Question

Find the shortest distance between the lines whose vector equation are $\vec r = \left( {\hat i + 2\hat j + 3\hat k} \right) + \lambda \left( {\hat i - 3\hat j + 2\hat k} \right)$ and $\vec r = \left( {4\hat i + 5\hat j + 6\hat k} \right) + \mu \left( {2\hat i + 3\hat j + \hat k} \right)$.

Answer 1

Hint: Check whether lines are parallel or not.
When lines are parallel, the direction ratios of both lines are in proportion.
Parallel lines are at the same distance throughout the line.
Neither parallel nor intersecting lines are known as skew lines.
Let a point $\left( {{x_1},{y_1},{z_1}} \right)$ , lie on the line, and have direction ratios: $\left( {a,b,c} \right)$
Cartesian equation of the line:
$\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c}$
Vector equation of the line:
$x\hat i + y\hat j + z\hat k = \left( {{x_1}\hat i + {y_1}\hat j + {z_1}\hat k} \right) + \lambda \left( {a\hat i + b\hat j + c\hat k} \right)$
$\vec r = x\hat i + y\hat j + z\hat k$ , \[{\vec a_1} = {x_1}\hat i + {y_1}\hat j + {z_1}\hat k\] , \[{\vec b_1} = a\hat i + b\hat j + c\hat k\]
Therefore, the vector equation of a line:
${\vec r_1} = {\vec a_1} + \lambda {\vec b_1}$
calculate the shortest distance between the lines using the defined formula.

Complete step-by-step answer:
Step 1: Given equations of lines.
Line 1: $\vec r = \left( {\hat i + 2\hat j + 3\hat k} \right) + \lambda \left( {\hat i - 3\hat j + 2\hat k} \right)$ , where $\vec r = \left( {x\hat i + y\hat j + z\hat k} \right)$
On comparing with the vector equation of the line: $x\hat i + y\hat j + z\hat k = \left( {{x_1}\hat i + {y_1}\hat j + {z_1}\hat k} \right) + \lambda \left( {a\hat i + b\hat j + c\hat k} \right)$
Given a point on line 1: $\left( {{x_1},{y_1},{z_1}} \right) = \left( {1,2,3} \right)$
Direction ratios of line 1: $\left( {{a_1},{b_1},{c_1}} \right) = \left( {1, - 3,2} \right)$ …… (1)
(Subscript 1 to show the direction ratios of ‘line 1’)
Line 2: $\vec r = \left( {4\hat i + 5\hat j + 6\hat k} \right) + \mu \left( {2\hat i + 3\hat j + \hat k} \right)$, where $\vec r = \left( {x\hat i + y\hat j + z\hat k} \right)$
On comparing with the vector equation of the line: $x\hat i + y\hat j + z\hat k = \left( {{x_1}\hat i + {y_1}\hat j + {z_1}\hat k} \right) + \lambda \left( {a\hat i + b\hat j + c\hat k} \right)$
Given a point on line 2: $\left( {{x_2},{y_2},{z_2}} \right) = \left( {4,5,6} \right)$
Direction ratios of line 2: $\left( {{a_2},{b_2},{c_2}} \right) = \left( {2,3,1} \right)$ …… (2)
(Subscript 2 to show the direction ratios of ‘line 2’)
Step 2: Check for parallel lines.
The ratio of direction ratios of line 1 and line 2 should be constant.
$
   \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{1}{2} \\
   \Rightarrow \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{ - 3}}{{ - 3}} = - 1 \\
 $
$\because \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$
This implies line 1 and line 2 are not parallel lines. Therefore, the given lines are two skew lines.
Step 3: Use the formula of the shortest distance between the lines.
Distance, \[d = \dfrac{{\left| {\begin{array}{*{20}{c}}
  {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}}
\end{array}} \right|}}{{\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{a_1}{c_2} - {a_2}{c_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}} }}\]
We have values of every variable involved in the formula for the shortest distance. Thus, substituting them from step 1.
  $
  d = \dfrac{{\left| {\begin{array}{*{20}{c}}
  {4 - 1}&{5 - 2}&{6 - 3} \\
  1&{ - 3}&2 \\
  2&3&1
\end{array}} \right|}}{{\sqrt {{{\left( { - 3 - 6} \right)}^2} + {{\left( {1 - 4} \right)}^2} + {{\left( {3 + 6} \right)}^2}} }} \\
   \Rightarrow \dfrac{{\left| {\begin{array}{*{20}{c}}
  3&3&3 \\
  1&{ - 3}&2 \\
  2&3&1
\end{array}} \right|}}{{\sqrt {{{\left( { - 9} \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( 9 \right)}^2}} }} \\
 $
Calculate the determinant of the matrix in the numerator.
$
   \Rightarrow \dfrac{{3\left( { - 9} \right) - 3\left( { - 3} \right) + 3\left( 9 \right)}}{{\sqrt {81 + 9 + 81} }} \\
   \Rightarrow \dfrac{{ - 27 + 9 + 27}}{{\sqrt {171} }} \\
   \Rightarrow \dfrac{9}{{3\sqrt {19} }} \\
 $
Rationalize the denominator, by multiplying $\sqrt {19} $to both numerator and denominator.
$\because d = \dfrac{{3\sqrt {19} }}{{19}}{\text{ }}units$
Final answer: The shortest distance between the given lines is $\dfrac{{3\sqrt {19} }}{{19}}{\text{ }}units$ .

Note: We know the Cartesian equation of the line: $\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c}$
Hence, the cartesian equation of line 1:
$ \Rightarrow \dfrac{{x - 1}}{1} = \dfrac{{y - 2}}{{ - 3}} = \dfrac{{z - 3}}{2}$
Hence, the cartesian equation of line 2:
$ \Rightarrow \dfrac{{x - 4}}{2} = \dfrac{{y - 5}}{3} = \dfrac{{z - 6}}{1}$
We have given equations of line in vector form, hence, we could have used the vector formula of distance. But students might make mistakes while using the formula, so it is easier to convert the vector equation of a line into the cartesian equation of a line and use the Cartesian formula of distance.
Two skew lines:
$
  \vec r = {{\vec a}_1} + \lambda {{\vec b}_1} \\
  \vec r = {{\vec a}_2} + \mu {{\vec b}_2} \\
 $
The shortest distance between two skew lines, $d = \dfrac{{\left( {{{\vec b}_1} \times {{\vec b}_2}} \right) \cdot \left( {{{\vec a}_2} - {{\vec a}_1}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}$
For two parallel lines:
$
  \vec r = {{\vec a}_1} + \lambda \vec b \\
  \vec r = {{\vec a}_2} + \mu \vec b \\
 $
The distance between two parallel lines, $d = \dfrac{{\left( {\vec b} \right) \cdot \left( {{{\vec a}_2} - {{\vec a}_1}} \right)}}{{\left| {\vec b} \right|}}$