Animated Solution for Physics - Kinematics: Three vectors $\mathbf{P}$, $\mathbf{Q}$ and $\mathbf{R}$ are shown in the figure. Let $S$ be any point on the vector $\mathbf{R}$. The distance between the points $P$ and $S$ is $b|\mathbf{R}|$. The general relation among vectors $\mathbf{P}$, $\mathbf{Q}$ and $\mathbf{S}$ is

Vectors are the language of space, and understanding how they combine is fundamental to mastering physics. Let's embark on a journey to decode the relationship between the position vectors $\mathbf{P}$, $\mathbf{Q}$, and a mysterious point $S$ lying between them.

Position Vectors

Imagine you are standing at the origin, point $O$. You look out into the coordinate plane and identify two distinct points, $P$ and $Q$. The vector pointing directly from you to point $P$ is its position vector, denoted as $\mathbf{P}$. Similarly, the vector pointing to $Q$ is $\mathbf{Q}$.

Mathematically, we write this as:

Now, suppose you want to travel directly from point $P$ to point $Q$. This straight-line path is represented by the vector $\mathbf{R}$. How do we express $\mathbf{R}$ in terms of our known position vectors?

We invoke the powerful Triangle Law of Vector Addition. To go from $P$ to $Q$, you can conceptually travel backward from $P$ to the origin $O$, and then forward from $O$ to $Q$.

Therefore, the vector $\mathbf{R}$ is simply the final position minus the initial position:

The problem introduces a new point, $S$, which lies exactly on the vector $\mathbf{R}$ (the line segment connecting $P$ and $Q$). We are given a crucial piece of information: the distance from $P$ to $S$ is a fraction $b$ of the total length of $\mathbf{R}$.

Because the segment $PS$ is just a collinear piece of the larger vector $\mathbf{R}$, the vector $\overrightarrow{PS}$ must point in the exact same direction as $\mathbf{R}$. Since its magnitude is scaled by a factor of $b$, we can confidently write:

Our ultimate goal is to find the position vector of $S$, which we will call $\mathbf{S}$. This is the vector $\overrightarrow{OS}$ originating from the origin.

Let's look at the triangle formed by points $O$, $P$, and $S$. Using the triangle law once more, the journey from the origin to $S$ can be broken down into two steps: first, go from $O$ to $P$, and then from $P$ to $S$.

Substituting the variables we've defined, we get our master equation:

We are almost at the finish line. The options provided in the question only contain $\mathbf{P}$ and $\mathbf{Q}$, so we must eliminate $\mathbf{R}$. We do this by substituting our earlier finding, $\mathbf{R}=\mathbf{Q}-\mathbf{P}$, into the master equation:

Now, we perform some simple algebraic expansion. Be careful with the signs!

Finally, we group the terms containing the vector $\mathbf{P}$ together and factor it out:

This elegant expression perfectly matches option (c).

Bonus Insight: If you look closely, this result is a direct derivation of the Section Formula for internal division. Point $S$ divides the line segment $PQ$ in the ratio of $b:(1-b)$. Physics and mathematics are truly two sides of the same coin!