Question

Two vectors \( \vec{P} \) and \( \vec{Q} \) have equal magnitudes. If \( | \vec{P} + \vec{Q} | = 5 \) and \( | \vec{P} - \vec{Q} | = 4 \), then the angle between \( \vec{P} \) and \( \vec{Q} \) is

• A. \( \cos^{-1} \left( \frac{12}{13} \right) \)
• B. \( \sin^{-1} \left( \frac{12}{13} \right) \)
• C. \( \cos^{-1} \left( \frac{3}{5} \right) \)
• D. \( \sin^{-1} \left( \frac{3}{5} \right) \)

Hint:

In problems involving vector magnitudes and angles, use the law of cosines to relate the magnitudes and the angle between the vectors.

Answer 1

The Correct Option is A

Step 1: Using the law of cosines.
The magnitudes of the vectors \( \vec{P} + \vec{Q} \) and \( \vec{P} - \vec{Q} \) are given, so we can use the law of cosines. For \( \vec{P} + \vec{Q} \), we have: \[ | \vec{P} + \vec{Q} |^2 = P^2 + Q^2 + 2PQ \cos \theta \] where \( \theta \) is the angle between the vectors. Since \( P = Q \), this simplifies to: \[ 25 = 2P^2 + 2P^2 \cos \theta \] Similarly, for \( \vec{P} - \vec{Q} \), we have: \[ 16 = 2P^2 - 2P^2 \cos \theta \] Step 2: Solving for \( \cos \theta \).
From the two equations, we solve for \( \cos \theta \) and find: \[ \cos \theta = \frac{12}{13} \] Step 3: Conclusion.
Thus, the angle between \( \vec{P} \) and \( \vec{Q} \) is \( \cos^{-1} \left( \frac{12}{13} \right) \), corresponding to option (A).