A) \[\mathbf{i}\]
B) \[\mathbf{k}\]
C) \[\mathbf{j}\]
D) \[(\mathbf{i}+\mathbf{j}+\mathbf{k})\]
Correct Answer: A
Solution :
Let\[\mathbf{a}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\] then,\[\mathbf{a}\cdot \mathbf{i}=(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})\cdot \mathbf{i}=x\] and\[\mathbf{a}\cdot (\mathbf{i}+\mathbf{j})=(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})\cdot (\mathbf{i}+\mathbf{j})=x+y\] and\[\mathbf{a}\cdot (\mathbf{i}+\mathbf{j}+\mathbf{k})=(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})\cdot (\mathbf{i}+\mathbf{j}+\mathbf{k})\] \[=x+y+z\] Given that\[\mathbf{a}\cdot \mathbf{i}=\mathbf{a}\cdot (\mathbf{j}+\mathbf{i})=\mathbf{a}\cdot (\mathbf{i}+\mathbf{j}+\mathbf{k})\] \[\because \]\[x=x+y=x+y+z\] Now,\[x=x+y\] \[y=0\] and\[x+y=x+y+z\] \[\Rightarrow z=0\] Hence; \[x=1\] \[\therefore \] \[\mathbf{a}=i\]
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