Question

Resultant of two vectors \(\overset{⃗}{P}\) and \(\overset{⃗}{Q}\) is of magnitude A. If \(\overset{⃗}{Q}\) is reversed, then the resultant is of magnitude B. The value of \(A^2+B^2\) is

• A. \(P^2+Q^2\)
• B. \(P^2-Q^2\)
• C. \(2(P^2+Q^2)\)
• D. \(2(P^2-Q^2)\)

Hint:

Write \(A^2\) and \(B^2\) with the cosine law and add.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The magnitude of the resultant of \(\vec P\) and \(\vec Q\) at angle \(\theta\) is given by the parallelogram law: \(R^2=P^2+Q^2+2PQ\cos\theta\).

Step 2: Key Formula or Approach
Reversing \(\vec Q\) changes the angle to \(180^{\circ}-\theta\), so \(\cos\theta\) changes sign.

Step 3: Detailed Explanation
\[ A^2=P^2+Q^2+2PQ\cos\theta \]
\[ B^2=P^2+Q^2-2PQ\cos\theta \]
Adding, the cross terms cancel:
\[ A^2+B^2=2(P^2+Q^2) \]

Final Answer:
\(A^2+B^2=2(P^2+Q^2)\), option (C). \[ \boxed{2(P^2+Q^2)\ \text{(C)}} \]