Which of the following sets are planes(in their various n-spaces)?

Question

A = { $\begin{pmatrix}0 \\1 \\-1\\0 \end{pmatrix}$ + s $\begin{pmatrix}-2 \\0 \\1\\0 \end{pmatrix}$ + t $\begin{pmatrix}-1 \\2 \\0\\1 \end{pmatrix}$ | s,t $\in$ R }

B = { $\begin{pmatrix}w \\x\\y\\z \end{pmatrix}$ | w+ 3x - y - 2z=1 }

C = { $\begin{pmatrix}w \\x\\y\\z \end{pmatrix}$ | w - 5y +z = 3, x+ 2y =1 }

D = { $\begin{pmatrix}w \\x \end{pmatrix}$ | w, x $\in$ R}

E = { $\begin{pmatrix}1 \\a\\b \end{pmatrix}$ | a = b $\in$ R }

Which of the following sets are planes(in their various n-spaces)?

I think A, C, D are the sets which are planes in their spaces, as they each have two free variables/arbitrary parameters. But, I am not sure if set E counts as a plane? Any help is appreciated!

Answer 1

E is not a plane, because there is only one free parameter, which means it is one dimensional. That same space can be written as:

\begin{equation} E = \{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix}0 \\ 1 \\ 1 \end{pmatrix} : \lambda \in \mathbb{R}\} \end{equation}

which is a line in $\mathbb{R}^3$.

However, strictly speaking, lines are $1-$planes, but I assume you want those that are $2-$planes (i.e. two-dimentional).

You are correct in that A, C and D are all $2-$planes, using the same argument.

Answer 2

Depends on the definition: A plane is a $2$-dimensional subspace; a subspace of dimension $n-1$ is called a hyperplane. In that sense (A) is a plane but no hyperplane, (B) isn't a plane but a hyperplane, (C)is a plane but not a hyperplane and (E) is neither.

Case (D) may be regarded a a plane, but no a hyperplane.