Question

Find the shortest distance between the lines $\overrightarrow r = \widehat i + 2\widehat j + 3\widehat k + \lambda \left( {2\widehat i + 3\widehat j + 4\widehat k} \right)$ and $\overrightarrow r = 2\widehat i + 4\widehat j + 5\widehat k + \mu \left( {4\widehat i + 6\widehat j + 8\widehat k} \right)$.

Answer 1

Hint- Here, we will proceed by proving that the given lines are parallel to each other and then we will apply the formula for the shortest distance between two parallel lines which is given by $d = \dfrac{{\left| {\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)} \right|}}{{\left| {\overrightarrow b } \right|}}$.

Complete step-by-step solution -
Given equations of lines in the vector form are $\overrightarrow r = \widehat i + 2\widehat j + 3\widehat k + \lambda \left( {2\widehat i + 3\widehat j + 4\widehat k} \right){\text{ }} \to {\text{(1)}}$ and $\overrightarrow r = 2\widehat i + 4\widehat j + 5\widehat k + \mu \left( {4\widehat i + 6\widehat j + 8\widehat k} \right) \\\ \Rightarrow \overrightarrow r = 2\widehat i + 4\widehat j + 5\widehat k + 2\mu \left( {2\widehat i + 3\widehat j + 4\widehat k} \right) \\\ \Rightarrow \overrightarrow r = 2\widehat i + 4\widehat j + 5\widehat k + \alpha \left( {2\widehat i + 3\widehat j + 4\widehat k} \right){\text{ }} \to {\text{(2)}} \\\$
where $\alpha = 2\mu$
As we know that the general equations of any two lines in the vector form are given by
$\overrightarrow r = \overrightarrow {{a_1}} + \lambda \left( {\overrightarrow {{b_1}} } \right){\text{ }} \to {\text{(3)}}$ and $\overrightarrow r = \overrightarrow {{a_2}} + \alpha \left( {\overrightarrow {{b_2}} } \right){\text{ }} \to {\text{(4)}}$ where $\overrightarrow {{a_1}} ,\overrightarrow {{a_2}}$ represents the position vectors of the points through which these lines are passing respectively and $\overrightarrow {{b_1}} ,\overrightarrow {{b_2}}$ represents the vectors which are parallel to these lines respectively.
For these two lines to be parallel to each other, $\overrightarrow {{b_1}} = \overrightarrow {{b_2}}$
Clearly from equations (1) and (2), we can see that
$\overrightarrow {{b_1}} = \overrightarrow {{b_2}} = 2\widehat i + 3\widehat j + 4\widehat k$
So, both the given lines are parallel to each other with $\overrightarrow b = 2\widehat i + 3\widehat j + 4\widehat k{\text{ }} \to {\text{(5)}}$
By comparing equations (1) and (3), we get
$\overrightarrow {{a_1}} = \widehat i + 2\widehat j + 3\widehat k{\text{ }} \to {\text{(6)}}$
By comparing equations (2) and (4), we get
$\overrightarrow {{a_2}} = 2\widehat i + 4\widehat j + 5\widehat k{\text{ }} \to {\text{(7)}}$
Also we know that the shortest distance between any two parallel lines whose equations in vector forms are $\overrightarrow r = \overrightarrow {{a_1}} + \lambda \left( {\overrightarrow b } \right)$ and $\overrightarrow r = \overrightarrow {{a_2}} + \alpha \left( {\overrightarrow b } \right)$ is given by
$d = \dfrac{{\left| {\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)} \right|}}{{\left| {\overrightarrow b } \right|}}$ where the modulus sign represents the magnitude of the inside vector
By substituting equations (5), (6) and (7) in the above equation, we get
Shortest distance between the given parallel lines, $\Rightarrow d = \dfrac{{\left| {\left( {2\widehat i + 3\widehat j + 4\widehat k} \right) \times \left( {2\widehat i + 4\widehat j + 5\widehat k - \left( {\widehat i + 2\widehat j + 3\widehat k} \right)} \right)} \right|}}{{\left| {2\widehat i + 3\widehat j + 4\widehat k} \right|}} \\\ \Rightarrow d = \dfrac{{\left| {\left( {2\widehat i + 3\widehat j + 4\widehat k} \right) \times \left( {2\widehat i + 4\widehat j + 5\widehat k - \widehat i - 2\widehat j - 3\widehat k} \right)} \right|}}{{\left| {2\widehat i + 3\widehat j + 4\widehat k} \right|}} \\\ \Rightarrow d = \dfrac{{\left| {\left( {2\widehat i + 3\widehat j + 4\widehat k} \right) \times \left( {\widehat i + 2\widehat j + 2\widehat k} \right)} \right|}}{{\left| {2\widehat i + 3\widehat j + 4\widehat k} \right|}}{\text{ }} \to {\text{(8)}} \\\$
Since, cross product of any two vectors $\overrightarrow {{a_1}} = {x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k$ and $\overrightarrow {{a_2}} = {x_2}\widehat i + {y_2}\widehat j + {z_2}\widehat k$ is given by

\overrightarrow {{a_1}} \times \overrightarrow {{a_2}} = \left| {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ {{x_1}}&{{y_1}}&{{z_1}} \\\ {{x_2}}&{{y_2}}&{{z_2}} \end{array}} \right| \\\ \Rightarrow \overrightarrow {{a_1}} \times \overrightarrow {{a_2}} = \left( {{y_1}{z_2} - {z_1}{y_2}} \right)\widehat i - \left( {{x_1}{z_2} - {z_1}{x_2}} \right)\widehat j + \left( {{x_1}{y_2} - {y_1}{x_2}} \right)\widehat k{\text{ }} \to {\text{(9)}} \\\

Also, the magnitude of any vector $\overrightarrow {{a_1}} = {x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k$ is given by $\begin{gathered} \left| {\overrightarrow {{a_1}} } \right| = \left| {{x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k} \right| \\\ \Rightarrow \left| {\overrightarrow {{a_1}} } \right| = \sqrt {{x_1}^2 + {y_1}^2 + {z_1}^2} {\text{ }} \to {\text{(10)}} \\\ \end{gathered}$
By using the formula given by equation (9) in equation (8), we get
Shortest distance between the given parallel lines,

$$\Rightarrow d = \dfrac{{\left| {\left( {3 \times 2 - 4 \times 2} \right)\widehat i - \left( {2 \times 2 - 4 \times 1} \right)\widehat j + \left( {2 \times 2 - 3 \times 1} \right)\widehat k} \right|}}{{\left| {2\widehat i + 3\widehat j + 4\widehat k} \right|}} \\\ \Rightarrow d = \dfrac{{\left| {\left( {6 - 8} \right)\widehat i - \left( {4 - 4} \right)\widehat j + \left( {4 - 3} \right)\widehat k} \right|}}{{\left| {2\widehat i + 3\widehat j + 4\widehat k} \right|}} \\\ \Rightarrow d = \dfrac{{\left| {-2\widehat i - 0\widehat j + \widehat k} \right|}}{{\left| {2\widehat i + 3\widehat j + 4\widehat k} \right|}} \\\ \Rightarrow d = \dfrac{{\left| {-2\widehat i + \widehat k} \right|}}{{\left| {2\widehat i + 3\widehat j + 4\widehat k} \right|}} \\\$$

By using the formula given by equation (10) in the above equation, we get
Shortest distance between the given parallel lines,

$$\Rightarrow d = \dfrac{{\sqrt {{2^2} + {0^2} + {1^2}} }}{{\sqrt {{2^2} + {3^2} + {4^2}} }} \\\ \Rightarrow d = \dfrac{{\sqrt {4 + 0 + 1} }}{{\sqrt {4 + 9 + 16} }} \\\ \Rightarrow d = \dfrac{{\sqrt 5 }}{{\sqrt {29} }} \\\$$

Therefore, the required shortest distance between the given lines is $\dfrac{{\sqrt 5 }}{{\sqrt {29} }}$.

Note- In any straight line whose equation in the vector form is $\overrightarrow r = \overrightarrow {{a_1}} + \lambda \left( {\overrightarrow {{b_1}} } \right)$, $\overrightarrow {{b_1}}$ have direction ratios of the vector which is parallel to $\overrightarrow r = \overrightarrow {{a_1}} + \lambda \left( {\overrightarrow {{b_1}} } \right)$ because any two parallel lines will have same direction ratios. Direction cosine of any vector can be obtained by dividing that vector by the magnitude of that vector.