A) \[\frac{1}{3}\,\,(\hat{i}-2\hat{j}+\hat{k})\]
B) \[\frac{1}{3}\,\,(-\,\hat{i}+2\hat{j}+5\hat{k})\]
C) \[\frac{1}{3}\,\,(\hat{i}+2\hat{j}-5\hat{k})\]
D) \[\frac{1}{3}\,\,(-\,\hat{i}+2\hat{j}-5\hat{k})\]
Correct Answer: B
Solution :
| \[\vec{a}\times \vec{b}=\vec{a}\times (\vec{a}\times \vec{c})=(\vec{a}\cdot \vec{c})\vec{a}-(\vec{a}\cdot \vec{a})\vec{c}=2\vec{a}-3\vec{c}\] |
| But \[\vec{a}\times \vec{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 1 & 1 \\ 1 & -\,2 & 1 \\ \end{matrix} \right|=3\hat{i}-3\hat{k}\] |
| Hence \[3\vec{c}=2\vec{a}-(3\hat{i}-3\hat{k})\] |
| \[=(2\hat{i}+2\hat{j}+2\hat{k})-(3\hat{i}-3\hat{k})=-\,\hat{i}+2\hat{j}+5\hat{k}\] |
| \[\Rightarrow \] \[\vec{c}=\frac{1}{3}(-\,\hat{i}+2\hat{j}+5\hat{k})\] |
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