Is there a geometrical way to obtain a relationship between these vectors? 
Suppose we have a setup like this. Here $a_1,a_2,b_1,b_2$ are acceleration magnitudes($b_1,b_2$ being relative) and $P,Q,R,S$ are not pulley/blocks but are points on the rope. If I use a geometrical constraint then ($k$ is constant)
$$PQ+QR+RS=k$$
$$\ddot {PQ}+\ddot{QR}+\ddot{RS}=0$$
And here since $\ddot {PQ}=b_1,\ddot {RS}=b_2, \ddot{QR}=-(a_1+a_2)$ we have
$$b_1+b_2=a_1+a_2$$
This is fine, but how do I get a relation between $\vec{a_1},\vec{a_2},\vec{b_1},\vec{b_2}$?  Can it be obtained in any geometrical way without listing out forces (assuming those triangular blocks are isosceles and have base angle $\theta$, base moving on ground)?
Thanks in advance.
 A: If you use direction vector $\hat{e}_1=(\sin\frac{\theta}{2},\cos\frac{\theta}{2})$ along PQ and $\hat{e}_2=(\sin\frac{\theta}{2},-\cos\frac{\theta}{2})$ along RS you get the positions:
$$ \vec{r}_Q = x_1 \hat{i} $$
 $$ \vec{r}_P = x_1 \hat{i} - PQ\, \hat{e}_1 $$
 $$ \vec{r}_R = -x_2 \hat{i} $$
 $$ \vec{r}_S = -x_2 \hat{i} + RS\, \hat{e}_2 $$
Differentiating twice you get
$$ \ddot{\vec{r}}_Q = a_1 \hat{i} $$
 $$ \ddot{\vec{r}}_P = a_1 \hat{i} - b_1\, \hat{e}_1 $$
 $$ \ddot{\vec{r}}_R = -a_2 \hat{i} $$
 $$ \ddot{\vec{r}}_S = -a_2 \hat{i} + b_2\, \hat{e}_2 $$
but you need the constraint of PQ + QR + RS = const to evaluate the componets. You can correctly shown that $(b_1+b_2)-(a_1+a_2)=0$ which allows you to find one acceleration in terms of the other three. So you will need 3 boundary conditions to fully solve the problem.
In this regard, the conservation of length is a fundamental principle and cannot be derived from other equations. Without it the physical cable connecting all the points cannot be described in this problem.
