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