Question

Find the shortest distance between the lines \[\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{4}\] and \[\dfrac{{x - 2}}{3} = \dfrac{{y - 4}}{4} = \dfrac{{z - 5}}{5}\].

Answer 1

Hint: We need to find the shortest distance between the given two lines. We will directly use the formula for the shortest distance between two lines. Let \[\dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}}\] and \[\dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{z - {z_2}}}{{{c_2}}}\] be two lines. Let \[d\] denote the shortest distance between these two lines. Then,
\[d = \left| {\dfrac{{\left| {\begin{array}{*{20}{c}}
  {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}}
\end{array}} \right|}}{{\sqrt {{{({b_1}{c_2} - {b_2}{c_1})}^2} + {{({c_1}{a_2} - {c_2}{a_1})}^2} + {{({a_1}{b_2} - {a_2}{b_1})}^2}} }}} \right|\],
Where,
\[\left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right| = {a_{11}}({a_{22}}{a_{33}} - {a_{32}}{a_{23}}) - {a_{12}}({a_{21}}{a_{33}} - {a_{31}}{a_{23}}) + {a_{13}}({a_{21}}{a_{32}} - {a_{31}}{a_{22}})\]
And the outer Lines indicate the modulus sign.

Complete step-by-step solution:
We need to find the distance between the lines \[\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{4}\] and \[\dfrac{{x - 2}}{3} = \dfrac{{y - 4}}{4} = \dfrac{{z - 5}}{5}\].
Comparing \[\dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}}\] and \[\dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{z - {z_2}}}{{{c_2}}}\] with \[\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{4}\] and \[\dfrac{{x - 2}}{3} = \dfrac{{y - 4}}{4} = \dfrac{{z - 5}}{5}\] respectively, we get
\[{x_1} = 1,{y_1} = 2,{z_1} = 3 - - - - - ()\]
\[{x_2} = 2,{y_2} = 4,{z_2} = 5 - - - - - (2)\]
\[{a_1} = 2,{b_1} = 3,{c_1} = 4 - - - - - (3)\]
\[{a_2} = 3,{b_2} = 4,{c_2} = 5 - - - - - (4)\]
Let \[d\] be the shortest distance between the lines \[\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{4}\] and \[\dfrac{{x - 2}}{3} = \dfrac{{y - 4}}{4} = \dfrac{{z - 5}}{5}\].
Substituting the values from (1), (2), (3) and (4) in the formula
\[d = \left| {\dfrac{{\left| {\begin{array}{*{20}{c}}
  {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}}
\end{array}} \right|}}{{\sqrt {{{({b_1}{c_2} - {b_2}{c_1})}^2} + {{({c_1}{a_2} - {c_2}{a_1})}^2} + {{({a_1}{b_2} - {a_2}{b_1})}^2}} }}} \right|\]
\[ \Rightarrow d = \left| {\dfrac{{\left| {\begin{array}{*{20}{c}}
  {2 - 1}&{4 - 2}&{5 - 3} \\
  2&3&4 \\
  3&4&5
\end{array}} \right|}}{{\sqrt {{{\left( {(3 \times 5) - (4 \times 4)} \right)}^2} + {{\left( {(4 \times 3) - (5 \times 2)} \right)}^2} + {{\left( {(2 \times 4) - (3 \times 3)} \right)}^2}} }}} \right|\]
Solving the brackets, we get
\[ \Rightarrow d = \left| {\dfrac{{\left| {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&3&4 \\
  3&4&5
\end{array}} \right|}}{{\sqrt {{{\left( {(15) - (16)} \right)}^2} + {{\left( {(12) - (10)} \right)}^2} + {{\left( {(8) - (9)} \right)}^2}} }}} \right|\]
\[ \Rightarrow d = \left| {\dfrac{{\left| {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&3&4 \\
  3&4&5
\end{array}} \right|}}{{\sqrt {{{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2}} }}} \right|\]
Now, solving the denominator using \[{a^2} = a \times a\]
\[ \Rightarrow d = \left| {\dfrac{{\left| {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&3&4 \\
  3&4&5
\end{array}} \right|}}{{\sqrt {1 + 4 + 1} }}} \right|\]
\[ \Rightarrow d = \left| {\dfrac{{\left| {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&3&4 \\
  3&4&5
\end{array}} \right|}}{{\sqrt 6 }}} \right| - - - - - - (5)\]
Now, solving the numerator part and substituting it in (5).
Solving \[\left| {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&3&4 \\
  3&4&5
\end{array}} \right|\] now,
\[\left| {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&3&4 \\
  3&4&5
\end{array}} \right| = 1\left( {(3 \times 5) - (4 \times 4)} \right) - 2\left( {(2 \times 5) - (3 \times 4)} \right) + 2\left( {(2 \times 4) - (3 \times 3)} \right)\]
Solving the brackets, we get
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&3&4 \\
  3&4&5
\end{array}} \right| = 1\left( {(15) - (16)} \right) - 2\left( {(10) - (12)} \right) + 2\left( {(8) - (9)} \right)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&3&4 \\
  3&4&5
\end{array}} \right| = 1\left( { - 1} \right) - 2\left( { - 2} \right) + 2\left( { - 1} \right)\]
Multiplying the brackets,
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&3&4 \\
  3&4&5
\end{array}} \right| = - 1 + 4 + - 2\]
Now solving the right hand side
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&3&4 \\
  3&4&5
\end{array}} \right| = - 3 + 4\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&3&4 \\
  3&4&5
\end{array}} \right| = 1 - - - - - - (6)\]
Now, substituting (6) in (5), we get
\[ \Rightarrow d = \left| {\dfrac{1}{{\sqrt 6 }}} \right|\]
Since \[\dfrac{1}{{\sqrt 6 }}\] is positive, \[\left| {\dfrac{1}{{\sqrt 6 }}} \right|\]is equal to \[\dfrac{1}{{\sqrt 6 }}\].
Hence, we get \[d = \dfrac{1}{{\sqrt 6 }}\].
Therefore, the shortest distance between the lines \[\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{4}\] and \[\dfrac{{x - 2}}{3} = \dfrac{{y - 4}}{4} = \dfrac{{z - 5}}{5}\] is \[\dfrac{1}{{\sqrt 6 }}\] units.

Note: First of all, we need to see the type of lines given in the question. We usually mix the formulas for the shortest distance between the two lines and the perpendicular distance from the origin to a line. Also, to find the shortest distance, we should compare the values with full concentration as if we compare the wrong values, we will get the wrong answer. In case, we have got our answer for the numerator to be negative, we will have to use modulus properties then as distance can never be negative.