In the previous section, we saw the shortest distance between two skew lines. In this section, we will see the distance between two parallel lines.
• If two lines in space intersect at a point, we can say that the shortest distance between the two lines is zero.
• If two lines in space are parallel, the shortest distance can be obtained in 3 steps:
(i) Mark any convenient point on one line.
(ii) Drop a perpendicular from that point onto the other line.
(iii) Length of that perpendicular is the shortest distance.
Now we will see the method to find the distance between parallel lines. The method can be explained in 10 steps:
1. In the fig.27.12 below, $\small{l_1~\text{and}~l_2}$ form a pair of parallel lines.
♦ Vector equation of $\small{l_1~\text{is:}~\vec{r}=\vec{u_1}+\lambda \vec{v}}$
♦ Vector equation of $\small{l_2~\text{is:}~\vec{r}=\vec{u_2}+\mu \vec{v}}$
• Note that both lines are parallel to $\small{\vec{v}}$. So the lines themselves are parallel to each other
 |
| Fig.27.12 |
2. Next, we want a point on each line.
• Mark the point S on $\small{l_1}$ such that, the position vector of S is $\small{\vec{u_1}}$
• Mark the point T on $\small{l_2}$ such that, the position vector of T is $\small{\vec{u_2}}$
3. Based on the above two position vectors, we can write:
$\small{\vec{ST}=\vec{u_2} - \vec{u_1}}$
4. Next we concentrate on PT. It is obtained by dropping a perpendicular from T onto $\small{l_1}$. Point P is the foot of the perpendicular. PT is the shortest line between $\small{l_1~\text{and}~l_2}$
So we can write:
♦ PT is perpendicular to $\small{l_1}$
♦ PT is perpendicular to $\small{l_2}$ also
5. We want the distance TP.
• From the triangle STP,we get: $\small{TP = ST \sin \theta}$.
• If we use vectors, the above calculation can be easily done.
6. The lines $\small{l_1,~l_2,~ST~\text{and}~TP}$ lie in the same plane.
The cross product $\small{\left(\vec{v}\times\vec{ST} \right)}$ will be a vector which is perpendicular to that plane.
7. We have:
$\small{\vec{v}\times\vec{ST} = \left(\left|\vec{v} \right|\,\left|\vec{ST} \right|\,\sin\theta \right)\hat{n}}$
• Where $\small{\hat{n}}$ is the unit vector perpendicular to the plane.
8. Substituting for $\small{\vec{ST}}$ from (3), we get:
$\small{\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) = \left(\left|\vec{v} \right|\,\left|\vec{ST} \right|\,\sin\theta \right)\hat{n}}$
• $\small{\left|\vec{ST} \right|}$ can be written as the length ST. So we get:
$\small{\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) =
\left(\left|\vec{v} \right|\,ST\,\sin\theta
\right)\hat{n}}$
• $\small{ST\,\sin\theta = PT}$. So we get:
$\small{\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) =
\left(\left|\vec{v} \right|PT
\right)\hat{n}}$
9. The above result is the equality of two vectors. Their magnitudes will be the same. So we can write:
$\small{\left|\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) \right| = \left|\vec{v} \right|\,PT}$
10. Thus we get:
$\small{PT = \frac{\left|\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) \right|}{\left|\vec{v} \right|}}$
Let us see a solved example
Solved example 27.29
Find the distance between the lines $\small{l_1~\text{and}~l_2}$ whose vector equations are
$\small{\vec{r} = \hat{i}+2\hat{j}-4\hat{k}+\lambda\left(2\hat{i}+3\hat{j}+6\hat{k} \right)}$
and $\small{\vec{r} = 3\hat{i}+3\hat{j}-5\hat{k}+\mu\left(2\hat{i}+3\hat{j}+6\hat{k} \right)}$
Solution:
The given two lines are parallel because, both the lines are parallel to the vector $\small{2\hat{i}+3\hat{j}+6\hat{k} }$
1. We have:
Shortest distance
= Length of the line $\small{PT}$
= $\small{\frac{\left|\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) \right|}{\left|\vec{v} \right|}}$
2. Based on the given equations of the lines, we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{u_1}} & {~=~} &{\hat{i}+2\hat{j}-4\hat{k}}
\\ {~\color{magenta} 2 } &{{}} &{\vec{u_2}} & {~=~} &{3\hat{i}+3\hat{j}-5\hat{k}}
\\ {~\color{magenta} 3 } &{{}} &{\vec{v}} & {~=~} &{2\hat{i}+3\hat{j}+6\hat{k}}
\\ \end{array}}$
3. $\small{\vec{u_2} - \vec{u_1} = 2\hat{i}+\hat{j}-\hat{k}}$
4. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{PT} & {~=~} &{\frac{\left|\vec{v}\times\left(\vec{u_2} - \vec{u_1} \right) \right|}{\left|\vec{v} \right|}}
\\ {~\color{magenta} 2 } &{{}} &{} & {~=~} &{\frac{\left|\left(2\hat{i}+3\hat{j}+6\hat{k} \right)\times\left(2\hat{i}+\hat{j}-\hat{k} \right) \right|}{\left|2\hat{i}+3\hat{j}+6\hat{k} \right|}}
\\ {~\color{magenta} 3 } &{{}} &{} & {~=~} &{\frac{\left|-9\hat{i}+14\hat{j}-4\hat{k} \right|}{\left|2\hat{i}+3\hat{j}+6\hat{k} \right|}}
\\ {~\color{magenta} 4 } &{{}} &{} & {~=~} &{\frac{\sqrt{293}}{\sqrt{49}}~=~\frac{\sqrt{293}}{7}}
\\ \end{array}}$
• So the shortest distance = $\small{\frac{\sqrt{293}}{7}}$ units
• Therefore, the shortest distance = $\small{\frac{8}{\sqrt{29}}}$ units
The link below gives the exercise questions. All those exercise questions are already answered in our discussions.
Exercise 27.2
In the next section, we will see Plane.
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