Writing geometric lines and planes as sets I am investigating some differences between geometry in $\mathbb R^2$ and $\mathbb R^3$. Define a line in the $\mathbb R^2$ as the set $\{(x,y)\in \mathbb R^2 : ax+by=c\}$ for fixed $a,b\in \mathbb R$ with $(a,b)\ne (0,0)$.  
Likewise, define a plane in $\mathbb R^3$ as $\{(x,y,z)\in \mathbb R^3 : ax+by+cz=d\}$ with $(a,b,c)\ne (0,0,0)$.
I want to show that any plane in space can be written as  $\{\vec{v}+\lambda \vec{w}+\mu \vec{u}:\lambda, \mu \in \mathbb R\}$ for some $\vec{v}, \vec w, \vec u$, and vice versa.
I also want to show that any line in $\mathbb R^2$ can be written as $\{\vec v+\lambda \vec w :\lambda \in \mathbb R\}$ but that the "vice versa" part does not hold in this case. For this last part, if we set $\vec w=\vec 0$, then the set would be a singleton, but I'm not sure how to show that a singleton can't be a line.
I would really appreciate some help with these problems.
 A: To get you started, here's the first part of the backward proof. The forward is essentially this reversed. You do a similar thing for lines.

Let plane $P=\{\vec{v}+\lambda \vec{w}+\mu \vec{u}:\lambda, \mu \in \mathbb R\}\,\,,\,\,\,\vec w \times \vec u \neq 0$. (This condition means $\vec w$ and $\vec u$ aren't parallel, so you have a plane for sure)
Then
$$P=\left\{\begin{pmatrix} v_1 \\ v_2 \\ v_3\end{pmatrix}+\lambda \begin{pmatrix} w_1 \\ w_2 \\ w_3\end{pmatrix}+\mu \begin{pmatrix} u_1 \\ u_2 \\ u_3\end{pmatrix}:\lambda, \mu \in \mathbb R\right\}$$
$$P=\left\{(x,y,z)\in \mathbb R^3:\begin{cases}x=v_1+\lambda w_1+\mu u_1 \\y=v_2+\lambda w_2+\mu u_2\\z=v_3+\lambda w_3+\mu u_3\end{cases}\,\,\,,\,\,\,\lambda, \mu \in \mathbb R\right\}$$

For the line in $\mathbb R^2$:
Let line $L=\{(x,y)\in \mathbb R^2 : ax+by=c\}\,\,\,,\,\,\,a,b,c\in\mathbb R\,\,\,,\,\, (a,b)\ne (0,0)$.
Cases:
$a=0\implies$
$$L=\left\{(x,y)\in \mathbb R^2 : (x\in \mathbb R) \land \left(y=\frac{c}{b}\right)\right\}\,\,\,,\,\,\,b,c\in\mathbb R\,\,\,,\,\, b\ne 0$$
$$=\left\{\begin{pmatrix}x\\y\end{pmatrix} \in \mathbb R^2 : \begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\\frac{c}{b}\end{pmatrix}+\lambda\begin{pmatrix}1\\0\end{pmatrix}\,\,\,,\,\,\,\lambda\in\mathbb{R}\right\}\,\,\,,\,\,\,b,c\in\mathbb R\,\,\,,\,\, b\ne 0$$
$$=\left\{\begin{pmatrix}0\\\frac{c}{b}\end{pmatrix}+\lambda\begin{pmatrix}1\\0\end{pmatrix}:\lambda\in\mathbb{R}\right\}\,\,\,,\,\,\,b,c\in\mathbb R\,\,\,,\,\, b\ne 0$$
$\implies$ $L$ is in the form $\{\vec v+\lambda\vec w :\lambda\in\mathbb R\}$ for some $\vec v,\vec w\in\mathbb R^2\,\,\,,\,\, \vec w\ne \vec0$.

$b=0\implies$ wlog, switching $x$ and $y$, use proof for case ($a=0$).
$\implies$ $L$ is in the form $\{\vec v+\lambda\vec w :\lambda\in\mathbb R\}$ for some $\vec v,\vec w\in\mathbb R^2\,\,\,,\,\, \vec w\ne \vec0$.

$ab\neq0\implies$
Let $\begin{pmatrix} v_1 \\ v_2\end{pmatrix}=\begin{pmatrix} c\over a \\  0\end{pmatrix}$ and $\begin{pmatrix} w_1 \\ w_2\end{pmatrix}=\begin{pmatrix} -b \\ a\end{pmatrix}\,\,\,,\,\,\,w_1w_2\neq0$.
Then
$$L=\left\{\begin{pmatrix} x \\ y\end{pmatrix} \in \mathbb R^2 : \frac{x}{w_1}-\frac{v_1}{w_1}=\frac{y}{w_2}-\frac{v_2}{w_2}\right\}\,\,\,,\,\,\,\begin{pmatrix} v_1 \\ v_2\end{pmatrix},\begin{pmatrix} w_1 \\ w_2\end{pmatrix}\in\mathbb R^2\,\,\,,\,\, w_1w_2\neq0$$
$$=\left\{\begin{pmatrix} x \\ y\end{pmatrix} \in \mathbb R^2 : \begin{cases}\lambda=\frac{x}{w_1}-\frac{v_1}{w_1} \\ \lambda=\frac{y}{w_2}-\frac{v_2}{w_2}\end{cases}\,\,\,,\,\,\,\lambda\in\mathbb R\right\}\,\,\,,\,\,\,\begin{pmatrix} v_1 \\ v_2\end{pmatrix},\begin{pmatrix} w_1 \\ w_2\end{pmatrix}\in\mathbb R^2\,\,\,,\,\, w_1w_2\neq0$$
$$=\left\{\begin{pmatrix} x \\ y\end{pmatrix} \in \mathbb R^2 : \begin{cases}x=v_1+\lambda w_1\\y=v_2+\lambda w_2\end{cases}\,\,\,,\,\,\,\lambda\in\mathbb R\right\}\,\,\,,\,\,\,\begin{pmatrix} v_1 \\ v_2\end{pmatrix},\begin{pmatrix} w_1 \\ w_2\end{pmatrix}\in\mathbb R^2\,\,\,,\,\, w_1w_2\neq0$$
$$=\left\{\begin{pmatrix}x\\y\end{pmatrix}\in\mathbb R^2: \begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}v_1\\v_2\end{pmatrix}+\lambda\begin{pmatrix}w_1\\w_2\end{pmatrix}:\lambda\in\mathbb{R}\right\}\,\,\,,\,\,\,\begin{pmatrix} v_1 \\ v_2\end{pmatrix},\begin{pmatrix} w_1 \\ w_2\end{pmatrix}\in\mathbb R^2\,\,\,,\,\, w_1w_2\neq0$$
$$=\left\{\begin{pmatrix}v_1\\v_2\end{pmatrix}+\lambda\begin{pmatrix}w_1\\w_2\end{pmatrix}:\lambda\in\mathbb{R}\right\}\,\,\,,\,\,\,\begin{pmatrix} v_1 \\ v_2\end{pmatrix},\begin{pmatrix} w_1 \\ w_2\end{pmatrix}\in\mathbb R^2\,\,\,,\,\, w_1w_2\neq0$$
$\implies$ $L$ is in the form $\{\vec v+\lambda\vec w :\lambda\in\mathbb R\}$ for some $\vec v,\vec w\in\mathbb R^2\,\,\,,\,\, \vec w\ne \vec0$.
A: Backwards proof in $\mathbb R^3$ with vectors. It requires knowledge of cross and dot products. This proof is easy to reverse.

Let plane $P=\{\vec{v}+\lambda \vec{w}+\mu \vec{u}:\lambda, \mu \in \mathbb R\}\,\,\,,\,\,\,\vec u,\vec v,\vec w \in\mathbb R^3\,\,,\,\,\,\vec w \times \vec u \neq 0$. (This condition means $\vec w$ and $\vec u$ aren't parallel, so you have a plane for sure)
Then $$P=\{\vec x \in \mathbb R^3: \vec x=\vec{v}+\lambda \vec{w}+\mu \vec{u}:\lambda, \mu \in \mathbb R\}\,\,\,,\,\,\,\vec u,\vec v,\vec w \in\mathbb R^3\,\,,\,\,\,\vec w \times \vec u \neq 0$$
$$=\{\vec x \in \mathbb R^3: (\vec w \times \vec u).\vec x=(\vec w \times \vec u).\vec{v}+\lambda (\vec w \times \vec u).\vec{w}+\mu (\vec w \times \vec u).\vec{u}:\lambda, \mu \in \mathbb R\}\,\,,\,\,\vec u,\vec v,\vec w \in\mathbb R^3\,\,,\,\,\,\vec w \times \vec u \neq 0$$
Now, in general, $\vec a \times \vec b$ is a vector perpendicular to $\vec a$ and $\vec b$, so $\vec a.(\vec a \times \vec b)=\vec b.(\vec a \times \vec b)=\vec0$.
Therefore $$P=\{\vec x \in \mathbb R^3 : (\vec w \times \vec u).\vec x=(\vec w \times \vec u).\vec{v}\}\,\,\,,\,\,\,\vec u,\vec v,\vec w \in\mathbb R^3\,\,,\,\,\,\vec w \times \vec u \neq 0$$
Let $\vec a=\vec w \times \vec u \neq \vec 0$ and $d=(\vec w \times \vec u).\vec{v}$. (Note that $\vec a$ and $d$ are constants as they are functions of constant vectors)
So
$$P=\{\vec x \in \mathbb R^3: \vec a.\vec x=d\}\,\,\,,\,\,\,\vec a \in \mathbb R^3\,\,\,,\,\,\, \vec a\neq\vec0\,\,\,,\,\,\,d\in\mathbb R$$
$$=\left\{\begin{pmatrix} x \\y \\ z \end{pmatrix}\in \mathbb R^3 : \begin{pmatrix} a \\b \\ c \end{pmatrix}.\begin{pmatrix} x \\y \\ z \end{pmatrix}=d\right\}\,\,\,,\,\,\,a,b,c,d\in\mathbb R\,\,\,,\,\, \begin{pmatrix} a \\b \\ c \end{pmatrix}\ne \begin{pmatrix} 0 \\0 \\ 0 \end{pmatrix}$$
$$=\{(x,y,z)\in \mathbb R^3 : ax+by+cz=d\}\,\,\,,\,\,\,a,b,c,d\in\mathbb R\,\,\,,\,\, (a,b,c)\ne (0,0,0)$$
