A) \[\frac{19}{2}\]
B) 8
C) \[\frac{17}{2}\]
D) 9
Correct Answer: A
Solution :
| \[\vec{a}\times \vec{c}+\vec{b}\,=\,\vec{0}\] \[\Rightarrow \]\[\vec{a}\times \,(\vec{a}\times \vec{c})+\vec{a}\times \vec{b}\,=\,\vec{0}\] |
| and \[\vec{a}.\vec{c}\,=\,4\,(\text{given})\] |
| \[{{\left| {\vec{a}} \right|}^{2}}\,\vec{c}\,=\,4\vec{a}+(\vec{a}\times \vec{b})\] |
| where \[\vec{a}\times \vec{b}\,=\,\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -1 & 0 \\ 1 & 1 & 1 \\ \end{matrix} \right|\,=\,-\,\hat{i}\,-\,\hat{j}+\,2\hat{k}\] |
| \[\left| {\vec{a}} \right|\,=\,\sqrt{{{1}^{2}}+\,{{(-1)}^{2}}}=\sqrt{2}\] |
| \[{{\left| {\vec{a}} \right|}^{2}}=\,2\] |
| \[4\vec{a}+(\vec{a}+\vec{b})=4(\hat{i}-\hat{j})+(-\hat{i}-\hat{j}+2\hat{k})\]\[=\,3\hat{i}-5\hat{j}+2\hat{k}\] |
| \[2\vec{c}\,=\,3\vec{i}-\,5\vec{j}+2\vec{k}\] |
| \[\Rightarrow \] \[2\left| {\vec{c}} \right|=\sqrt{9+25+4}=\sqrt{38}\] |
| \[\Rightarrow \] \[{{\left| {\vec{c}} \right|}^{2}}=\frac{38}{4}=\frac{19}{2}.\] |
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