Isn't the dual space to R^3 isomorphic to R^3? And planes going through the origin have equations Ax + By + Cz = 0 for various (but not all) real A, B and C.
Isn't the dual space to R^3 isomorphic to R^3? And planes going through the origin have equations Ax + By + Cz = 0 for various (but not all) real A, B and C.
suppose we have an element of (R^{3})*.
this is a linear function f:R^{3}-->R. since f is linear, f(x,y,z) = ax + by + cz.
now, consider the mapping φ:P-->(R^{3})*, where P is the set of planes through the origin, given by:
if p is the plane defined by Ax + By + Cz = 0, φ(p) = f, where f(x,y,z) = Ax + By + Cz.
the question is now: why is this not a bijection?
it might be instructive to note that:
1) the planes x + y + z = 0 and 2x + 2y + 2z = 0 are the same plane, but the linear functionals f(x,y,z) = x + y + z and g(x,y,z) = 2x + 2y + 2z are different.
2) the 0-functional does not correspond to any plane (why?).
if you are clever enough, can you think of a way to express φ(p) as an inner product of the vector (A,B,C) and something? how does this relate to the normal vector to a plane p?
since 3 = 1 + 2, can you see some connection here?