I was reading some geometry problems, and I have noticed something interesting about projections. Say we have a plane $\pi_{1}$, and a point A (a,b,c). We want to find the shortest distance between the plane and the point, which is straightforward:
Compute a "difference vector" between an arbitrary point in the plane and the point A (a,b,c) in question. Then, we can deduce a vector normal to the plane $\pi_{1}$ from the given equation of the plane. Now, we calculate the projection of this "difference vector" onto the normal vector, which will yield a vector that can give us the minimum distance.
However, what thrills me is the fact, can't we say that as well a scalar multiple of the normal vector is as well normal to this plane? Say we have a normal vector $n=(\alpha,\beta ,\omega)$, can't we choose a really small scalar multiple of this normal vector? Say $n=(0.0001\alpha,0.0001\beta,0.0001\omega)$, and then calculate the projection of the difference vector onto this small normal vector? We would then get a really small vector which would not represent a "distance vector". How do we now that the normal vector $n$ is actually "larger" to the minimum distance between a point and a plane? I have encountered this as well when dealing with minimum distance between two perpendicular lines.
I do understand how the process works, but I want to be sure HOW can we assume that the normal vector is a suitable one to do the projection.
Best Answer
You just compute the scalar product between any vector $\vec{AP}$ with $P\in\pi$ and the normal vector renormalized to have length 1.