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 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 :
Updated On: Apr 4, 2026
- $0$
- $\left|\vec{A}\right|\left(1-\frac{\Delta\theta^{2}}{2}\right)$
- $\left|\vec{A}\right|\Delta\theta$
- $\left|\vec{B}\right|\Delta\theta-\left|\vec{A}\right|$
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The Correct Option is C
Solution and Explanation
\(|\vec{B}-\vec{A}|=\sqrt{A^{2}+B^{2}+2 A B \cos (\pi-\Delta \theta)}\)
\(\Rightarrow\) since \(|\vec{A}|=|\vec{B}|=A,\)
\(\cos (\pi-\Delta \theta)=-\cos \Delta \theta\)
\(=\sqrt{2 A^{2}(1+\cos (\pi-\Delta \theta))}\)
\(=\sqrt{2 A^{2} 2 \cos ^{2}\left(\frac{\pi-\Delta \theta}{2}\right)}=2 A \cos \left(\frac{\pi}{2}-\frac{\Delta \theta}{2}\right)=2 A \sin \left(\frac{\Delta \theta}{2}\right)\)
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration