Q102 / 150 · MHT CET 2026 (17 April Shift 1) · Physics
Resultant of two vectors $\vec{P}$ and $\vec{Q}$ is of magnitude A. If $\vec{Q}$ is reversed, then the resultant is of magnitude B. The value of $A^2 + B^2$ is
- A$P^2 + Q^2$
- B$P^2 - Q^2$
- Correct answerC$2(P^2 + Q^2)$
- D$2(P^2 - Q^2)$
Solution
Let the angle between vectors $\vec{P}$ and $\vec{Q}$ be $\theta$.
The magnitude of the resultant of $\vec{P}$ and $\vec{Q}$ is $A$.
$A^2 = P^2 + Q^2 + 2PQ \cos\theta$
When $\vec{Q}$ is reversed, the angle between $\vec{P}$ and $-\vec{Q}$ becomes $180^\circ - \theta$.
The magnitude of the new resultant is $B$.
$B^2 = P^2 + Q^2 + 2PQ \cos(180^\circ - \theta)$
$B^2 = P^2 + Q^2 - 2PQ \cos\theta$
Adding the two equations, we get:
$A^2 + B^2 = (P^2 + Q^2 + 2PQ \cos\theta) + (P^2 + Q^2 - 2PQ \cos\theta)$
$A^2 + B^2 = 2(P^2 + Q^2)$
The magnitude of the resultant of $\vec{P}$ and $\vec{Q}$ is $A$.
$A^2 = P^2 + Q^2 + 2PQ \cos\theta$
When $\vec{Q}$ is reversed, the angle between $\vec{P}$ and $-\vec{Q}$ becomes $180^\circ - \theta$.
The magnitude of the new resultant is $B$.
$B^2 = P^2 + Q^2 + 2PQ \cos(180^\circ - \theta)$
$B^2 = P^2 + Q^2 - 2PQ \cos\theta$
Adding the two equations, we get:
$A^2 + B^2 = (P^2 + Q^2 + 2PQ \cos\theta) + (P^2 + Q^2 - 2PQ \cos\theta)$
$A^2 + B^2 = 2(P^2 + Q^2)$
Hence, the correct answer is option: 'C'.