A) \[\hat{i}+\hat{j}\]
B) \[\hat{i}-\hat{k}\]
C) \[\hat{i}\]
D) \[\hat{i}+\hat{j}-\hat{k}\]
Correct Answer: C
Solution :
Let \[\vec{a}={{a}_{1}}\,\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}\] Given, \[\vec{a}.\hat{i}=\vec{a}.(\hat{i}+\hat{j})=\vec{a}.(\hat{i}+\hat{j}+\hat{k})=1\] \[\therefore \] \[{{a}_{1}}={{a}_{1}}+{{a}_{2}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}=1\] \[\Rightarrow \] \[{{a}_{1}}=1,{{a}_{2}}=0,{{a}_{3}}=0\] \[\therefore \] s\[\vec{a}=\hat{i}\]
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