Now, while I understand that vectors are moveable and that we can move the position of a vector, the plane itself is modeled with a position vector : X = P0+tv+su
Cant it cause any miscalculations that the plane is bounded and constrained by a position vector while the normal is not?
If vectors are moveable why is the plane have a position vector while the normal does not? Why cant we simply move both vectors that define the plane to to the origin?
Am I missing or misunderstanding something ?
Perhaps my entire approach is wrong ?
Thank you and regards ,
James
Normal vectors are always referred to the origin even if the plane is in the form
$$ax+by+cz=d\neq 0, \quad P(s,t)=P_0+s\vec u + t \vec v, \, P_o\neq \vec 0$$
since it is parallel to the plane through the origin
$$ax+by+cz=0,\quad P(s,t)=s\vec u + t \vec v$$
with $$\vec n=\lambda(a,b,c), \quad \vec n=\lambda \vec u \times \vec v$$