A) 3
B) 1
C) 2
D) 0
Correct Answer: D
Solution :
The Given lines are \[{{\mathbf{r}}_{1}}={{\mathbf{a}}_{1}}+\lambda \,{{\mathbf{b}}_{1}},\,\,\,\,{{\mathbf{r}}_{2}}={{\mathbf{a}}_{2}}+\mu {{\mathbf{b}}_{2}}\] Where \[{{\mathbf{a}}_{1}}=4\mathbf{i}-3\mathbf{j}-\mathbf{k};\,\,\,\,{{\mathbf{b}}_{1}}=\mathbf{i}-4\mathbf{j}+7\mathbf{k}\] \[{{\mathbf{a}}_{2}}=\mathbf{i}-\mathbf{j}-10\mathbf{k};\,\,\,\,{{\mathbf{b}}_{2}}=2\mathbf{i}-3\mathbf{j}+8\mathbf{k}\] \[|{{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{2}}|=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -4 & 7 \\ 2 & -3 & 8 \\ \end{matrix} \right|=-11\mathbf{i}+6\mathbf{j}+5\mathbf{k}\] Now \[[({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}})\,\,{{\mathbf{b}}_{1}}\,\,{{\mathbf{b}}_{2}}]=({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}}).({{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{2}})\] \[=(-3\mathbf{i}+2\mathbf{j}-9\mathbf{k})(-11\mathbf{i}+6\mathbf{j}+5\mathbf{k})=0\] Therefore, shortest distance \[=\frac{[({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}})\,\,{{\mathbf{b}}_{1}}\,\,{{\mathbf{b}}_{2}}]}{|{{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{2}}|}=0\].
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