A) \[7\hat{i}+5\hat{j}+\hat{k}\]
B) \[-7\hat{i}-5\hat{j}-\hat{k}\]
C) \[\hat{i}+\hat{j}-\hat{k}\]
D) \[\hat{i}+\hat{j}+\hat{k}\]
Correct Answer: A
Solution :
Since, \[\vec{c}\]is perpendicular to the vectors \[\vec{a}=(2,-3,1)\]and\[\vec{b}=(1,-2,3),\] therefore\[\vec{c}\] is parallel to \[\vec{a}\times \vec{b}.\] \[\therefore \] \[\vec{c}=\lambda (\vec{a}\times \vec{b})\] \[\Rightarrow \] \[\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & -3 & 1 \\ 1 & -2 & 3 \\ \end{matrix} \right|\] \[\Rightarrow \] \[\vec{c}=\lambda (-7\hat{i}-5\hat{j}-\hat{k})\] Also, it is given that \[\vec{c}.(\hat{i}+2\hat{j}-7\hat{k})=10\] \[\Rightarrow \]\[\lambda (-7\hat{i}-5\hat{j}-\hat{k}).(\hat{i}+2\hat{j}-7\hat{k})=10\] \[\Rightarrow \]\[\lambda (-7-10+7)=10\] \[\Rightarrow \]\[-10\lambda =10\] \[\Rightarrow \]\[\lambda =-1\] Hence, \[\vec{c}=(-1)(-7\hat{i}-5\hat{j}-\hat{k})\] \[=(7\hat{i}+5\hat{j}+\hat{k})\]
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