question_answer

A vector \[\vec{A}\] is rotated by a small angle\[\Delta \theta \]radians \[(\Delta \theta <<1)\] to get a new vector \[\vec{B}.\] In that case\[|\vec{B}-\vec{A}|\]is:  [JEE MAIN 11-04-2015]

A) 0

B) \[|\vec{A}|\left( 1-\frac{{{\Delta }_{\theta }}2}{2} \right)\]

C) \[|\vec{A}|\Delta \theta \]

D) \[|\vec{B}|\Delta \theta -|\vec{A}|\]

Correct Answer: C

Solution :

[c]

By triangle rule

\[\vec{A}+\vec{C}=\vec{B}\]

\[\vec{B}-\vec{A}=\vec{C}\]

\[\left| \vec{B}-\vec{A} \right|=\left| {\vec{C}} \right|=\left| {\vec{B}} \right|\sin \Delta \theta \]

\[\left| \vec{B}-\vec{A} \right|=\left| {\vec{B}} \right|\Delta \theta \] \[(\because \sin \Delta \theta \simeq \Delta \theta )\]

again \[\left| {\vec{B}} \right|\cos \Delta \theta =\vec{A}\]

\[\therefore \] \[\cos \Delta \theta \simeq 1\]                         \[\left| {\vec{B}} \right|=\left| {\vec{A}} \right|\]

So, \[\left| \vec{B}-\vec{A} \right|=\left| {\vec{B}} \right|\Delta \theta =\left| {\vec{A}} \right|\Delta \theta \]