A) \[(-1,\,-1/2)\]
B) \[(1,\,-6)\]
C) \[(-1,\,0)\]
D) \[(1,\,1/2)\]
Correct Answer: C
Solution :
Vector along the line is \[\vec{V}=6\hat{i}+0.\hat{j}+(2-b)\hat{k}\] It is perpendicular to vector \[{{\vec{V}}_{1}}=\hat{i}+3\hat{j}-2\hat{k}\] \[\therefore \,\,\vec{V}\cdot {{\vec{V}}_{1}}=0\] \[\Rightarrow \] \[6-2(a-b)=0\] \[\Rightarrow \] \[b=-1\] Now the vector perpendicular to the vector \[\hat{i}+c\hat{k}\] and \[c\hat{j}+b\hat{k}\] is \[\vec{n}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 0 & c \\ 0 & c & b \\ \end{matrix} \right|=-{{c}^{2}}\hat{i}-b\hat{j}+c\hat{k}\]. \[\therefore \,\,\vec{u}.\vec{n}=0\] \[\Rightarrow \,-6{{c}^{2}}+3c=0\] \[\Rightarrow \,c=0,\,\frac{1}{2}\]. \[\therefore \,\,(b,c)=(-1,\,0)\] or \[\left( -1,\,\frac{1}{2} \right)\]
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