Describe what is (almost almost almost) a bijection between (R^{3})* [* signifying dual space] and planes in R^{3}through the origin

Any help with this, I am completely stumped.

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- February 20th 2013, 03:12 PMTimsBobby2Bijection?
Describe what is (almost almost almost) a bijection between (R

^{3})* [* signifying dual space] and planes in R^{3}through the origin

Any help with this, I am completely stumped. - February 20th 2013, 03:25 PMemakarovRe: Bijection?
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.

- February 21st 2013, 11:36 AMDevenoRe: Bijection?
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?