A) (2, 3]
B) [0, 1)
C) (3, 4]
D) [1, 2)
Correct Answer: A
Solution :
\[\frac{x}{z}=\frac{y}{z}\frac{z}{1}\]and\[\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\]shortest distance \[=({{\vec{a}}_{2}}-{{\vec{a}}_{1}}).\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)\] here\[{{\vec{b}}_{1}}-{{\vec{b}}_{2}}=(2i+2j+k)\times (-i+8j+4k)\] \[=-9j+18k\] \[\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)=\frac{-j+2k}{\sqrt{5}}\] \[{{\vec{a}}_{2}}-{{\vec{a}}_{1}}=-2i+4j+5k\] \[\therefore \]S.D.\[(-2i+4j+6k).\]\[\frac{(-j+2k)}{\sqrt{5}}=\frac{6}{\sqrt{5}}\]which lies in (2,3]
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