All the vectors interior to a parallelogram Consider two vectors $a$ and $b$, then their linear combinations
$c_1a + c_2b$
where $c_1+c_2=1$ and $c_1, c_2\ge 0$.  
We clearly get the line segment $ab$.
Next I have two questions.

1) Is there a nice expression to represent all the vectors interior to the triangle formed by the vectors $a$, $b$ and the origin?
(I know one ugly looking expression : $c_3*[c_1a + c_2b]$)
2) How to get an expression to represent all the vectors interior to the parallelogram formed by the vectors $a,b$ ?
 A: The answer to both of your questions is a convex combination.
For your triangle: take point $a$, $b$, and the origin $c=(0,0)$, then all points
$$\alpha_1 a+\alpha_2 b+\alpha_3 c$$
lie in the triangle provided $\alpha_i\geq 0$ and $\sum \alpha_i=1$
Similarly, the vertices of the parallelogram are $c=(0,0)$, $a$, $b$, and $a+b$, and all points in the parallelogram are given by
$$\alpha_1 a+\alpha_2 b+\alpha_3 c+\alpha_4 (a+b)$$
again with $\alpha_i\geq 0$ and $\sum \alpha_i=1$
A: For the triangle, 
$$(1-\lambda_a-\lambda_b)0+\lambda_a a+\lambda_b b=\lambda_a a+\lambda_b b$$
where $\lambda_a, \lambda_b \ge 0$ and $\lambda_a+\lambda_b \le 1$.
For the parallelogram formed by $0, a, b, c=a+b$.
\begin{align}(1-\lambda_a-\lambda_b-\lambda_c)0+\lambda_a a+\lambda_b b+\lambda_cc&=\lambda_a a+\lambda_b b+\lambda_cc\\
&=(\lambda_a+\lambda_c)a+(\lambda_b+\lambda_c)b\end{align}
where $\lambda_a, \lambda_b, \lambda_c \ge 0$ and $\lambda_a+\lambda_b+\lambda_c \le 1$.
A: The vectors strictly in the interior of the parallelogram with vertices at $0$, $a$, $b$ and $a+b$ are
$c_1a+c_2b$
where $0 < c_1 < 1$ and $0 < c_2 < 1$.
If you include vectors on the boundary of the parellogram then the constraints are $0 \le c_1 \le 1$ and $0 \le c_2 \le 1$.
