The shortest distance between the lines {x \over 2} = {y \over 2} = {z \over 1} and {{x + 2} \over { - 1}} = {{y - 4} \over 8} = {{z - 5} \over 4} lies in the interval :

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Answer: C. (2, 3] Shortest distance between the lines {{x - {x_1}} \over {{a_1}}} = {{y - {y_1}} \over {{b_1}}} = {{z - {z_1}} \over {{c_1}}} and {{x - {x_2}} \over {{a_2}}} = {{y - {y_2}} \over {{b_2}}} = {{z - {z_2}} \over {{c_2}}} is \left| {{{\left| {\begin{matrix} {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \ {{a_1}} & {{b_1}} & {{c_1}} \ {{a_2}} & {{b_2}} & {{c_2}} \ \end{matrix} } ight|} \over {\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} ight)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} ight)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} ight)}^2}} }}} ight| 	herefore Shortest distance between two given lines are, \left| {{{\left| {\begin{matrix} { - 2} & 4 & 5 \ 2 & 2 & 1 \ { - 1} & 8 & 4 \ \end{matrix} } ight|} \over {\sqrt {{{\left( {8 - 8} ight)}^2} + {{\left( { - 1 - 8} ight)}^2} + {{\left( {16 + 2} ight)}^2}} }}} ight| = \left| {{{ - 36 + 90} \over {\sqrt {405} }}} ight| = {{54} \over {20.1}} = 2.68

The shortest distance between the lines $${x \over 2} = {y \over 2} = {z \over 1}$$ and $${{x + 2} \over { - 1}} = {{y - 4} \over 8} = {{z - 5} \over 4}$$ lies in the interval :

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