Using vector notation describe a triangle

[mathmari]

Hey!

Using vector notation describe a triangle ( in space ) that has as vertices the origin and the endpoints of the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$.

Could you tell me what I am supposed to do?? (Wondering)

 

[Prove It]

Hey!

Using vector notation describe a triangle ( in space ) that has as vertices the origin and the endpoints of the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$.

Could you tell me what I am supposed to do?? (Wondering)

$\displaystyle \begin{align*} \mathbf{a} \end{align*}$ is a vector starting at (0, 0, 0) and ending at $\displaystyle \begin{align*} \left( a_x , a_y , a_z \right) \end{align*}$, and $\displaystyle \begin{align*} \mathbf{b} \end{align*}$ is a vector starting at (0, 0, 0) and ending at $\displaystyle \begin{align*} \left( b_x, b_y, b_z \right) \end{align*}$.

Obviously two of the sides of the triangle are the vectors $\displaystyle \begin{align*} \mathbf{a} \end{align*}$ and $\displaystyle \begin{align*} \mathbf{b} \end{align*}$. What is the final side? i.e. can you write a vector that starts at $\displaystyle \begin{align*} \left( a_x, a_y, a_z \right) \end{align*}$ and ends at $\displaystyle \begin{align*} \left( b_x, b_y, b_z \right) \end{align*}$?

 

[mathmari]

$\displaystyle \begin{align*} \mathbf{a} \end{align*}$ is a vector starting at (0, 0, 0) and ending at $\displaystyle \begin{align*} \left( a_x , a_y , a_z \right) \end{align*}$, and $\displaystyle \begin{align*} \mathbf{b} \end{align*}$ is a vector starting at (0, 0, 0) and ending at $\displaystyle \begin{align*} \left( b_x, b_y, b_z \right) \end{align*}$.

Obviously two of the sides of the triangle are the vectors $\displaystyle \begin{align*} \mathbf{a} \end{align*}$ and $\displaystyle \begin{align*} \mathbf{b} \end{align*}$. What is the final side? i.e. can you write a vector that starts at $\displaystyle \begin{align*} \left( a_x, a_y, a_z \right) \end{align*}$ and ends at $\displaystyle \begin{align*} \left( b_x, b_y, b_z \right) \end{align*}$?

The final side is $\displaystyle \begin{align*} \left( b_x-a_x, b_y-a_y, b_z-a_z \right) \end{align*}$, right?? (Wondering)

 

[Prove It]

The final side is $\displaystyle \begin{align*} \left( b_x-a_x, b_y-a_y, b_z-a_z \right) \end{align*}$, right?? (Wondering)

Yes, and can you write this in terms of $\displaystyle \begin{align*} \mathbf{a} \end{align*}$ and $\displaystyle \begin{align*} \mathbf{b} \end{align*}$?

 

[mathmari]

Yes, and can you write this in terms of $\displaystyle \begin{align*} \mathbf{a} \end{align*}$ and $\displaystyle \begin{align*} \mathbf{b} \end{align*}$?

Is it $\displaystyle \begin{align*} \mathbf{b}-\mathbf{a} \end{align*}$ ?? (Wondering)

 

[Prove It]

Is it $\displaystyle \begin{align*} \mathbf{b}-\mathbf{a} \end{align*}$ ?? (Wondering)

Yes :)

 

[mathmari]

I see... Thanks! (Smile)

The prof showed us an other way to solve it...

View attachment 4038

$\overrightarrow{v}=\overrightarrow{OK}=t\overrightarrow{OM}, 0 \leq t \leq 1$

$\overrightarrow{OM}=s \overrightarrow{a}+(1-s)\overrightarrow{b}, 0 \leq s \leq 1$

So, $\overrightarrow{v}=(ts)\overrightarrow{a}+t(1-s)\overrightarrow{b}=x \overrightarrow{a}+y \overrightarrow{b}$
where $x=ts, y=t(1-s)$
and therefore $x \geq 0, y \geq 0, x+y \leq 1$. Conversely, if $\overrightarrow{v}=x\overrightarrow{a}+y\overrightarrow{b}, x \geq 0, y \geq 0, x+y \leq 1$ then we set $t=x+y$ then $0 \leq t \leq 1$, and $s=\frac{x}{t}$, so $1-s=\frac{y}{t}$, and so $0 \leq s \leq 1$ and $\overrightarrow{v}=t(s\overrightarrow{a}+(1-s)\overrightarrow{b})$
So, $K$ belongs to the triangle. Could you explain to me this way?? (Wondering)

Why do we have to show both directions?? (Wondering)

 

[mathmari]

Why do we take the points $K$ and $M$ ?? (Wondering)