Find a basis for all vectors perpendicular to $x-2y+3z=0$. I am looking at the following question:
Find a basis for all vectors perpendicular to $x-2y+3z=0$.
Clearly $<1,-2,3>$ is a vector that must be in the basis or a scalar multiple of $<1,-2,3>$ must be in the basis. But the solution given says that this vector is enough to describe a basis for all orthogonal vectors to the plane.
However, my initial intuition when solving this problem told me that this was not enough to describe the basis of all perpendicular vectors. My understanding is that any basis of a vector space is a minimal set of vectors whose linear combinations span the space.
My answer was that we would need to take a basis of the plane, say, $<2,1,0>$ and $<-3,0,1>$ and then combine that with $<1,-2,3>$. The reason for this is that a perpendicular vector need not be fixed. That is, we can move anywhere on the plane, and then move in the orthogonal dimension to that plane, by taking any scalar multiple of $<1,-2,3>$ and adding it to a linear combination of $<2,1,0>$ and $<-3,0,1>$. How does the basis $<1,-2,3>$ capture the orthogonal vector $<2,1,0>+<-3,0,1>+<1,-2,3>=<0,-1,4>$? Perhaps my plane/vector geometry is off. Also, my answer seems to suggest that the basis for the set of perpendicular vectors is also a basis for $\mathbb{R}^3$, which also seems odd.

It appears that you’re confusing yourself a bit by considering what are often called bound vectors: geometrical (a.k.a. Euclidean) vectors that have magnitude, direction, and a particular starting point. Linear Algebra concerns itself with vectors as abstract objects—elements of vector spaces—which means that when you’re dealing with geometrical vectors in this context, you’re generally dealing with free vectors for which the starting point is of no importance. If you like, these free vectors are equivalence classes of bound vectors that have the same magnitude and direction. The set of free vectors orthogonal to the given plane is one-dimensional.
