Question

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

Answer 1

Hint: Make use of the formula of the perpendicular distance between the two lines and try to find the distance

Formula used:
$\vec r = {\vec a_1} + \mu \vec b$and $\vec r = {\vec a_2} + \mu {\vec b_2}$ is given by $d = \left| {\dfrac{{\left( {{b_1} \times {b_2}} \right).\left( {{a_2} - {a_1}} \right)}}{{\left| {{b_1} \times {b_2}} \right|}}} \right|$ .


Complete step-by-step answer:
If two lines ${l_1}$ and ${l_2}$ are two skew lines, then there is only one line which is perpendicular to the two lines, which is the shortest distance between the two skew lines.

Equation of line first ${l_1}$ in vector form,
$\vec r = \left( {\hat i + 2\hat j + 3\hat k} \right) + \beta (\hat i - 3\hat j + 2\hat k)$

Equation of line second ${l_2}$ in vector form,
$\vec r = \left( {4\hat i + 5\hat j + 6\hat k} \right) + \mu (2\hat i + 3\hat j + \hat k)$

The shortest distance between the two lines $\vec r = {\vec a_1} + \mu \vec b$and $\vec r = {\vec a_2} + \mu {\vec b_2}$
$d = \left| {\dfrac{{\left( {{{\vec b}_1} \times {{\vec b}_2}} \right).\left( {{{\vec a}_2} - {{\vec a}_1}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right|$

Calculation of $\left( {{{\vec a}_2} - {{\vec a}_1}} \right)$ requires subtracting the terms which are in same direction, subtraction of two vectors give a vector quantity.
${a_1} = (\hat i + 2\hat j + 3\hat k)$ and ${a_2} = (4\hat i + 5\hat j + 6\hat k)$
\[
  \left( {{a_2} - {a_1}} \right) = \left( {4\hat i + 5\hat j + 6\hat k} \right) - \left( {\hat i - 3\hat j + 2\hat k} \right) \\
  \left( {{a_2} - {a_1}} \right) = \left( {4 - 1} \right)\hat i + \left( {5 - 3} \right)\hat j + \left( {6 - 2} \right)\hat k \\
  \left( {{a_2} - {a_1}} \right) = 3\hat i + 2\hat j + 4\hat k \\
 \]

Calculation of ${\vec b_1} \times {\vec b_2}$. It is a cross product of 2 vector quantities and it gives a vector quantity which is perpendicular to both the vectors.
Here, ${\vec b_1} = \left( {\hat i - 3\hat j + 2\hat k} \right)$ and ${\vec b_2} = \left( {2\hat i + 3\hat j + \hat k} \right)$
\[{b_1} \times {b_2} = \left( {\hat i + 2\hat j + 3\hat k} \right) \times \left( {2\hat i + 3\hat j + \hat k} \right)\]

The cross product of two vectors is calculated by the determinant.
$
  {b_1} \times {b_2} = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  1&2&3 \\
  2&3&1
\end{array}} \right| \\
  {b_1} \times {b_2} = \hat i\left[ {1\left( {2 \times 1 - 3 \times 3} \right) - \hat j\left( {1 \times 1 - 3 \times 2} \right) + \hat k\left( {1 \times 3 - 2 \times 2} \right)} \right] \\
  {b_1} \times {b_2} = \hat i\left( {2 - 9} \right) - \hat j\left( {1 - 6} \right) + \hat k\left( {3 - 4} \right) \\
  {b_1} \times {b_2} = - 7\hat i + 5\hat j - \hat k \\
 $

Calculate the modulus of \[\left| {{b_1} \times {b_2}} \right|\] ,
Using the formula for $\vec a = x\hat i + y\hat j + z\hat k$ as $\left| {\vec a} \right| = \sqrt {{x^2} + {y^2} + {z^2}} $
$
  \left| {{b_1} \times {b_2}} \right| = \sqrt {{{\left( { - 7} \right)}^2} + {{\left( 5 \right)}^2} + {{\left( { - 1} \right)}^2}} \\
  \left| {{b_1} \times {b_2}} \right| = \sqrt {49 + 25 + 1} \\
  \left| {{b_1} \times {b_2}} \right| = \sqrt {75} \\
    \\
 $

Now, calculate the shortest distance between the two lines , by substituting the calculated values in equation(1),
$d = \left| {\dfrac{{\left( { - 7\hat i + 5\hat j - \hat k} \right).\left( {3\hat i + 2\hat j + 4\hat k} \right)}}{{\sqrt {75} }}} \right|$

Now, use the dot product of the two vectors in the numerator of equation (2). The dot product of the vector is the scalar product as it gives only the magnitude.
\[
  d = \left| {\dfrac{{\left( { - 7\hat i + 5\hat j - \hat k} \right).\left( {3\hat i + 2\hat j + 4\hat k} \right)}}{{\sqrt {75} }}} \right| \\
  d = \left| {\dfrac{{\left( { - 7} \right)\left( 3 \right) + \left( 5 \right)\left( 2 \right) - \left( 1 \right)\left( 4 \right)}}{{5\sqrt 3 }}} \right| \\
  d = \left| {\dfrac{{ - 21 + 10 - 4}}{{5\sqrt 3 }}} \right| \\
  d = \left| {\dfrac{{ - 15}}{{3\sqrt 3 }}} \right| \\
  d = \left| {\dfrac{{ - 5}}{{\sqrt 3 }}} \right| \\
  d = \dfrac{5}{{\sqrt 3 }} \\
 \]
Hence, the shortest distance between the two lines is $d = \dfrac{5}{{\sqrt 3 }}$

Note: The important points which are supposed to be remembered are:
The vector multiplication can be done in two ways,
1)Dot product: It is the product which gives a scalar quantity.
2)Cross product: It gives the vector quantity.
Skew lines are the lines which are neither parallel nor intersect each other.