Vector Addition and Point Partitioning along Directed Segment

To find the general relationship linking the vectors $\vec{P}$, $\vec{Q}$, and $\vec{S}$, we analyze the position vectors and vector additions from the given diagram.

1. Position Vectors of Points: Let the origin be denoted by $O$. The position vectors of points $P$, $Q$, and $S$ with respect to the origin $O$ are: $\vec{OP} = \vec{P}$ $\vec{OQ} = \vec{Q}$ $\vec{OS} = \vec{S}$

2. Vector Representation of $\vec{R}$: The vector $\vec{R}$ represents the directed line segment from point $P$ to point $Q$: $\vec{R} = \vec{PQ} = \vec{OQ} - \vec{OP} = \vec{Q} - \vec{P}$

3. Displacement Vector from $P$ to $S$: Since point $S$ lies along the directed line segment $\vec{R}$ (from $P$ to $Q$) and the distance between $P$ and $S$ is given as $b|\vec{R}|$, the vector $\vec{PS}$ is parallel to $\vec{R}$ and its magnitude is $b$ times the magnitude of $\vec{R}$. Therefore, we can write: $\vec{PS} = b\vec{R}$

4. Determining the Vector $\vec{S}$: Using vector addition for the triangle $OPS$: $\vec{OS} = \vec{OP} + \vec{PS}$ $\vec{S} = \vec{P} + b\vec{R}$

5. Substituting $\vec{R} = \vec{Q} - \vec{P}$: $\vec{S} = \vec{P} + b(\vec{Q} - \vec{P})$ $\vec{S} = \vec{P} + b\vec{Q} - b\vec{P}$ $\vec{S} = (1 - b)\vec{P} + b\vec{Q}$

Thus, the correct relation among the vectors is $\vec{S} = (1 - b)\vec{P} + b\vec{Q}$, which corresponds to option A.