A) \[\vec{a}\]
B) \[\vec{a}\,\times \,\hat{k}\]
C) \[-2\vec{a}\]
D) \[-\,\vec{a}\]
Correct Answer: C
Solution :
| [c] Suppose \[\vec{a}\,={{a}_{1}}\,\hat{i}+\,{{a}_{2}}\hat{j}+\,{{a}_{3}}\hat{k}\] |
| Now, \[(\hat{i}\times \,\vec{a})\,={{a}_{2}}\,\hat{k}\,-{{a}_{3}}\hat{j}\] |
| Now, \[\hat{i}\times \,(\hat{i}\times \,\vec{a})\,=-{{a}_{2}}\,\hat{j}-{{a}_{2}}\hat{k}\] |
| Similarly, \[\hat{j}\times \,(\hat{j}\times \,\vec{a})=-{{a}_{1}}i6-{{a}_{3}}\hat{k}\,\] |
| and \[\hat{k}\,\times \,(\hat{k}\times \,\vec{a})\,=-{{a}_{1}}\,\hat{i}-{{a}_{2}}\hat{j}\] |
| \[\therefore \] \[\hat{i}\,\times \,(\hat{i}\times \vec{a})\,+\,\hat{j}\times \,(\hat{j}\times \,\vec{a})\,+\vec{k}\,\times \,(\hat{k}\times \,\vec{a})=-2\vec{a}.\] |
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