Question

If \[ | \vec{P} + \vec{Q} | = | \vec{P} | = | \vec{Q} |, \] then the angle between \( \vec{P} \) and \( \vec{Q} \) is:

• A. \( 0^\circ \)
• B. \( 120^\circ \)
• C. \( 60^\circ \)
• D. \( 90^\circ \)

Answer 1

The Correct Option is B

Step 1: Given Conditions

Let the magnitude of \( \vec{P} \) and \( \vec{Q} \) be \( P \). So, we are told: \[ | \vec{P} + \vec{Q} | = P \]

Step 2: Use the vector addition formula 

From the formula for the magnitude of the sum of two vectors: \[ | \vec{P} + \vec{Q} | = \sqrt{P^2 + Q^2 + 2PQ \cos \theta} \] Since \( | \vec{P} | = | \vec{Q} | = P \), we have: \[ | \vec{P} + \vec{Q} | = \sqrt{P^2 + P^2 + 2P^2 \cos \theta} = \sqrt{2P^2 (1 + \cos \theta)} \]

Step 3: Equating with the given magnitude

Since \( | \vec{P} + \vec{Q} | = P \), we equate: \[ \sqrt{2P^2 (1 + \cos \theta)} = P \] Squaring both sides: \[ 2P^2 (1 + \cos \theta) = P^2 \Rightarrow 2(1 + \cos \theta) = 1 \Rightarrow 1 + \cos \theta = \frac{1}{2} \Rightarrow \cos \theta = -\frac{1}{2} \Rightarrow \theta = 120^\circ \]

Answer:

120°

Answer 2

We are given that: \[ | \vec{P} + \vec{Q} | = | \vec{P} | = | \vec{Q} |. \] Let the magnitudes of \( \vec{P} \) and \( \vec{Q} \) be \( r \), so we have: \[ | \vec{P} | = | \vec{Q} | = r. \] Also, from the given condition, we know: \[ | \vec{P} + \vec{Q} | = r. \] The magnitude of the vector sum \( \vec{P} + \vec{Q} \) is given by the formula: \[ | \vec{P} + \vec{Q} | = \sqrt{|\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos \theta}, \] where \( \theta \) is the angle between \( \vec{P} \) and \( \vec{Q} \). Substituting the values \( | \vec{P} | = | \vec{Q} | = r \), we get: \[ r = \sqrt{r^2 + r^2 + 2r^2 \cos \theta}. \] Simplifying: \[ r = \sqrt{2r^2 + 2r^2 \cos \theta}. \] Squaring both sides: \[ r^2 = 2r^2 + 2r^2 \cos \theta. \] Rearranging: \[ 0 = r^2 (1 + \cos \theta). \] For \( r \neq 0 \), we have: \[ 1 + \cos \theta = 0, \] which gives: \[ \cos \theta = -1. \] Therefore, \( \theta = 120^\circ \). Thus, the correct answer is option (2).