When do the linear combinations of three vectors fill out the triangle? Let $P$, $Q$, and $R$ be any three (non-collinear) points in the 3D space. Let $O$ denote the origin. Let $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ denote the vectors $\vec{OP}$, $\vec{OQ}$, and $\vec{OR}$, respectively. Let $\alpha$, $\beta$, and $\gamma$ be any three scalars. 
Then under what condition(s) on $\alpha$, $\beta$, and $\gamma$ do the end points of the linear combinations $\alpha \mathbf{u} + \beta \mathbf{v} + \gamma \mathbf{w}$ fill out and stay within the triangle $PQR$? 
I know intuitively that, for this to happen, we must have $\alpha \geq 0$, $\beta \geq 0$,  $\gamma \geq 0$, and $\alpha + \beta + \gamma = 1$. 
I got this problem from "Prob. 20, Sec. 1.1 in Strang's Intro to Linear Algebra, 4th ed".
How do we prove this rigorously? 
 A: The triangle $\triangle$ in the $(x,y)$-plane with vertices $O=(0,0)$, $E=(1,0)$, $F=(0,1)$ can as a point set be described as
$$\triangle=\{(x,y)\in{\mathbb R}^2\>|\>x\geq0, \ y\geq 0,\ x+y\leq1\}\ .\tag{1}$$
Introducing the vectors ${\bf e}:=\vec{OE}$, $\>{\bf f}:=\vec{OF}$ we can rewrite $(1)$ as
$$\triangle=\{\beta {\bf e}+\gamma{\bf f}\>|\>\beta\geq0, \ \gamma\geq0,\ \beta+\gamma\leq1\}\ .$$
It follows that your triangle $T$ in space can be described as
$$T=\{{\bf u}+\beta( {\bf v}-{\bf u})+\gamma({\bf w}-{\bf u})\>|\>\beta\geq0, \ \gamma\geq0,\ \beta+\gamma\leq1\}\ .$$
If we now put $\alpha=1-\beta-\gamma$ then $\alpha$, $\beta$, $\gamma$ exactly fulfill your conditions, and one has
$$T=\{\alpha {\bf u}+\beta {\bf v}+\gamma{\bf w}\>|\>\alpha\geq0, \ \beta\geq0, \ \gamma\geq0,\ \alpha+\beta+\gamma=1\}\ .$$
A: Choose ${\bf v} - {\bf u}$ and ${\bf w} - {\bf u}$ as your base elements of 2 dimensional space (you should explain why this is possible), now any vector $\bf z$ can be expressed as:
$${\bf z}={\bf u}+a({\bf v} - {\bf u})+b({\bf w} - {\bf u})$$
So you want this arbitrary $\bf z$ be inside your triangle, draw a picture, it is easy to see:
$$a>0,b>0,a<1,b<1,a+b<1$$
Are necessary conditions (proof is so easy by contradiction method), But sufficiency is more complicated. I've done it also by contradiction method but you maybe prove it better than me. Now proof is complete because you can write such a $\bf z$ this way:
$${\bf z}=(1-a-b){\bf u}+a{\bf v}+b{\bf w}$$
$$a>0,b>0,a+b<1$$
A: A nice proof, Enjoy :)
Theorem:
Suppose ${\bf{v}}_1,{\bf{v}}_2,{\bf{v}}_3$ are 3 vectors in $\Bbb R^2$ such that ${\bf{v}}_1-{\bf{v}}_3$ and ${\bf{v}}_2-{\bf{v}}_3$ are linearly independent. Then for each ${\bf{v}}\in \Bbb R^2$ there exists a unique triple ${\bf{\lambda}}=(\lambda_1,\lambda_2,\lambda_3)$ such that
$${\bf{v}}=\lambda_1 {\bf{v}}_1+\lambda_2 {\bf{v}}_2+\lambda_3 {\bf{v}}_3\,\,\,\text{and}\,\,\,\lambda_1+\lambda_2+\lambda_3=1$$
Proof:
We have:
$$\begin{bmatrix}
        {\begin{bmatrix}{\bf{v}}\end{bmatrix}_{2×1}} \\
        1 \\
        \end{bmatrix}=
\begin{bmatrix}
{\begin{bmatrix}{\bf{v}}_1\end{bmatrix}_{2×1}}&{\begin{bmatrix}{\bf{v}}_2\end{bmatrix}_{2×1}}&{\begin{bmatrix}{\bf{v}}_3\end{bmatrix}_{2×1}} \\
1&1&1 \\
        \end{bmatrix}\begin{bmatrix}{\bf{\lambda}}\end{bmatrix}_{3×1}$$
This equation always has a unique solution because the above $3 \times 3$ matrix is invertable. We can simply show this as follows
$$c_1\begin{bmatrix}
        {\begin{bmatrix}{\bf{v}}_1\end{bmatrix}_{2×1}} \\
        1 \\
        \end{bmatrix}+c_2\begin{bmatrix}
        {\begin{bmatrix}{\bf{v}}_2\end{bmatrix}_{2×1}} \\
        1 \\
        \end{bmatrix}+c_3\begin{bmatrix}
        {\begin{bmatrix}{\bf{v}}_3\end{bmatrix}_{2×1}} \\
        1 \\
        \end{bmatrix}=0\\
\Rightarrow c_1{\bf {v}_1}+c_2{\bf {v}_2}+c_3{\bf {v}_3}={\bf 0}\,\,\,\text{and}\,\,\,c_1+c_2+c_3=0\\
\Rightarrow c_1({\bf {v}_1}-{\bf {v}_3})+c_2({\bf {v}_2}-{\bf {v}_3})={\bf 0}\\
\Rightarrow c_1=c_2=0\,\, , \,\, c_3=0\,\,\,\square$$
Exercise 1:
The line segment ${\bf{v}}_1{\bf{v}}_2=\{{\bf{v}}\in \Bbb R^2:{\lambda}_3({\bf{v}})=0\}$
The line segment ${\bf{v}}_1{\bf{v}}_3=\{{\bf{v}}\in \Bbb R^2:{\lambda}_2({\bf{v}})=0\}$
The line segment ${\bf{v}}_2{\bf{v}}_3=\{{\bf{v}}\in \Bbb R^2:{\lambda}_1({\bf{v}})=0\}$
Exercise 2:
Identify these sets:
$P_3=\{{\bf{v}}\in \Bbb R^2:{\lambda}_3({\bf{v}})\gt 0\}\\\\
P_2=\{{\bf{v}}\in \Bbb R^2:{\lambda}_2({\bf{v}})\gt 0\}\\\\
P_1=\{{\bf{v}}\in \Bbb R^2:{\lambda}_1({\bf{v}})\gt 0\}$
Exercise 3:
The triangle ${\bf{v}}_1{\bf{v}}_2{\bf{v}}_3=P_1\cap P_2\cap P_3$
The End :)
We also have this theorem for n vectors (polygon). Scalers $\lambda_i\,\,(i=1,2,...,n)$ called Barycentric Coordinates.
