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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.

CuriousKid7
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    The "vice versa" part does not hold in $\Bbb R^3$ either : if one vector is zero, you get a line instead of a plane, if the two are zero, you get again a singleton. The important concept is linear independence. – Arnaud D. Jul 07 '17 at 11:48
  • @Curious: It's easy to show every line (according to your definition) contains infinitely many points. – Andrew D. Hwang Jul 07 '17 at 11:56
  • @ArnaudD. Right, that's true. I'll have to reconsider that one. – CuriousKid7 Jul 07 '17 at 12:01
  • @AndrewD.Hwang I don't see it unfortunately. – CuriousKid7 Jul 07 '17 at 12:02
  • @CuriousKid7 Parameterize the line. Also parameterizing the plane helps solve the first problem I believe. The intuition for this can be seen if you write out the result you are trying to get to explicitly. (ie. write $\vec v$ as $\begin{pmatrix} v_1 \v_2 \ v_3 \end{pmatrix}$ and such) – Shuri2060 Jul 07 '17 at 12:09
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    @CuriousKid7: If $b = 0$, then $(c/a, y)$ satisfies $ax + by = c$ for all real $y$. Otherwise, $(x, (c - ax)/b)$ satisfies $ax + by = c$ for all real $x$. (To prove a singleton is not a line you need even less, just two points on each line, not infinitely many. ;) – Andrew D. Hwang Jul 07 '17 at 12:25
  • @AndrewD.Hwang I see, thanks. If we force $\vec w$ to be nonzero, will we definitely get a line? – CuriousKid7 Jul 07 '17 at 12:59
  • Yes. To prove this, it will be convenient to write $w = (-b, a)$, in which case you'll find the components of $w$ give the coefficients of your line. (Together with $w$, the vector $v$ determines the constant $c$.) – Andrew D. Hwang Jul 07 '17 at 13:08
  • The set of solutions to a system of one equation with variables in $\mathbb{R}^3$ equals the set of vectors $x_0+x_1$ where $x_1$ is a particular solution (your v) and $x_0$ runs over all homogeneous solutions (which is the span of the basic solutions of the associated system, which is your $\lambda w + \mu u$). To go the other way, take the cross-product $w\times u$ to get a normal vector (if w,u are independent) and solve for your constant term. – AnonymousCoward Jul 07 '17 at 14:02

2 Answers2

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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 $\vec v,\vec w\in\mathbb R^2\,\,\,,\,\, \vec w\ne \vec0$.

Shuri2060
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    The condition $\vec w \times \vec u \ne 0$ is equivalent to the statement that ${\vec w,\vec u}$ is a linearly independent set of vectors, the important concept mentioned in the comment of @ArnaudD. – Lee Mosher Jul 07 '17 at 12:43
  • Doing the question this way involves quite a bit of algebra tbh. I'll include a separate vectors answer which is much easier, but requires understanding of a few concepts (cross and dot product and how they interact with each other). However, the proof will give more understanding and can be generalized to higher dimensions easier. – Shuri2060 Jul 07 '17 at 12:51
  • I appreciate your answers. But I actually am not seeing how to do the forward direction as the reverse. Could you please show how you would do the forward direction for the $\mathbb R^2$ case? – CuriousKid7 Jul 07 '17 at 13:15
  • @CuriousKid7 For $\mathbb R^3$, you can just reverse the proof I've given in the other answer. For $\mathbb R^2$, give me a moment – Shuri2060 Jul 07 '17 at 13:16
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    @CuriousKid7 I'm also going to say that doing the backwards direction first and the reversing it is probably easier than the forwards direction as it'll indicate what substitutions you need to make. However, I'll just edit in the plain forwards – Shuri2060 Jul 07 '17 at 13:34
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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

$$P=\{\vec x \in \mathbb R^3: \vec a.\vec x=d\}\,\,\,,\,\,\,\vec a \in \mathbb R^3\,\,\,,\,\,\, \vec a\neq\vec0\,\,\,,\,\,\,d\in\mathbb R$$

$$=\left\{\begin{pmatrix} x \\y \\ z \end{pmatrix}\in \mathbb R^3 : \begin{pmatrix} a \\b \\ c \end{pmatrix}.\begin{pmatrix} x \\y \\ z \end{pmatrix}=d\right\}\,\,\,,\,\,\,a,b,c,d\in\mathbb R\,\,\,,\,\, \begin{pmatrix} a \\b \\ c \end{pmatrix}\ne \begin{pmatrix} 0 \\0 \\ 0 \end{pmatrix}$$

$$=\{(x,y,z)\in \mathbb R^3 : ax+by+cz=d\}\,\,\,,\,\,\,a,b,c,d\in\mathbb R\,\,\,,\,\, (a,b,c)\ne (0,0,0)$$

Shuri2060
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