Question

Resultant of two vectors \( \vec{P} \) and \( \vec{Q} \) is of magnitude \( R_1 \). If the direction of \( \vec{Q} \) is reversed, the resultant is of magnitude \( R_2 \). The value of \( (R_1^2 + R_2^2) \) is \( \cos (\pi - \theta) = - \cos \theta \)

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

Hint:

When reversing the direction of one vector, use the law of cosines to find the resultant. Adding the squares of the magnitudes will give the desired result.

Answer 1

The Correct Option is B

Step 1: Using the law of cosines.
The magnitude of the resultant vector \( R_1 \) is given by the law of cosines: \[ R_1^2 = P^2 + Q^2 + 2PQ \cos \theta \] Similarly, when the direction of \( \vec{Q} \) is reversed, the magnitude of the resultant \( R_2 \) is: \[ R_2^2 = P^2 + Q^2 - 2PQ \cos \theta \]
Step 2: Adding the equations.
Adding \( R_1^2 \) and \( R_2^2 \), we get: \[ R_1^2 + R_2^2 = (P^2 + Q^2 + 2PQ \cos \theta) + (P^2 + Q^2 - 2PQ \cos \theta) = 2(P^2 + Q^2) \]
Step 3: Conclusion.
Thus, the value of \( R_1^2 + R_2^2 \) is \( 2(P^2 + Q^2) \), which is option (B).