Why is $\overrightarrow{OM}$ in that form?

Question

We have the following:

We have that $M$ is on the line segment $AB$.

$\overrightarrow{OA}=\overrightarrow{a}$

$\overrightarrow{OB}=\overrightarrow{b}$

Could you explain to me why it stands that $$\overrightarrow{OM}=s \overrightarrow{a}+(1-s)\overrightarrow{b}, 0 \leq s \leq 1$$ ??

Answer 1

HINT: Write $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \vec a + \dots$.

Answer 2

In this linear-combination

$\overset{\rightharpoonup }{\text{OM}}=\lambda \overset{\rightharpoonup }{\text{OA}}+\mu \overset{\rightharpoonup }{\text{OB}}$

with

$\lambda +\mu =1$

one let

$\mu =1-\lambda$

which gives

$\overset{\rightharpoonup }{\text{OM}}=\lambda \overset{\rightharpoonup }{\text{OA}}+(1-\lambda ) \overset{\rightharpoonup }{\text{OB}}$

with

$0\leq \lambda \leq 1$