3.6 Distance between parallel lines

2.1 Vectorial form of a line passing through a given point and parallel to a given vector
2.2 Vectorial form of a line passing through two given points
2.3 Cartesian form of a line passing through a given point and parallel to a given vector
2.4 Cartesian form of a line passing through two given points

3.1 Condition for perpendicularity
3.2 Condition for parallelism
3.3 Co-planarity of two lines
3.4 Shortest distance between two lines
3.5 Distance between two skew lines
3.6 Distance between parallel lines
3.7 Perpendicular distance of a point from a line
3.8 Image of a point in a straight line

4.1 Equation of plane in normal form
4.2 Equation of a plane perpendicular to a given vector and passing through a given point
4.3 Equation of plane passing through a given point and parallel to the two given vectors
4.4 Equation of plane passing through three non-collinear points
4.5 Intercept form of plane
4.6 Equation of plane passing through the intersection of two given planes

5.1 Angle between two planes
5.2 Angle between a line and a plane
5.3 Distance of a point from a plane
5.4 Distance between two parallel planes

Let us assume the equation of two parallel lines be $$\begin{equation} \begin{aligned} {l_1} \equiv \overrightarrow r = \overrightarrow {{a_1}} + \lambda \overrightarrow b \\ {l_2} \equiv \overrightarrow r = \overrightarrow {{a_2}} + \mu \overrightarrow b \\\end{aligned} \end{equation} $$

where $\overrightarrow {{a_1}} $ represents the position vector of a point $A$ on line ${l_1}$ and $\overrightarrow {{a_2}} $ represents the position vector of a point $B$ on line ${l_2}$ as shown in figure.

As both the lines are parallel which means they are coplanar also and if the foot of perpendicular from $B$ on the line ${l_1}$ is $P$, then the distance between parallel lines is $BP$.

Let us assume the angle between line $AB$ and line ${l_1}$ be $\theta $ which is also the angle between $\overrightarrow {AB} $ and $\overrightarrow b $. Therefore,

$$\overrightarrow b \times \overrightarrow {AB} = \left( {\left| {\overrightarrow b } \right|\left| {\overrightarrow {AB} } \right|\sin \theta } \right)\widehat n$$ where $\widehat n$ represents the unit vector perpendicular to the plane of both the lines ${l_1}$ and ${l_2}$. But $$\overrightarrow {AB} = \overrightarrow {{a_2}} - \overrightarrow {{a_1}} $$ Therefore, we get

$$\begin{equation} \begin{aligned} \overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) = \left( {\left| {\overrightarrow b } \right|PB} \right)\widehat n{\text{ }}\left( {\because \left| {\overrightarrow {AB} } \right|\sin \theta } \right) \\ \therefore \left| {\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)} \right| = \left| {\overrightarrow b } \right|PB.1{\text{ }}\left( {\because \left| {\widehat n{\text{ }}} \right| = 1} \right) \\\end{aligned} \end{equation} $$ Hence, the distance between two parallel lines is $$d = \left| {\overrightarrow {PB} } \right| = \left| {\frac{{\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)}}{{\left| {\overrightarrow b } \right|}}} \right|$$

Question 7. Find the shortest distance between the lines $$\frac{{x - 1}}{2} = \frac{{y - 2}}{3} = \frac{{z - 3}}{4}$$ and $$\frac{{x - 2}}{3} = \frac{{y - 4}}{4} = \frac{{z - 5}}{5}$$

Solution: On comparing the given equations with the cartesian form of line, we get $$\begin{equation} \begin{aligned} {x_1} = 1,{y_1} = 2,{z_1} = 3 \\ {a_1} = 2,{b_1} = 3,{c_1} = 4 \\ {x_2} = 2,{y_2} = 4,{z_2} = 5 \\ {a_2} = 3,{b_2} = 4,{c_2} = 5 \\\end{aligned} \end{equation} $$

As from the given equations, we can say that the lines are skew lines, so apply the formulae to find the distance between skew lines if the equation is given in cartesian form i.e.,

\[\left| {\frac{{\begin{array}{c} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}}}{{\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}} }}} \right|\]

Now, putting the values in the formulae we get,

\[\begin{gathered} \Rightarrow \frac{{\left| {\begin{array}{c} {2 - 1}&{4 - 2}&{5 - 3} \\ 2&3&4 \\ 3&4&5 \end{array}} \right|}}{{\sqrt {{{\left( {8 - 9} \right)}^2} + {{\left( {15 - 16} \right)}^2} + {{\left( {10 - 12} \right)}^2}} }} \hspace{1em} \\ \Rightarrow \frac{{\left| {\begin{array}{c} 1&2&2 \\ 2&3&4 \\ 3&4&5 \end{array}} \right|}}{{\sqrt 6 }} \hspace{1em} \\ \end{gathered} \]

$$\begin{equation} \begin{aligned} \Rightarrow \frac{{1(15 - 16) - 2(10 - 12) + 2(8 - 9)}}{{\sqrt 6 }} = \frac{1}{{\sqrt 6 }} \\ \\\end{aligned} \end{equation} $$

Question 8. Find the distance between the lines ${l_1}$ and ${l_2}$ given by $${l_1} \equiv \overrightarrow r = \widehat i + 2\widehat j - 4\widehat k + \lambda \left( {2\widehat i + 3\widehat j + 6\widehat k} \right)$$ and $${l_2} \equiv \overrightarrow r = 3\widehat i + 3\widehat j - 5\widehat k + \mu \left( {2\widehat i + 3\widehat j + 6\widehat k} \right)$$

Solution: On comparing the given equations of line with the vector form i.e., $$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b $$ We get $$\overrightarrow {{a_1}} = \widehat i + 2\widehat j - 4\widehat k{\text{ }}\overrightarrow {{a_2}} = 3\widehat i + 3\widehat j - 5\widehat k{\text{ }}\overrightarrow b = 2\widehat i + 3\widehat j + 6\widehat k$$

Now, we can say that the lines are parallel as the parallel vector $\overrightarrow b $ is similar for both the lines. Now, we apply the formulae to find the distance between two parallel lines i.e., $$d = \left| {\frac{{\vec b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)}}{{\left| {\vec b} \right|}}} \right|$$

Put the values in the equation we get,

\[d = \left| {\frac{{\left| {\begin{array}{c} {\widehat i}&{\widehat j}&{\widehat k} \\ 2&3&6 \\ 2&1&{ - 1} \end{array}} \right|}}{{\sqrt {4 + 9 + 36} }}} \right|\]$$d = \frac{{\left| { - 9\widehat i + 14\widehat j - 4\widehat k} \right|}}{{\sqrt {49} }} = \frac{{\sqrt {293} }}{7}$$