Question

In the given diagram, if \( \overrightarrow{PQ} = \vec{A} \), \( \overrightarrow{QR} = \vec{B} \) and \( \overrightarrow{RS} = \vec{C} \) then \( \overrightarrow{PS} \) equals

• A. $\vec{A} - \vec{B} + \vec{C}$
• B. $\vec{A} + \vec{B} - \vec{C}$
• C. $\vec{A} + \vec{B} + \vec{C}$
• D. $\vec{A} - \vec{B} - \vec{C}$
• E. $-\vec{A} - \vec{B} - \vec{C}$

Hint:

Follow head-to-tail rule: just add vectors in order of motion.

Answer 1

The Correct Option is C

Concept: Vector addition along a path
If multiple vectors are connected head-to-tail: \[ \text{Resultant vector} = \text{sum of all vectors} \] ---

Step 1: Understand the path
Given: \[ \overrightarrow{PQ} = \vec{A}, \quad \overrightarrow{QR} = \vec{B}, \quad \overrightarrow{RS} = \vec{C} \] We need: \[ \overrightarrow{PS} \] ---

Step 2: Traverse from P to S
Path: \[ P \rightarrow Q \rightarrow R \rightarrow S \] So: \[ \overrightarrow{PS} = \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RS} \] ---

Step 3: Substitute vectors
\[ \overrightarrow{PS} = \vec{A} + \vec{B} + \vec{C} \] ---

Step 4: Physical interpretation

• Moving along edges step-by-step
• Net displacement is direct vector from P to S
• Independent of path shape --- Final Answer: \[ \boxed{\vec{A} + \vec{B} + \vec{C}} \]