Question

The shortest distance between the lines $ \dfrac{{x - 3}}{3} = \dfrac{{y - 8}}{{ - 1}} = \dfrac{{z - 3}}{1} $ and $ \dfrac{{x + 3}}{{ - 3}} = \dfrac{{y + 7}}{2} = \dfrac{{z - 6}}{4} $
A. $ \sqrt {30} $
B. $ 2\sqrt {30} $
C. $ 5\sqrt {30} $
D. $ 3\sqrt {30} $

Answer 1

Hint: We are provided with two equations and we have to find the shortest distance between these lines. The formula for the shortest distance between any two lines is
\[\dfrac{{\left| {\begin{array}{*{20}{c}}
  {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}}
\end{array}} \right|}}{{\sqrt {{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2} + {{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2}} }}\]
Firstly calculate the numerator value and then denominator and divide the two values.

Complete step by step solution:
The general equations are of the form $ \dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}} $ and $ \dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{z - {z_2}}}{{{c_2}}} $
Compare the terms with the given equations which are $ \dfrac{{x - 3}}{3} = \dfrac{{y - 8}}{{ - 1}} = \dfrac{{z - 3}}{1} $ and $ \dfrac{{x + 3}}{{ - 3}} = \dfrac{{y + 7}}{2} = \dfrac{{z - 6}}{4} $
So, the terms become
 $
  {x_1} = 3,{y_1} = 8,{z_1} = 3 \\
  {a_1} = 3,{b_1} = - 1,{c_1} = 1 \\
  $ and $
  {x_2} = - 3,{y_2} = - 7,{z_2} = 6 \\
  {a_2} = - 3,{b_2} = 2,{c_2} = 4 \\
  $
Calculating the numerator of the formula of the shortest distance
\[\left| {\begin{array}{*{20}{c}}
  {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}}
\end{array}} \right|\]
Substituting the values
\[\left| {\begin{array}{*{20}{c}}
  { - 3 - 3}&{ - 7 - 8}&{6 - 3} \\
  3&{ - 1}&1 \\
  { - 3}&2&4
\end{array}} \right|\]
Calculating the determinant
 $
   = - 6\left( { - 4 - 2} \right) + 15\left( {12 + 3} \right) + 3\left( {6 - 3} \right) \\
   = - 6\left( { - 6} \right) + 15\left( {15} \right) + 3\left( 3 \right) \\
   = 36 + 225 + 9 \\
   = 270 \;
  $
Now, calculating the denominator for the formula
 $ \sqrt {{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2} + {{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2}} $
 $
   = \sqrt {{{\left( {3.2 - \left( { - 3} \right)\left( { - 1} \right)} \right)}^2} + {{\left[ {\left( { - 1} \right)4 - 2.1} \right]}^2} - {{\left[ {1\left( { - 3} \right) - 4.3} \right]}^2}} \\
   = \sqrt {{{\left( {6 - 3} \right)}^2} + {{\left( { - 4 - 2} \right)}^2} + {{\left( { - 3 - 12} \right)}^2}} \\
   = \sqrt {9 + 36 + 225} \\
   = \sqrt {270} \;
  $
  The formula for calculating the shortest distance is
\[\dfrac{{\left| {\begin{array}{*{20}{c}}
  {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}}
\end{array}} \right|}}{{\sqrt {{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2} + {{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2}} }}\]
Substituting the values
 $ = \dfrac{{270}}{{\sqrt {270} }} = \sqrt {270} $ $ = 3\sqrt {30} $
Hence, the shortest distance between the two given lines is d $ = 3\sqrt {30} $
So, the correct answer is “Option D”.

Note: Learn the formula for the shortest distance between the two lines carefully. For our convenience calculate the numerator first and then calculate the denominator and divide the two values. Do the calculations carefully.