A) \[\mathbf{a}=(\mathbf{a}\,.\,\mathbf{i})\,\mathbf{i}+\,(\mathbf{a}\,.\,\mathbf{j})\,\mathbf{j}+\,(\mathbf{a}\,.\,\mathbf{k})\,\mathbf{k}\]
B) \[\mathbf{a}=(\mathbf{a}\,\times \,\mathbf{i})+\,(\mathbf{a}\,\times \,\mathbf{j})\,+\,(\mathbf{a}\,\times \,\mathbf{k})\,\]
C) \[\mathbf{a}=\mathbf{j}\,(\mathbf{a}\,.\,\mathbf{i})\,+\mathbf{k}\,(\mathbf{a}\,.\,\mathbf{j})\,+\,\mathbf{i}\,(\mathbf{a}\,.\,\mathbf{k})\,\]
D) \[\mathbf{a}=(\mathbf{a}\,\times \,\mathbf{i})\times \mathbf{i}+\,(\mathbf{a}\,\times \,\mathbf{j})\times \mathbf{j}\,+\,(\mathbf{a}\,\times \,\mathbf{k})\times \mathbf{k}\,\]
Correct Answer: A
Solution :
Let\[\mathbf{a}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k},\]then \[\mathbf{a}.\mathbf{i}={{a}_{1}},\] \[\mathbf{a}.\mathbf{j}={{a}_{2}},\] \[\mathbf{a}.\mathbf{k}={{a}_{3}}\] \[\therefore \,\,\,\mathbf{a}=(\mathbf{a}\,.\,\mathbf{i})\mathbf{i}+(\mathbf{a}\,.\,\mathbf{j})\mathbf{j}+(\mathbf{a}\,.\,\mathbf{k})\mathbf{k}\].
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