A) \[\mathbf{u}=0\]
B) \[\mathbf{u}=\mathbf{i}+\mathbf{j}+\mathbf{k}\]
C) \[\mathbf{u}=2\mathbf{a}\]
D) \[\mathbf{u}=\mathbf{a}\]
Correct Answer: C
Solution :
Let \[\mathbf{a}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\] \[\mathbf{i}\times (\mathbf{a}\times \mathbf{i})+\mathbf{j}\times (\mathbf{a}\times \mathbf{j})+\mathbf{k}\times (\mathbf{a}\times \mathbf{k})\] \[=(\mathbf{i}.\mathbf{i})\mathbf{a}-\mathbf{i}(\mathbf{a}.\mathbf{i})+(\mathbf{j}.\mathbf{j})\mathbf{a}-\mathbf{j}(\mathbf{a}.\mathbf{j})+(\mathbf{k}.\mathbf{k})\mathbf{a}-\mathbf{k}(\mathbf{a}.\mathbf{k})\] \[=3\mathbf{a}-\mathbf{a}=2\mathbf{a}.\]You need to login to perform this action.
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