A) \[\vec{a}.\hat{j}=0\]
B) \[\vec{a}.\hat{i}=0\]
C) \[\vec{a}.\hat{k}=0\]
D) \[{{\vec{v}}_{1}}.{{\vec{v}}_{2}}={{(\vec{a}.\hat{j})}^{2}}\]
Correct Answer: A
Solution :
[a] \[{{\vec{v}}_{1}}=(\vec{a}\times \hat{i})\times \hat{i}=(\vec{a}.\hat{i})\hat{i}-\vec{a}\] \[{{\vec{v}}_{2}}=(\vec{a}\times \hat{j})\times \hat{j}=(\vec{a}.\hat{j})\hat{j}-\vec{a}\] \[{{\vec{v}}_{3}}=(\vec{a}\times \hat{k})\times \hat{k}=(\vec{a}.\hat{k})\hat{k}-\vec{a}\] \[{{\vec{v}}_{1}}={{\vec{v}}_{2}}-{{\vec{v}}_{3}}\] \[\Rightarrow \,\,\,(\vec{a}.\hat{i})\hat{i}-\vec{a}=(\vec{a}.\hat{j})\hat{j}-\vec{a}-(\vec{a}.\hat{k})\hat{k}+\vec{a}\] \[\Rightarrow \,\,\,\,(\vec{a}.\hat{i})\hat{i}-\{(\vec{a}.\hat{i})\hat{i}+(\vec{a}.\hat{j})\hat{j}+(\vec{a}.\hat{k})\hat{k}\}=(\vec{a}.\hat{j})\hat{j}-(\vec{a}.\hat{k})\hat{k}\] \[\Rightarrow \,\,\,\,\,\,\,\,2(\vec{a}.\hat{j})\hat{j}=0\] \[\Rightarrow \,\,\,\,\,\vec{a}.\hat{j}=0\]You need to login to perform this action.
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