Centre of mass - definition or equation? Please consider the following extract from my lecturer's notes:

He writes that the position vector $\vec R$ is defined by equation 2.11, but, evidently, this vector lies on the line between $m_1$ and $m_2$. I'd be grateful if someone might prove that this vector lies on the line - or perhaps suggests a motivation for defining $\vec R$ as it is in equation 2.11.
 A: The motivation becomes clear when you get to the dynamics. It will turn out, that the sum of all forces acting on all particles of the system result in the motion of the center of mass according to
$$\sum_i \vec{F}_i = \ddot{\vec{R}} \sum_i m_i$$
As if the system was a single point-like mass with $M=\sum_i m_i$ located at $\vec{R}$.
The proof that it lies on the line is straightforward. Just observe that
$$\vec{r}_1-\vec{R}=\frac{m_2}{m_1+m_2}(\vec{r}_1 - \vec{r}_2)$$
is collinear with $$\vec{r}_1 - \vec{r}_2$$.
A: The previous answer by @John is correct, but I want to add a bit more details. We can derive the center of mass, before we even consider the dynamics of a body. We look at momentum to find where it is.
Consider a rigid body consisting of $n$ particles, each with mass $m_i$.
We have the motivation that we need some way to describe the total momentum of the body without having to track the motion of each particle.
$$ \vec{p} = \sum_{i=1}^n (m_i \vec{v}_i) \tag{1}$$
what we want to do is instead track the motion of one point only on the rigid body such that
$$ \vec{p} = \left( \sum_{i=1}^n m_i \right) \dot{\vec{R}} \tag{2}$$
Now if all the points moved parallel to each other with $\vec{v}_i = \vec{v}$ then there isn't one special point, but all points have the above property.
So the situation we should be considering is when all the points are rotating about this special point. Let's call the position of each particle relative to this special point $\vec{d}_i$ such that the position of each particle is
$$ \vec{r}_i = \vec{R} + \vec{d}_i \tag{3}$$
and the kinematics of rotation are
$$ \vec{v}_i = \dot{\vec{R}} + \vec{\omega} \times \vec{d}_i \tag{4} $$
which makes total momentum equal to
$$ \begin{aligned} 
 \vec{p} & = \sum_{i=1}^n (m_i \left( \dot{\vec{R}} + \vec{\omega} \times \vec{d}_i \right)) \\
 & = \left( \sum_{i=1}^n m_i \right) \dot{\vec{R}} + \sum_{i=1}^n m_i ( \left( \vec{\omega} \times \vec{d}_i \right))
\end{aligned}$$
$$ \vec{p} = \left( \sum_{i=1}^n m_i \right) \dot{\vec{R}} +  \vec{\omega} \times \left(  \sum_{i=1}^n (m_i \vec{d}_i)  \right) \tag{5}$$
In order to make (2) equal to (5), you must have
$$ \sum_{i=1}^n (m_i \vec{d}_i) = 0 $$
Now take (3), multiply by $m_i$ and sum up all the particles
$$ \require{cancel} \begin{aligned}
m_i \vec{r}_i & = m_i \vec{R} + m_i \vec{d}_i \\
\sum_{i=1}^n(m_i \vec{r}_i) & = \sum_{i=1}^n(m_i \vec{R}) + \sum_{i=1}^n(m_i \vec{d}_i) \\
\sum_{i=1}^n(m_i \vec{r}_i) & = \left( \sum_{i=1}^n m_i \right) \vec{R} + \cancel{ \sum_{i=1}^n(m_i \vec{d}_i)} \\
\end{aligned} $$
$$ \sum_{i=1}^n(m_i \vec{r}_i) = \left( \sum_{i=1}^n m_i \right) \vec{R} \tag{6}$$
which is solved for $\vec{R}$ as
$$ \boxed{ \vec{R} = \frac{ \sum_{i=1}^n(m_i \vec{r}_i)} { \sum_{i=1}^n m_i  } } 
 \tag{7}$$
which gives us the definition for the center of mass.
For example for two particles
$$ \vec{R} = \tfrac{m_1}{m_1 + m_2} \vec{r}_1 + \tfrac{m_2}{m_2 + m_2} \vec{r}_1 $$
The above is a linear combination of the two positions $\vec{r}_1$ and $\vec{r}_2$ where the center of mass is the weighted average of those positions. It is no coincidence this language is used, as the center of mass lies on the line connecting the two particles (hence linear combination) and the position is determined by how much weight each point carries.
