Using vector notation describe the triangle ( in space ) that has as vertices the origin and the endpoints of the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$.
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The solution is the following:
$\overrightarrow{v}=\overrightarrow{OK}=t\overrightarrow{OM}, 0 \leq t \leq 1$
$\overrightarrow{OM}=s \overrightarrow{a}+(1-s)\overrightarrow{b}, 0 \leq s \leq 1$
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.
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Could someone explain it to me ?? Why do we have to prove both directions??
The first bit shows that every point in the triangle (as defined geometrically) can be written as $tsa + t(1-s)b$. We then need to show that every such vector satisfies the geometric definition.
That is, to show that $A = B$, we show that $a \in A \implies a \in B$ and then $b \in B \implies b \in A$, so that $A \subseteq B$ and $B \subseteq A$, as desired.