A) \[\sqrt{21}\]
B) \[\sqrt{102}\]
C) 4
D) 3
Correct Answer: C
Solution :
The given lines are\[\mathbf{r}={{\mathbf{a}}_{1}}+\lambda {{\mathbf{b}}_{1}},\mathbf{r}={{\mathbf{a}}_{2}}+\mu {{\mathbf{b}}_{2}}\], Where \[{{\mathbf{a}}_{1}}=3\mathbf{i}-2\mathbf{j}-2\mathbf{k},\,\,\,{{\mathbf{b}}_{1}}=\mathbf{i}\] \[{{\mathbf{a}}_{2}}=\mathbf{i}-\mathbf{j}+2\mathbf{k},\,\,\,\,\,{{\mathbf{b}}_{2}}=\mathbf{j}\] \[|{{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{2}}|\,=\,|\mathbf{i}\times \mathbf{j}|\,=\,|\mathbf{k}|=1\] Now, \[[({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}})\ {{\mathbf{b}}_{1}}\ {{\mathbf{b}}_{2}}]=({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}}).({{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{2}})\] \[=(-2\mathbf{i}+\mathbf{j}+4\mathbf{k})(\mathbf{k})=4\] \ Shortest distance\[=\frac{[({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}})({{\mathbf{b}}_{1}}-{{\mathbf{b}}_{2}})]}{|{{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{2}}|}=\frac{4}{1}=4\].
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