Question

A vector \(\vec{A}\) is rotated by a small angle \(\Delta\theta\) radians \(\left(\Delta\theta<<1\right)\) to get a new vector \(\vec{B}.\) In that case \(\left|\vec{B}-\vec{A}\right|\) is :

• A. $0$
• B. $\left|\vec{A}\right|\left(1-\frac{\Delta\theta^{2}}{2}\right)$
• C. $\left|\vec{A}\right|\Delta\theta$
• D. $\left|\vec{B}\right|\Delta\theta-\left|\vec{A}\right|$

Answer 1

The Correct Option is C

The problem involves finding the magnitude of the difference between two vectors, \( \vec{A} \) and \( \vec{B} \), where \( \vec{B} \) is obtained by rotating \( \vec{A} \) by a small angle \( \Delta\theta \). We can use the concept of small angle approximations in rotational dynamics to solve this.

1. **Understanding Vector Rotation:** When a vector \( \vec{A} \) is rotated by a small angle \( \Delta\theta \), where \( \Delta\theta \ll 1 \), the resulting vector \( \vec{B} \) forms an arc of a circle with initial vector \( \vec{A} \) as one of its radii.
2. **Magnitude of the Difference:**
   • The angle \( \Delta\theta \) defines a small segment of the circle. The arc length (which represents the path of rotation) can be approximated using the small-angle formula:
   • The length of this arc for small angles is approximately \( \left|\vec{A}\right|\Delta\theta \).
3. **Calculations:**
   • For small angles, \(\sin(\Delta\theta) \approx \Delta\theta\) and \(\cos(\Delta\theta) \approx 1\). Therefore, \( \vec{B} = \vec{A} \cos(\Delta\theta) + \vec{k} \times \vec{A} \sin(\Delta\theta) \), where \( \vec{k} \) is a unit vector perpendicular to \( \vec{A} \).
   • \( |\vec{B} - \vec{A}| \approx |\vec{A}| \Delta\theta \) due to the arc of rotation.
4. **Conclusion:** The magnitude of the difference between the vectors \( \vec{B} \) and \( \vec{A} \) is approximately \( \left|\vec{A}\right|\Delta\theta \).

The correct answer is therefore $\left|\vec{A}\right|\Delta\theta$.