Question

Find the distance between the lines \[{l_1}\] and \[{l_2}\] given by
\[\overrightarrow r = \widehat i + 2\widehat j - 4\widehat k + \lambda \left( {2\widehat i + 3\widehat j + 6\widehat k} \right)\] and \[\overrightarrow r = 3\widehat i + 3\widehat j - 5\widehat k + \mu \left( {2\widehat i + 3\widehat j + 6\widehat k} \right)\]

Answer 1

Hint:
Here, we will compare the equations of the given lines with the general vector equation of a line. We will then substitute the values in the vector formula of shortest distance between two lines to find the required distance.

Formula Used:
Shortest distance between two lines \[ = \left| {\dfrac{{\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)}}{{\overrightarrow b }}} \right|\]

Complete step by step solution:
Given equations of the lines are: \[\overrightarrow r = \widehat i + 2\widehat j - 4\widehat k + \lambda \left( {2\widehat i + 3\widehat j + 6\widehat k} \right)\] and \[\overrightarrow r = 3\widehat i + 3\widehat j - 5\widehat k + \mu \left( {2\widehat i + 3\widehat j + 6\widehat k} \right)\].
Comparing these with the general vector equation of a line \[\overrightarrow r = \overrightarrow {{a_1}} + \lambda \overrightarrow b \] and \[\overrightarrow r = \overrightarrow {{a_2}} + \mu \overrightarrow b \], we get the position vectors of the lines as:
\[\overrightarrow {{a_1}} = \widehat i + 2\widehat j - 4\widehat k\]
\[\overrightarrow {{a_2}} = 3\widehat i + 3\widehat j - 5\widehat k\]
Also, in these lines,
\[\overrightarrow b = 2\widehat i + 3\widehat j + 6\widehat k\]
Hence, in order to find the distance between the given lines, we use the formula \[\left| {\dfrac{{\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)}}{{\overrightarrow b }}} \right|\].
So, first we will calculate different terms of the formula.
Now,
\[\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) = 3\widehat i + 3\widehat j - 5\widehat k - \widehat i - 2\widehat j + 4\widehat k\]
Adding and subtracting the like terms, we get
\[ \Rightarrow \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) = 2\widehat i + \widehat j - \widehat k\]
Now, we will find \[\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \times \overrightarrow b \]. So,
\[\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \times \overrightarrow b = \left( {2\widehat i + \widehat j - \widehat k} \right) \times \left( {2\widehat i + 3\widehat j + 6\widehat k} \right)\]
We will find the cross product of the above vector using determinant. Therefore,
\[ \Rightarrow \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \times \overrightarrow b = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\2&1&{ - 1}\\2&3&6\end{array}} \right|\]
Solving this further, we get,
\[ \Rightarrow \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \times \overrightarrow b = \widehat i\left( {6 + 3} \right) - \widehat j\left( {12 + 2} \right) + \widehat k\left( {6 - 2} \right)\]
Adding and subtracting the like terms, we get
\[ \Rightarrow \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \times \overrightarrow b = 9\widehat i - 14\widehat j + 4\widehat k\]
Taking modulus on both sides, we get
\[ \Rightarrow \left| {\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \times \overrightarrow b } \right| = \left| {9\widehat i - 14\widehat j + 4\widehat k} \right|\]
Simplifying further, we get
\[ \Rightarrow \left| {\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \times \overrightarrow b } \right| = \sqrt {{{\left( 9 \right)}^2} + {{\left( { - 14} \right)}^2} + {{\left( 4 \right)}^2}} \]
Applying the exponent on the terms and adding them, we get
\[ \Rightarrow \left| {\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \times \overrightarrow b } \right| = \sqrt {81 + 196 + 16} = \sqrt {293} \]
Now we will find \[\left| {\overrightarrow b } \right|\].
\[\left| {\overrightarrow b } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 6 \right)}^2}} \]
Applying the exponent on the terms, we get
\[ \Rightarrow \left| {\overrightarrow b } \right| = \sqrt {4 + 9 + 36} \]
Adding the terms, we get
\[ \Rightarrow \left| {\overrightarrow b } \right| = \sqrt {49} = 7\]
Now substituting the obtained values in the formula \[\left| {\dfrac{{\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)}}{{\overrightarrow b }}} \right|\], we get
The distance between the lines \[{l_1}\] and \[{l_2}\]\[ = \left| {\dfrac{{\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)}}{{\overrightarrow b }}} \right| = \dfrac{{\sqrt {293} }}{7}\] units

Therefore, the distance between the lines \[{l_1}\] and \[{l_2}\] is \[\dfrac{{\sqrt {293} }}{7}\] units.

Note:
A scalar is a quantity that has a magnitude whereas; a vector is a mathematical quantity that has both magnitude and direction. A line of given length and pointing along a given direction, such as an arrow, is a typical representation of a vector. Now, position vector is also known as location vector, it is a straight line having one end fixed and the other end attached to a moving point, it is used to describe the position of a certain point, which turns out to be its respective coordinates.