MHB What is the geometric description of a set of vectors in $\mathbb{R}^2$?

[mathmari]

Hey!

We have two vectors $\vec{u}, \vec{v}\in \mathbb{R}^2$. I want to describe geometrically the set of vectors $\vec{z}$, for which it holds that $$\vec{z}=\lambda{u}+(1-\lambda)\vec{v}$$ with $0\leq \lambda \leq 1$.

Does this set describe all the points that are on the line that connects $\vec{u}$ and $\vec{v}$ ? Or is the geometric description something else? (Wondering)
If we have also $\vec{w}\in \mathbb{R}^2$, I want to describe geometrically the set $$A=\{\vec{z} \mid \vec{z}=a\vec{u}+b\vec{v}+c\vec{w}; \ a, b, c\geq 0; \ a+b+c=1\}$$

Which is the geometric description in this case? Could you give me a hint? (Wondering)

 

[I like Serena]

Hi mathmari! (Smile)

Hey!

We have two vectors $\vec{u}, \vec{v}\in \mathbb{R}^2$. I want to describe geometrically the set of vectors $\vec{z}$, for which it holds that $$\vec{z}=\lambda{u}+(1-\lambda)\vec{v}$$ with $0\leq \lambda \leq 1$.

Does this set describe all the points that are on the line that connects $\vec{u}$ and $\vec{v}$ ? Or is the geometric description something else? (Wondering)

Yes.
More specifically, it could also be a point if $\vec{u}$ and $\vec{v}$ happen to be the same vector. (Nerd)

If we have also $\vec{w}\in \mathbb{R}^2$, I want to describe geometrically the set $$A=\{\vec{z} \mid \vec{z}=a\vec{u}+b\vec{v}+c\vec{w}; \ a, b, c\geq 0; \ a+b+c=1\}$$

Which is the geometric description in this case? Could you give me a hint? (Wondering)

This is any point that is a linear combination between those 3 vectors.
If they are independent, it's a triangle in the plane defined by those 3 vectors, and with those 3 vectors as corners.
If they are dependent, it's either a line segment or a point. (Nerd)

 

[mathmari]

Yes.
More specifically, it could also be a point if $\vec{u}$ and $\vec{v}$ happen to be the same vector. (Nerd)

Ah yes. If $\vec{u}=\vec{v}$, we have $\vec{z}=\lambda{u}+(1-\lambda)\vec{u}=\vec{u}$, which is a point.

We have that:

[DESMOS=-5,10,-4,8]y=x\ \left\{y\ge0\right\}\ \left\{x\le4\right\};y=-x\ \left\{y\ge0\right\}\ \left\{x\ge-4\right\};y=4\ \left\{-4\le x\le4\right\}[/DESMOS]

So, if $\vec{u}$ is the green vector and $\vec{v}$ the blue one, then $\vec{z}$ is the orange one, right? (Wondering)

This is any point that is a linear combination between those 3 vectors.
If they are independent, it's a triangle in the plane defined by those 3 vectors, and with those 3 vectors as corners.

If they are independent, we know that the end of the vectors are not collinear, right? (Wondering)

Then we have the following:

[DESMOS=-10,10,-10,10]y=x\ \left\{y\ge0\right\}\ \left\{x\le4\right\};y=-x\ \left\{y\ge0\right\}\ \left\{x\ge-5\right\};y=5x\ \left\{0\le y\le8\right\}\ ;y=\frac{3}{1.6+5}\left(x+5\right)+5\ \left\{-5\le x\le1.6\right\};y=-\frac{4}{4-1.6}\left(x-4\right)+4\ \left\{1.6\le x\le4\right\};y=-\frac{1}{9}\left(x-4\right)+4\ \left\{-5\le x\le4\right\}[/DESMOS]

where the green, orange and blue lines are the vectors $\vec{u}, \vec{v}, \vec{w}$. If we connect the endpoints of these vectors we get a triangle.

If they are dependent, it's either a line segment or a point. (Nerd)

If they are dependent, for example $\vec{w}=\beta \vec{u}+\gamma \vec{v}$, then we get $$\vec{z}=a\vec{u}+b\vec{v}+c\left (\beta \vec{u}+\gamma \vec{v}\right )=(a+c\beta )\vec{u}+(b+c\gamma )\vec{v}$$

A line segment is defined as $\vec{x}=s\vec{a}+t\vec{b}$, with $s,t\geq 0, \ s+t=1$. Or without $s+t=1$ ? (Wondering)

In this case we have that $a+c\beta +b+c\gamma=a+b+c(\beta+\gamma)$. From the definition of the set $A$, we have that $a+b+c=1$. So, can we conclude that $a+b+c(\beta+\gamma)=1$ ? Or isn't this necessary? (Wondering) We get a point if $\vec{u}=\vec{v}=\vec{w}$, right? (Wondering)

 

[I like Serena]

So, if $\vec{u}$ is the green vector and $\vec{v}$ the blue one, then $\vec{z}$ is the orange one, right? (Wondering)

If they are independent, we know that the end of the vectors are not collinear, right? (Wondering)

Right.
Nice drawing! (Smile)

A line segment is defined as $\vec{x}=s\vec{a}+t\vec{b}$, with $s,t\geq 0, \ s+t=1$. Or without $s+t=1$ ? (Wondering)

That would be with the $s+t=1$. Otherwise it's part of a plane.

In this case we have that $a+c\beta +b+c\gamma=a+b+c(\beta+\gamma)$. From the definition of the set $A$, we have that $a+b+c=1$. So, can we conclude that $a+b+c(\beta+\gamma)=1$ ? Or isn't this necessary? (Wondering)

We get a point if $\vec{u}=\vec{v}=\vec{w}$, right? (Wondering)

I have just realized that if the vectors are dependent, it can still be a triangle.
It only becomes a line segment if 2 vectors are identical, or if all 3 vectors are on the same line (but not all identical). (Thinking)

And yes, we only get a point if all vectors are identical.

 

[mathmari]

Ah ok! (Malthe)

A vector is a line segment, that has a specific direction and length. Does it describe its endpoint? (Wondering)

Because at the case where all the vectors are identical we get $\vec{z}=\vec{u}$, and we say that it is a point. So, we have the following:
\begin{equation*}A=\{\vec{z} \mid \vec{z}=\alpha \vec{u}+\beta \vec{v}+\gamma \vec{w} ; \ \alpha, \beta , \gamma \geq 0 ; \ \alpha +\beta +\gamma=1\}\end{equation*}

• If not all the endpoints of the vectors $\vec{u}, \vec{v}, \vec{w}$ are on the same line segment, then $A$ describes a triangle with endpoints the endpoints of the vectors.
• If all the endpoints of the vectors $\vec{u}, \vec{v}, \vec{w}$ are on the same line segment, then $A$ describes that line segment.
• If two of the three vectors are identical, for example $\vec{u}=\vec{v}$, then we get:
  \begin{equation*}\vec{z}=(\alpha +\beta )\vec{u}+\vec{w} \ \text{ with } \ (\alpha +\beta )+\gamma=1\end{equation*} which describes the line segment between the endpoints of $\vec{u}$ and $\vec{w}$.
• If all the three vectors are identical, $\vec{u}=\vec{v}=\vec{w}$, then we get:
  \begin{equation*}\vec{z}=(\alpha +\beta+\gamma )\vec{u}\end{equation*}
  which describes the endpoint of $\vec{u}$.

right? (Wondering)

 

[I like Serena]

Ah ok! (Malthe)

A vector is a line segment, that has a specific direction and length. Does it describe its endpoint? (Wondering)

Because at the case where all the vectors are identical we get $\vec{z}=\vec{u}$, and we say that it is a point.

In this case, yes.
More generally we use vectors for basically 2 purposes:

• A vector can define the direction and speed (or acceleration or some other direction related quantity) of a particle in space.
  In this case it does not describe the end point.
• A vector can define the position of a particle in space.
  In this case the vector identifies the direction and distance of the end point from the origin.

So, we have the following:
\begin{equation*}A=\{\vec{z} \mid \vec{z}=\alpha \vec{u}+\beta \vec{v}+\gamma \vec{w} ; \ \alpha, \beta , \gamma \geq 0 ; \ \alpha +\beta +\gamma=1\}\end{equation*}

...

right? (Wondering)

Yes.
Although perhaps we should make it more explicit that each of those cases are supposed to exclude the cases that come after. (Nerd)

 

[mathmari]

In this case, yes.
More generally we use vectors for basically 2 purposes:

• A vector can define the direction and speed (or acceleration or some other direction related quantity) of a particle in space.
  In this case it does not describe the end point.
• A vector can define the position of a particle in space.
  In this case the vector identifies the direction and distance of the end point from the origin.

Yes.
Although perhaps we should make it more explicit that each of those cases are supposed to exclude the cases that come after. (Nerd)

Ok! Thank you very much! (Sun)