Question

The resultant magnitude of two vectors of same magnitude is equal to magnitude of either. The angle between the two vectors is

• A. $30^\circ$
• B. $60^\circ$
• C. $90^\circ$
• D. $120^\circ$

Hint:

Vector Addition Rule:

• $\vecR = \vecA + \vecB$
• $|\vecR| = \sqrtA^2 + B^2 + 2AB\cos\theta$
• If $|\vecR| = A = B$, solve for $\cos\theta$

Answer 1

The Correct Option is D

Let $|\vec{A}| = |\vec{B}| = P$, and $|\vec{A} + \vec{B}| = P$.
Using vector addition formula: \[ |\vec{R}|^2 = A^2 + B^2 + 2AB\cos\theta = P^2 \] \[ P^2 = 2P^2 + 2P^2 \cos\theta \Rightarrow -P^2 = 2P^2\cos\theta \Rightarrow \cos\theta = -\frac{1}{2} \Rightarrow \theta = 120^\circ \]

Answer 2

Step 1: Understand the problem
Two vectors have the same magnitude \(A\). The magnitude of their resultant is equal to the magnitude of either vector.

Step 2: Use the formula for the magnitude of the resultant of two vectors
If the angle between the vectors is \(\theta\), then the magnitude of the resultant \(R\) is given by:
\[ R = \sqrt{A^2 + A^2 + 2A \times A \cos \theta} = \sqrt{2A^2 (1 + \cos \theta)} = A \sqrt{2(1 + \cos \theta)} \]

Step 3: Set the resultant equal to the magnitude of one vector
\[ R = A \]
\[ A = A \sqrt{2(1 + \cos \theta)} \implies 1 = \sqrt{2(1 + \cos \theta)} \]
Square both sides:
\[ 1 = 2(1 + \cos \theta) \implies 1 = 2 + 2 \cos \theta \implies 2 \cos \theta = -1 \implies \cos \theta = -\frac{1}{2} \]

Step 4: Find \(\theta\)
\[ \cos \theta = -\frac{1}{2} \implies \theta = 120^\circ \]

Step 5: Conclusion
The angle between the two vectors is \(120^\circ\).