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For vectors $\vec{v},\vec{w}\in\mathbb{R}^2$, prove that the combinations $c\vec{v}+d\vec{w}$ with $0 ≤ c ≤ 1$ and $0 ≤ d ≤ 1$ fill the parallelogram with sides $\vec{v}$ and $\vec{w}$

If I take a simple examples like $\vec{v}=(1,0)$ and $\vec{w}=(0,1)$ $$ c\begin{bmatrix} 1\\0 \end{bmatrix}+d\begin{bmatrix} 0\\1 \end{bmatrix}=\begin{bmatrix} c\\d \end{bmatrix}=\begin{bmatrix} x\in[0,1]\\y\in[0,1] \end{bmatrix} $$

But how do I prove it mathematically for the general case ?

This is stated by @Jack D'Aurizio in the post Decide if a point is inside parallelogram, but no proof is given.

My Attempt

pic $$ \vec{r}=c\vec{v}+d\vec{w}\\ c,d\in[0,1]\implies c+d\in[0,2]\\ \text{when }c+d\in[0,1]\implies c\leq 1-d\\ \vec{r}\leq (1-d)\vec{v}+d\vec{w}=\vec{v}+d(\vec{w}-\vec{v}) \\ \implies\vec{r} \text{ fills the triangle }\Delta OAB\\ $$ When $c+d\in[1,2]$ how do I prove that $\vec{r}$ span the other half of the parallelogram ?

Sooraj S
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    How to you define the filled parallelogram? – mathcounterexamples.net Aug 09 '20 at 19:56
  • @mathcounterexamples.net I think that is where the confusion lies, could you please put some insights into it ? – Sooraj S Aug 09 '20 at 21:30
  • @mathcounterexamples.net The boundary lines are $\vec{r}=\vec{u}+t\vec{v}$ and $\vec{r}=\vec{v}+n\vec{u}$, right ? Thats the only thing I am able to define. – Sooraj S Aug 10 '20 at 04:34
  • Note $c+d\in[1,2]$ doesn't imply $c,d\in[0,1]$ – JMP Aug 11 '20 at 11:45
  • I think what mathcounterexamples is getting at is that this equation can be the definition of the parallelogram (and its interior). But if you question that then there must be some other definition you are using, and providing that definition is more-or-less essential to deriving the equation form from it. – zibadawa timmy Aug 11 '20 at 12:12

2 Answers2

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enter image description here

By the figure above, $$\vec u = \vec v+ \alpha \vec w$$ with $\alpha \in [0,1]$. See that the vector $\vec u$ takes the "border" of the parallelogram. Furthermore, if we take $\beta \vec u$ (green vector) with $\beta \in [0,1]$ we get interior points of the parallelogram.

In this case, the vector $$\beta \vec u = \beta \vec v + \beta \alpha \vec w$$ when we vary $\alpha$ and $\beta$, the vector $\beta \vec u$ takes all interior points of the upper triangle shaped by $\vec v$ and $\alpha \vec w$. You just need to make the necessary changes to get the lower triangle.

Can you finish?

Arnaldo
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  • you are taking some vector $\vec{u}$ such that $\vec{w}=\vec{u}+\alpha\vec{v}$, after that I don't understand. – Sooraj S Aug 11 '20 at 13:25
  • @ss1729: I updated the answer. Please, see if it satisfies you. – Arnaldo Aug 11 '20 at 14:22
  • @ss1729, be carefull, there is no inequality for vectors. – Arnaldo Aug 11 '20 at 16:40
  • A general vector inside $\sqsubset ! \sqsupset OADB$ is $\vec{r}=a\vec{v}+b(\vec{v}+\vec{w})$ or $\vec{r}=\alpha\vec{w}+\beta(\vec{v}+\vec{w})$ with $a+b,\alpha+\beta\in[0,1]$ and $a,b,\alpha,\beta\in[0,1]$. $\vec{r}=(a+b)\vec{v}+b\vec{w}=c\vec{v}+d\vec{w}$ or $\vec{r}=\beta\vec{v}+(\alpha+\beta)\vec{w}=c\vec{v}+d\vec{w}$ with $c,d\in[0,1]$, is it right ? – Sooraj S Aug 12 '20 at 07:39
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    @ss1729: That is right! – Arnaldo Aug 12 '20 at 12:55
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To solve by triangles, rotate by $180^\circ$ about either $\frac{\vec v+\vec w}{2}$ or the origin (i.e. consider the parallelogram $(-\vec v, -\vec w)$), and the transforms $c\to 1-c, d\to 1-d$.

$\vec r = (\vec v+\vec w) - c\vec v - d\vec w = (1-c)\vec v+(1-d)\vec w$

To continue your method, try:

$$ \vec{r}=c\vec{v}+d\vec{w}\\ c,d\in[0,1]\implies c+d\in[0,2]\\ \text{when }c+d\in[1,2]\implies 1-d\leq c\leq 1\leq 2-d\\ (1-d)\vec{v}+d\vec{w}\leq\vec{r}\leq \vec{v}+d\vec{w}\\ \text{when }c+d\in[1,2]\implies 1-c\leq d\leq 1\leq 2-c\\ c\vec{v}+(1-c)\vec{w}\leq\vec{r}\leq c\vec{v}+\vec{w}\\ \implies\vec{r} \text{ fills the triangle }\triangle ABD\\ $$

I think you need to add lower bounds to your original argument for completeness.

JMP
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  • how come $c\vec u+d\vec v = c(u_1+v_1)\begin{bmatrix}1\0\end{bmatrix}+d(u_2+v_2)\begin{bmatrix}0\1\end{bmatrix}$ ? – Sooraj S Aug 11 '20 at 10:57
  • $c\vec u+d\vec v =c\bigg{u_1\begin{bmatrix}1\0\end{bmatrix}+u_2\begin{bmatrix}0\1\end{bmatrix}\bigg}+d\bigg{v_1\begin{bmatrix}1\0\end{bmatrix}+v_2\begin{bmatrix}0\1\end{bmatrix}\bigg}\neq c(u_1+v_1)\begin{bmatrix}1\0\end{bmatrix}+d(u_2+v_2)\begin{bmatrix}0\1\end{bmatrix}$, right ? – Sooraj S Aug 11 '20 at 11:04