In the previous section, we saw the conditions for two lines to be parallel or perpendicular. In this section, we will see the shortest distance between two 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 skew lines. It can be explained in 3 steps:
1. Consider two lines which satisfy the following two conditions:
(i) The two lines do not intersect at any point
(ii) The two lines are not parallel.
2. Two lines which satisfy both the conditions are called skew lines.
3. Skew lines will be lying in different planes. In other words, skew lines are non coplanar.
Let us try to visualize a pair of skew lines. It can be done in 2 steps:
1. Fig.27.10 below shows a room of size:
♦ 2 units width along the x-axis
♦ 5 units length along the y-axis
♦ 3 units height along the z-axis.
 |
| Fig.27.10 |
2. Two lines are shown in the fig.
♦ The yellow line is aligned with the diagonal BD of the wall ABED
♦ The green line is aligned with the diagonal EG of the ceiling EFGD
• Those two lines do not intersect at any point. Also, they are not parallel. So the yellow and green lines form a pair of skew lines
Now we will see the method to find the shortest distance between skew lines. This method make use of the fact that, the line of shortest distance between two skew lines, will be perpendicular to both the lines. The method can be explained in 8 steps:
1. In the fig.27.11 below, $\small{l_1~\text{and}~l_2}$ form a pair of skew lines.
♦ Vector equation of $\small{l_1~\text{is:}~\vec{r}=\vec{u_1}+\lambda \vec{v_1}}$
♦ Vector equation of $\small{l_2~\text{is:}~\vec{r}=\vec{u_2}+\mu \vec{v_2}}$
 |
| Fig.27.11 |
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 PQ. It is the shortest line between $\small{l_1~\text{and}~l_2}$
So we can write:
♦ PQ is perpendicular to $\small{l_1}$
♦ PQ is perpendicular to $\small{l_2}$ also
5. Imagine that the vector $\small{\vec{PQ}}$ is present between the points P and Q.
• This vector will be perpendicular to both $\small{l_1~\text{and}~l_2}$
6. Next we want a vector with the same direction as $\small{\vec{PQ}}$. It can be obtained in 4 steps:
(i) Based on the vector equations of $\small{l_1~\text{and}~l_2}$, we can write:
♦ $\small{l_1}$ is parallel to $\small{\vec{v_1}}$
♦ $\small{l_2}$ is parallel to $\small{\vec{v_2}}$
(ii) So $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$ will be a vector perpendicular to both $\small{\vec{v_1}~\text{and}~\vec{v_2}}$
(iii) Consequently, $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$ will be a vector perpendicular to both $\small{l_1~\text{and}~l_2}$
(iv) That is., $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$ will have the same direction as $\small{\vec{PQ}}$
7. Imagine that, $\small{\vec{ST}}$ is shifted in such a way that, the initial point S of $\small{\vec{ST}}$ coincides with the initial point P of $\small{\vec{PQ}}$
• In this situation, we can write:
The projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
= Length of the line $\small{PQ}$
8. We can easily calculate the projection. See section 26.9.
• We can write:
Shortest distance
= Length of the line $\small{PQ}$
= $\small{\vec{ST}.\hat{PQ}}$
♦ We can obtain $\small{\vec{ST}}$ from (3)
♦ $\small{\hat{PQ}}$ is the unit vector in the direction of $\small{\vec{PQ}}$. We can obtain it from 6(iv)
Let us see some solved examples:
Solved example 27.24
Find the shortest distance between the lines $\small{l_1~\text{and}~l_2}$ whose vector equations are
$\small{\vec{r} = \hat{i}+\hat{j}+\lambda\left(2\hat{i}-\hat{j}+\hat{k} \right)}$
and $\small{\vec{r} = 2\hat{i}+\hat{j}-\hat{k}+\mu\left(3\hat{i}-5\hat{j}+2\hat{k} \right)}$
Solution:
1. We have:
Shortest distance
= Length of the line $\small{PQ}$
= Projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
• The projection is given by: $\small{\vec{ST}.\hat{PQ}}$
2. Based on the given equations of the lines, we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{u_1}} & {~=~} &{\hat{i}+\hat{j}}
\\ {~\color{magenta} 2 } &{{}} &{\vec{u_2}} & {~=~} &{2\hat{i}+\hat{j}-\hat{k}}
\\ {~\color{magenta} 3 } &{{}} &{\vec{v_1}} & {~=~} &{2\hat{i}-\hat{j}+\hat{k}}
\\ {~\color{magenta} 4 } &{{}} &{\vec{v_2}} & {~=~} &{3\hat{i}-5\hat{j}+2\hat{k}}
\\ \end{array}}$
3. $\small{\vec{ST} = \vec{u_2} - \vec{u_1} = \hat{i}-\hat{k}}$
4. $\small{\hat{PQ} = \frac{\vec{v_1}\times\vec{v_2}}{\left|\vec{v_1}\times\vec{v_2} \right|}}$
= $\small{\frac{3\hat{i}-\hat{j}-7\hat{k}}{\sqrt{3^2 + (-1)^2 + 7^2}}}$
= $\small{\frac{3\hat{i}-\hat{j}+7\hat{k}}{\sqrt{59}}}$
• The reader may write all the steps related to the cross product
5. Substituting (3) and (4) in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\text{Projection}} & {~=~} &{\vec{ST}.\hat{PQ}}
\\ {~\color{magenta} 2 } &{{}} &{} & {~=~} &{\left(\hat{i}-\hat{k} \right).\left(\frac{3\hat{i}-\hat{j}-7\hat{k}}{\sqrt{59}} \right)}
\\ {~\color{magenta} 3 } &{{}} &{} & {~=~} &{\frac{(1)(3)+(0)(-1)+(-1)(-7)}{\sqrt{59}}}
\\ {~\color{magenta} 4 } &{{}} &{} & {~=~} &{\frac{10}{\sqrt{59}}}
\\ \end{array}}$
• So the shortest distance = $\small{\frac{10}{\sqrt{59}}}$ units
Solved example 27.25
Find the shortest distance between the lines $\small{l_1~\text{and}~l_2}$ whose vector equations are
$\small{\vec{r} = \hat{i}+2\hat{j}+\hat{k}+\lambda\left(\hat{i}-\hat{j}+\hat{k} \right)}$
and $\small{\vec{r} = 2\hat{i}-\hat{j}-\hat{k}+\mu\left(2\hat{i}+\hat{j}+2\hat{k} \right)}$
Solution:
1. We have:
Shortest distance
= Length of the line $\small{PQ}$
= Projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
• The projection is given by: $\small{\vec{ST}.\hat{PQ}}$
2. Based on the given equations of the lines, we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{u_1}} & {~=~} &{\hat{i}+2\hat{j}+\hat{k}}
\\ {~\color{magenta} 2 } &{{}} &{\vec{u_2}} & {~=~} &{2\hat{i}-\hat{j}-\hat{k}}
\\ {~\color{magenta} 3 } &{{}} &{\vec{v_1}} & {~=~} &{\hat{i}-\hat{j}+\hat{k}}
\\ {~\color{magenta} 4 } &{{}} &{\vec{v_2}} & {~=~} &{2\hat{i}+\hat{j}+2\hat{k}}
\\ \end{array}}$
3. $\small{\vec{ST} = \vec{u_2} - \vec{u_1} = \hat{i}-3\hat{j}-2\hat{k}}$
4. $\small{\hat{PQ} = \frac{\vec{v_1}\times\vec{v_2}}{\left|\vec{v_1}\times\vec{v_2} \right|}}$
= $\small{\frac{-3\hat{i}+3\hat{k}}{\sqrt{(-3)^2 + 3^2}}}$
= $\small{\frac{-3\hat{i}+3\hat{k}}{\sqrt{18}}}$
• The reader may write all the steps related to the cross product
5. Substituting (3) and (4) in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\text{Shortest distance}} & {~=~} &{\vec{ST}.\hat{PQ}}
\\ {~\color{magenta} 2 } &{{}} &{} & {~=~} &{\left(\hat{i}-3\hat{j}-2\hat{k} \right).\left(\frac{-3\hat{i}+3\hat{k}}{\sqrt{18}} \right)}
\\ {~\color{magenta} 3 } &{{}} &{} & {~=~} &{\frac{(1)(-3)+(-3)(0)+(-2)(3)}{\sqrt{18}}}
\\ {~\color{magenta} 4 } &{{}} &{} & {~=~} &{\frac{-9}{\sqrt{18}}=\frac{(-1)\sqrt{9}\,\sqrt{9}}{\sqrt{2}\,\sqrt{9}} = \frac{(-1)(3)}{\sqrt{2}}}
\\ {~\color{magenta} 5 } &{{}} &{} & {~=~} &{\frac{-3\sqrt{2}}{2}}
\\ \end{array}}$
• Projection is a distance. It cannot be −ve. So we need to take the absolute value.
• Therefore, the shortest distance = $\small{\left|\frac{-3\sqrt{2}}{2} \right|~=~\frac{3\sqrt{2}}{2}}$ units
Solved example 27.26
Find the shortest distance between the lines $\small{l_1~\text{and}~l_2}$ whose vector equations are
$\small{\vec{r} = \hat{i}+2\hat{j}+3\hat{k}+\lambda\left(\hat{i}-3\hat{j}+2\hat{k} \right)}$
and $\small{\vec{r} = 4\hat{i}+5\hat{j}+6\hat{k}+\mu\left(2\hat{i}+3\hat{j}+\hat{k} \right)}$
Solution:
1. We have:
Shortest distance
= Length of the line $\small{PQ}$
= Projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
• The projection is given by: $\small{\vec{ST}.\hat{PQ}}$
2. Based on the given equations of the lines, we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{u_1}} & {~=~} &{\hat{i}+2\hat{j}+3\hat{k}}
\\ {~\color{magenta} 2 } &{{}} &{\vec{u_2}} & {~=~} &{4\hat{i}+5\hat{j}+6\hat{k}}
\\ {~\color{magenta} 3 } &{{}} &{\vec{v_1}} & {~=~} &{\hat{i}-3\hat{j}+2\hat{k}}
\\ {~\color{magenta} 4 } &{{}} &{\vec{v_2}} & {~=~} &{2\hat{i}+3\hat{j}+\hat{k}}
\\ \end{array}}$
3. $\small{\vec{ST} = \vec{u_2} - \vec{u_1} = 3\hat{i}-3\hat{j}+3\hat{k}}$
4. $\small{\hat{PQ} = \frac{\vec{v_1}\times\vec{v_2}}{\left|\vec{v_1}\times\vec{v_2} \right|}}$
= $\small{\frac{-9\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{(-9)^2 + 3^2 + 9^2}}}$
= $\small{\frac{-9\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{171}}}$
• The reader may write all the steps related to the cross product
5. Substituting (3) and (4) in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\text{Shortest distance}} & {~=~} &{\vec{ST}.\hat{PQ}}
\\ {~\color{magenta} 2 } &{{}} &{} & {~=~} &{\left(3\hat{i}-3\hat{j}+3\hat{k} \right).\left(\frac{-9\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{171}} \right)}
\\ {~\color{magenta} 3 } &{{}} &{} & {~=~} &{\frac{(3)(-9)+(-3)(-3)+(3)(9)}{\sqrt{171}}}
\\ {~\color{magenta} 4 } &{{}} &{} & {~=~} &{\frac{9}{\sqrt{171}}=\frac{9}{3\,\sqrt{19}} }
\\ {~\color{magenta} 5 } &{{}} &{} & {~=~} &{\frac{3}{\sqrt{19}}}
\\ \end{array}}$
• Therefore, the shortest distance = $\small{\frac{3}{\sqrt{19}}}$ units
Solved example 27.27
Find the shortest distance between the lines $\small{l_1~\text{and}~l_2}$ whose vector equations are
$\small{\vec{r} = (1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k}}$
and $\small{\vec{r} = (s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}}$
Solution:
1. Let us convert the given vector equations to standard form:
• The vector equation of $\small{l_1}$ is:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{r}} & {~=~} &{(1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k}}
\\ {~\color{magenta} 2 } &{{}} &{} & {~=~} &{\hat{i}-t\hat{i}+t\hat{j}-2\hat{j}+3\hat{k}-2t\hat{k}}
\\ {~\color{magenta} 3 } &{{}} &{} & {~=~} &{\hat{i}-2\hat{j}+3\hat{k}-t\hat{i}+t\hat{j}-2t\hat{k}}
\\ {~\color{magenta} 4 } &{{}} &{} & {~=~} &{\hat{i}-2\hat{j}+3\hat{k}+t\left(-\hat{i}+\hat{j}-2\hat{k} \right)}
\\ \end{array}}$
• The vector equation of $\small{l_2}$ is:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{r}} & {~=~} &{(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}}
\\ {~\color{magenta} 2 } &{{}} &{} & {~=~} &{s\hat{i}+\hat{i}+2s\hat{j}-\hat{j}-2s\hat{k}-\hat{k}}
\\ {~\color{magenta} 3 } &{{}} &{} & {~=~} &{\hat{i}-\hat{j}-\hat{k}+s\hat{i}+2s\hat{j}-2s\hat{k}}
\\ {~\color{magenta} 4 } &{{}} &{} & {~=~} &{\hat{i}-\hat{j}-\hat{k}+s\left(\hat{i}+2\hat{j}-2\hat{k} \right)}
\\ \end{array}}$
2. We have:
Shortest distance
= Length of the line $\small{PQ}$
= Projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
• The projection is given by: $\small{\vec{ST}.\hat{PQ}}$
3. Based on the given equations of the lines, we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{u_1}} & {~=~} &{\hat{i}-2\hat{j}+3\hat{k}}
\\ {~\color{magenta} 2 } &{{}} &{\vec{u_2}} & {~=~} &{\hat{i}-\hat{j}-\hat{k}}
\\ {~\color{magenta} 3 } &{{}} &{\vec{v_1}} & {~=~} &{-\hat{i}+\hat{j}-2\hat{k}}
\\ {~\color{magenta} 4 } &{{}} &{\vec{v_2}} & {~=~} &{\hat{i}+2\hat{j}-2\hat{k}}
\\ \end{array}}$
4. $\small{\vec{ST} = \vec{u_2} - \vec{u_1} = \hat{j}-4\hat{k}}$
5. $\small{\hat{PQ} = \frac{\vec{v_1}\times\vec{v_2}}{\left|\vec{v_1}\times\vec{v_2} \right|}}$
= $\small{\frac{2\hat{i}-4\hat{j}-3\hat{k}}{\sqrt{2^2 + (-4)^2 + (-3)^2}}}$
= $\small{\frac{2\hat{i}-4\hat{j}-3\hat{k}}{\sqrt{29}}}$
• The reader may write all the steps related to the cross product
6. Substituting (4) and (5) in (2), we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\text{Shortest distance}} & {~=~} &{\vec{ST}.\hat{PQ}}
\\ {~\color{magenta} 2 } &{{}} &{} & {~=~} &{\left(\hat{j}-4\hat{k} \right).\left(\frac{2\hat{i}-4\hat{j}-3\hat{k}}{\sqrt{29}} \right)}
\\ {~\color{magenta} 3 } &{{}} &{} & {~=~} &{\frac{(0)(2)+(1)(-4)+(-4)(-3)}{\sqrt{29}}}
\\ {~\color{magenta} 4 } &{{}} &{} & {~=~} &{\frac{8}{\sqrt{29}} }
\\ {~\color{magenta} 5 } &{{}} &{} & {~=~} &{\frac{3}{\sqrt{19}}}
\\ \end{array}}$
• Therefore, the shortest distance = $\small{\frac{8}{\sqrt{29}}}$ units
• In the above discussion, we were given the equations of the lines in vector form. If the equations were given in the Cartesian form, we can quickly convert them into vector form. See the "easy method" mentioned in Solved examples 27.11 and 27.22 of section 27.2
Let us see a solved example
Solved example 27.28
Find the shortest distance between the lines $\small{l_1~\text{and}~l_2}$ whose Cartesian equations are
$\small{\frac{x+1}{7}~=~\frac{y+1}{-6}~=~\frac{z+1}{1}}$
and $\small{\frac{x-3}{1}~=~\frac{y-5}{-2}~=~\frac{z-7}{1}}$
Solution:
1. Let us convert the given Cartesian equations to vector equations:
• The vector equation of $\small{l_1}$ is:
$\small{\vec{r} = -\hat{i}-\hat{j}-\hat{k}+\lambda\left(7\hat{i}-6\hat{j}+\hat{k} \right)}$
• The vector equation of $\small{l_2}$ is:
$\small{\vec{r} = 3\hat{i}+5\hat{j}+7\hat{k}+\mu\left(\hat{i}-2\hat{j}+\hat{k} \right)}$
2. We have:
Shortest distance
= Length of the line $\small{PQ}$
= Projection of $\small{\vec{ST}~\text{on}~\vec{PQ}}$
• The projection is given by: $\small{\vec{ST}.\hat{PQ}}$
3. Based on the given equations of the lines, we can write:
\\ {~\color{magenta} 2 } &{{}} &{\vec{u_2}} & {~=~} &{3\hat{i}+5\hat{j}+7\hat{k}}
\\ {~\color{magenta} 3 } &{{}} &{\vec{v_1}} & {~=~} &{7\hat{i}-6\hat{j}+\hat{k}}
\\ {~\color{magenta} 4 } &{{}} &{\vec{v_2}} & {~=~} &{\hat{i}-2\hat{j}+\hat{k}}
\\ \end{array}}$
4. $\small{\vec{ST} = \vec{u_2} - \vec{u_1} = 4\hat{i}+6\hat{j}+8\hat{k}}$
5. $\small{\hat{PQ} = \frac{\vec{v_1}\times\vec{v_2}}{\left|\vec{v_1}\times\vec{v_2} \right|}}$
= $\small{\frac{-4\hat{i}-6\hat{j}-8\hat{k}}{\sqrt{(-4)^2 + (-6)^2 + (-8)^2}}}$
= $\small{\frac{-4\hat{i}-6\hat{j}-8\hat{k}}{\sqrt{116}}}$
• The reader may write all the steps related to the cross product
6. Substituting (4) and (5) in (2), we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\text{Shortest distance}} & {~=~} &{\vec{ST}.\hat{PQ}}
\\ {~\color{magenta} 2 } &{{}} &{} & {~=~} &{\left(4\hat{i}+6\hat{j}+8\hat{k} \right).\left(\frac{-4\hat{i}-6\hat{j}-8\hat{k}}{\sqrt{116}} \right)}
\\ {~\color{magenta} 3 } &{{}} &{} & {~=~} &{\frac{(4)(-4)+(6)(-6)+(8)(-8)}{\sqrt{116}}}
\\ {~\color{magenta} 4 } &{{}} &{} & {~=~} &{\frac{-116}{\sqrt{116}}=(-1)\sqrt{116}}
\\ {~\color{magenta} 5 } &{{}} &{} & {~=~} &{-2\sqrt{29}}
\\ \end{array}}$
• Projection is a distance. It cannot be −ve. So we need to take the absolute value.
• Therefore, the shortest distance = $\small{\left|-2\sqrt{29} \right|~=~2\sqrt{29}}$ units
In the next section, we will see distance between parallel lines.
Previous
Contents
Next
Copyright©2026 Higher secondary mathematics.blogspot.com
