A) \[\overset{\to }{\mathop{A}}\,\]
B) \[2\overset{\to }{\mathop{A}}\,\]
C) \[3\overset{\to }{\mathop{A}}\,\]
D) 0
Correct Answer: B
Solution :
[b] We have \[\hat{i}\times (\vec{A}\times \hat{i})+\hat{j}\times (\vec{A}\times \hat{j})+\hat{k}\times (\vec{A}\times \hat{k})\] \[\hat{i}\times (\vec{A}\times \hat{i})=(\hat{i}.\hat{i})\vec{A}-(\hat{i}.\vec{A})\hat{i}=\vec{A}-(\hat{i}.\vec{A})\hat{i}\] ?. (i) \[\hat{j}\times (\vec{A}\times \hat{j})=(\hat{j}.\hat{j})\vec{A}-(\hat{j}.\vec{A})\hat{j}=\vec{A}-(\hat{j}.\vec{A})\hat{j}\] ? (ii) And \[\hat{k}\times (\vec{A}\times \hat{k})=(\hat{k}.\hat{k})\vec{A}-(\hat{k}.\vec{A})\hat{k}\] \[=\vec{A}-(\hat{k}.\vec{A})\hat{k}\] ? (iii) Now, eqn (i) + eqn (ii) + eqn (iii): \[\hat{i}\times (\vec{A}\times \hat{i})+\hat{j}\times (\vec{A}\times \hat{j})+\hat{k}(\vec{A}\times \hat{k})=3\vec{A}\] \[-[(\hat{i}\cdot \vec{A})\hat{i}+(\hat{j}\cdot \vec{A})\hat{j}+(\hat{k}\cdot \vec{A})\hat{k}]=3\vec{A}-\vec{A}=2\vec{A}.\]
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