27.6 - Distance Between Two Skew Lines

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.

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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.

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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

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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)

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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

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• 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:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u_1}}    & {~=~}    &{-\hat{i}-\hat{j}-\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{u_2}}    & {~=~}    &{3\hat{i}+5\hat{j}+7\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{v_1}}    & {~=~}    &{7\hat{i}-6\hat{j}+\hat{k}}
\\ {~\color{magenta}    4    }    &{{}}    &{\vec{v_2}}    & {~=~}    &{\hat{i}-2\hat{j}+\hat{k}}
\\ \end{array}}$

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

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In the next section, we will see distance between parallel lines.

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27.5 - Two Lines Parallel or Perpendicular

In the previous section, we saw the angle between lines in 3D. In this section, we will see the conditions for two lines to be parallel or perpendicular to each other.

First we will see angle in terms of direction cosines. It can be explained in 7 steps:
1. For $\small{L_1}$:
    ♦ Let the direction ratios be $\small{a_1,~b_1,~c_1}$
    ♦ Let the direction cosines be $\small{l_1,~m_1,~n_1}$
• Then we can write:
$\small{\frac{l_1}{a_1}~=~\frac{m_1}{b_1}~=~\frac{n_1}{c_1}~=~\lambda_1}$

2. For $\small{L_2}$:
    ♦ Let the direction ratios be $\small{a_2,~b_2,~c_2}$
    ♦ Let the direction cosines be $\small{l_2,~m_2,~n_2}$
• Then we can write:
$\small{\frac{l_2}{a_2}~=~\frac{m_2}{b_2}~=~\frac{n_2}{c_2}~=~\lambda_2}$

3. Now we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1 a_2 ~+~ b_1 b_2~+~c_1 c_2}    & {~=~}    &{\left(\frac{l_1}{\lambda_1} \right)\left(\frac{l_2}{\lambda_2} \right)+\left(\frac{l_1}{\lambda_1} \right)\left(\frac{l_2}{\lambda_2} \right)+\left(\frac{l_1}{\lambda_1} \right)\left(\frac{l_2}{\lambda_2} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{l_1 l_2~+~m_1 m_2~+~n_1 n_2}{\lambda_1\lambda_2}}
\\ \end{array}}$

4. Also we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1^2 ~+~ b_1^2~+~c_1^2}    & {~=~}    &{\left(\frac{l_1}{\lambda_1} \right)^2~+~\left(\frac{m_1}{\lambda_1} \right)^2~+~\left(\frac{n_1}{\lambda_1} \right)^2}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{l_1^2~+~m_1^2~+~n_1^2}{\lambda_1^2}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\sqrt{a_1^2 ~+~ b_1^2~+~c_1^2}}    & {~=~}    &{\frac{\sqrt{l_1^2~+~m_1^2~+~n_1^2}}{\lambda_1}}
\\ \end{array}}$

5. Also we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_2^2 ~+~ b_2^2~+~c_2^2}    & {~=~}    &{\left(\frac{l_2}{\lambda_2} \right)^2~+~\left(\frac{m_2}{\lambda_2} \right)^2~+~\left(\frac{n_2}{\lambda_2} \right)^2}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{l_2^2~+~m_2^2~+~n_2^2}{\lambda_2^2}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\sqrt{a_2^2 ~+~ b_2^2~+~c_2^2}}    & {~=~}    &{\frac{\sqrt{l_2^2~+~m_2^2~+~n_2^2}}{\lambda_2}}
\\ \end{array}}$

6. In the previous section, we derived the formula:
$\small{\cos\theta~=~\frac{a_1 a_2 ~+~ b_1 b_2~+~c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\,\sqrt{a_2^2 + b_2^2 + c_2^2}}}$

7. Substituting from (3), (4) and (5), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\theta}    & {~=~}    &{\frac{\frac{l_1 l_2~+~m_1 m_2~+~n_1 n_2}{\lambda_1\lambda_2}}{\left(\frac{\sqrt{l_1^2~+~m_1^2~+~n_1^2}}{\lambda_1} \right)\left(\frac{\sqrt{l_2^2~+~m_2^2~+~n_2^2}}{\lambda_2} \right)}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{l_1 l_2~+~m_1 m_2~+~n_1 n_2}{\left(\sqrt{l_1^2~+~m_1^2~+~n_1^2} \right)\left(\sqrt{l_2^2~+~m_2^2~+~n_2^2} \right)}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{l_1 l_2~+~m_1 m_2~+~n_1 n_2}{\left(1 \right)\left(1 \right)}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{l_1 l_2~+~m_1 m_2~+~n_1 n_2}
\\ \end{array}}$

◼ Remarks:
3 (magenta color): Here we use the fact that:
$\small{l^2 + m^2 +n^2 = 1}$

---

Now we will see the condition for two lines to be perpendicular. It can be explained in 4 steps:
1. We have seen that $\small{\cos \theta = l_1 l_2~+~m_1 m_2~+~n_1 n_2}$ 

2. When the two lines are perpendicular, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\left(\frac{\pi}{2} \right)}    & {~=~}    &{l_1 l_2~+~m_1 m_2~+~n_1 n_2}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{0}    & {~=~}    &{l_1 l_2~+~m_1 m_2~+~n_1 n_2}
\\ \end{array}}$

3. So the condition for two lines to be perpendicular is:
$\small{l_1 l_2~+~m_1 m_2~+~n_1 n_2~=~0}$

4. The above condition is in terms of direction cosines. We can write it in terms of direction ratios also:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l_1 l_2~+~m_1 m_2~+~n_1 n_2}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left(\lambda_1 a_1 \right)\left(\lambda_2 a_2 \right)~+~\left(\lambda_1 b_1 \right)\left(\lambda_2 b_2 \right)~+~\left(\lambda_1 c_1 \right)\left(\lambda_2 c_2 \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\lambda_1 \lambda_2\left(a_1 a_2~+~b_1 b_2~+~c_1 c_2 \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{a_1 a_2~+~b_1 b_2~+~c_1 c_2}    & {~=~}    &{0}
\\ \end{array}}$

---

We have already seen the condition for two lines to be parallel in a previous section 27.1.
1. The condition in terms of direction cosines is:
$\small{\frac{l_1}{l_2}~=~\frac{m_1}{m_2}~=~\frac{n_1}{n_2}}$

2. The condition in terms of direction ratios is:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$

---

Now we will see some solved examples

Solved example 27.19
Show that the three lines with direction cosines
$\small{\frac{12}{13},~\frac{-3}{13},~\frac{-4}{13};~~~~~~\frac{4}{13},~\frac{12}{13},~\frac{3}{13};~~~~~~\frac{3}{13},~\frac{-4}{13},~\frac{12}{13}}$
are mutually perpendicular.
Solution:
1. Let us name the three lines and classify the direction cosines:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L_1}    & {~:~}    &{\frac{12}{13},~\frac{-3}{13},~\frac{-4}{13}}
\\ {~\color{magenta}    2    }    &{}    &{L_2}    & {~:~}    &{\frac{4}{13},~\frac{12}{13},~\frac{3}{13}}
\\ {~\color{magenta}    3    }    &{}    &{L_3}    & {~:~}    &{\frac{3}{13},~\frac{-4}{13},~\frac{12}{13}}
\\ \end{array}}$

2. If two lines $\small{L_1~\text{and}~L_2}$ are perpendicular, we have:
$\small{l_1 l_2~+~m_1 m_2~+~n_1 n_2~=~0}$

3. Let us check the given $\small{L_1~\text{and}~L_2}$:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l_1 l_2~+~m_1 m_2~+~n_1 n_2}    & {~=~}    &{\left(\frac{12}{13} \right)\left(\frac{4}{13} \right)+\left(\frac{-3}{13} \right)\left(\frac{12}{13} \right)+\left(\frac{-4}{13} \right)\left(\frac{3}{13} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{48 -36 - 12}{169}~=~\frac{0}{169}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$
• So $\small{L_1~\text{and}~L_2}$ are mutually perpendicular.

4. Let us check the given $\small{L_2~\text{and}~L_3}$:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l_2 l_3~+~m_2 m_3~+~n_2 n_3}    & {~=~}    &{\left(\frac{4}{13} \right)\left(\frac{3}{13} \right)
+\left(\frac{12}{13} \right)\left(\frac{-4}{13} \right)
+\left(\frac{3}{13} \right)\left(\frac{12}{13} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{12 -48 + 36}{169}~=~\frac{0}{169}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$
• So $\small{L_2~\text{and}~L_3}$ are mutually perpendicular.

5. Let us check the given $\small{L_3~\text{and}~L_1}$:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l_3 l_1~+~m_3 m_1~+~n_3 n_1}    & {~=~}    &{\left(\frac{3}{13} \right)\left(\frac{12}{13} \right)
+\left(\frac{-4}{13} \right)\left(\frac{-3}{13} \right)
+\left(\frac{12}{13} \right)\left(\frac{-4}{13} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{36 +12 - 48}{169}~=~\frac{0}{169}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$
• So $\small{L_3~\text{and}~L_1}$ are mutually perpendicular.

6. Based on (3), (4) and (5), we can write:
The three lines are mutually perpendicular.

Solved example 27.20
Show that the line through the points (1,−1,2) and (3,4,−2) is perpendicular to the line through the points (0,3,2) and (3,5,6)
Solution:
1. For a line through two points $\small{U\left(x_1,y_1,z_1 \right)~\text{and}~V\left(x_2,y_2,z_2 \right)}$, we can consider $\small{\left(x_2 - x_1 \right),~\left(y_2 - y_1 \right)~\text{and}~\left(z_2 - z_1 \right)}$ as a set of direction ratios.

2. So for the first line, the direction ratios $\small{a_1,b_1,c_1}$ are:
2, 5 and −4

3. Similarly, for the second line, the direction ratios $\small{a_2,b_2,c_2}$ are:
3, 2 and 4

4. If two lines $\small{L_1~\text{and}~L_2}$ are perpendicular, we have:
$\small{a_1 a_2~+~b_1 b_2~+~c_1 c_2~=~0}$

5. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1 a_2~+~b_1 b_2~+~c_1 c_2}    & {~=~}    &{\left(2 \right)\left(3 \right)+\left(5 \right)\left(2 \right)+\left(-4 \right)\left(4 \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{6+10-16}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$

6. So the two lines are perpendicular to each other.

Solved example 27.21
Show that the line through the points (4,7,8) and (2,3,4) is  parallel to the line through the points (−1,−2,1) and (1,2,5)
Solution:
1. For a line through two points $\small{U\left(x_1,y_1,z_1 \right)~\text{and}~V\left(x_2,y_2,z_2 \right)}$, we can consider $\small{\left(x_2 - x_1 \right),~\left(y_2 - y_1 \right)~\text{and}~\left(z_2 - z_1 \right)}$ as a set of direction ratios.

2. So for the first line, the direction ratios $\small{a_1,b_1,c_1}$ are:
−2, −4 and −4

3. Similarly, for the second line, the direction ratios $\small{a_2,b_2,c_2}$ are:
2, 4 and 4

4. If two lines $\small{L_1~\text{and}~L_2}$ are parallel, we have:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$

5. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{-2}{2}}& {~=~}    &{\frac{-4}{4}}    & {~=~}    &{\frac{-4}{4}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{-1}& {~=~}    &{-1}    & {~=~}    &{-1}
\\ \end{array}}$ 

6. So the two lines are parallel.

Solved example 27.22
Show that the lines
$\small{\frac{x-5}{7}~=~\frac{y+2}{-5}~=~\frac{z}{1}}$ and
$\small{\frac{x}{1}~=~\frac{y}{2}~=~\frac{z}{3}}$
are perpendicular to each other
Solution:
1. When the equation of the line is in Cartesian form, the denominators can be considered as a set of direction ratios.

2. So for the first line, the direction ratios $\small{a_1,b_1,c_1}$ are:
7, −5 and 1

3. Similarly, for the second line, the direction ratios $\small{a_2,b_2,c_2}$ are:
1, 2 and 3

4. If two lines $\small{L_1~\text{and}~L_2}$ are perpendicular, we have:
$\small{a_1 a_2~+~b_1 b_2~+~c_1 c_2~=~0}$

5. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1 a_2~+~b_1 b_2~+~c_1 c_2}    & {~=~}    &{\left(7 \right)\left(1\right)+\left(-5 \right)\left(2 \right)+\left(1 \right)\left(3 \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{7-10+3}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$

6. So the two lines are perpendicular to each other.

Solved example 27.23
Find the value of $\small{p}$ so that the lines
$\small{\frac{1-x}{3}~=~\frac{7y-14}{2p}~=~\frac{z-3}{2}}$ and
$\small{\frac{7-7x}{3p}~=~\frac{y-5}{1}~=~\frac{6-z}{5}}$
are perpendicular to each other
Solution:
1. When the equation of the line is in Cartesian form, the denominators can be considered as a set of direction ratios.
• But we need to convert the equations into standard form.
• For the first line, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{1-x}{3}}& {~=~}    &{\frac{7y-14}{2p}}    & {~=~}    &{\frac{z-3}{2}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{(-1)(1-x)}{(-1)(3)}}& {~=~}    &{\frac{(7y-14)/7}{(2p)/7}}    & {~=~}    &{\frac{z-3}{2}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x-1}{-3}}& {~=~}    &{\frac{y-2}{(2/7)p}}    & {~=~}    &{\frac{z-3}{2}}
\\ \end{array}}$
• For the second line, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{7-7x}{3p}}& {~=~}    &{\frac{y-5}{1}}    & {~=~}    &{\frac{6-z}{5}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{(-1)(1/7)(7-7x)}{(-1)(1/7)(3p)}}& {~=~}    &{\frac{y-5}{1}}    & {~=~}    &{\frac{(-1)(6-z)}{(-1)(5)}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x-1}{(-3/7)p}}& {~=~}    &{\frac{y-5}{1}}    & {~=~}    &{\frac{z-6}{-5}}
\\ \end{array}}$ 

2. So for the first line, the direction ratios $\small{a_1,b_1,c_1}$ are:
−3, (2/7)p and 2

3. Similarly, for the second line, the direction ratios $\small{a_2,b_2,c_2}$ are:
(−3/7)p, 1 and −5

4. If two lines $\small{L_1~\text{and}~L_2}$ are perpendicular, we have:
$\small{a_1 a_2~+~b_1 b_2~+~c_1 c_2~=~0}$

5. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1 a_2~+~b_1 b_2~+~c_1 c_2}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left(-3 \right)\left((-3/7)p\right)+\left((2/7)p \right)\left(1 \right)+\left(2 \right)\left(-5 \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\frac{9p}{7} + \frac{2p}{7} - 10}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\frac{11p - 70}{7}}    & {~=~}    &{0}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{11p - 70}    & {~=~}    &{0}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{p}    & {~=~}    &{\frac{70}{11}}
\\ \end{array}}$

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In the next section, we will see the shortest distance between two lines.

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27.4 - Angle Between Two Lines

In the previous section, and in the section before that, we saw the basic details about the equation of a line in 3D space. In this section, we will see the relation between 3D and 2D. Later in this section, we will see angle between two lines in 3D.

First we will see the relation. It can be written in 4 steps:
1. We have seen one type of equation of a line:
$\small{\frac{x-x_1}{a}~=~\frac{y-y_1}{b}~=~\frac{z-z_1}{c}}$
• Here, the line passes through a point $\small{U\left(x_1,y_1,z_1 \right)}$ 

2. Suppose that, we are considering the XY-plane only. Then all the points will lie on that plane. There is no z-axis.
• In such a situation, the equation becomes:
$\small{\frac{x-x_1}{a}~=~\frac{y-y_1}{b}}$
• This can be rearranged as shown below:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x-x_1}{a}}    & {~=~}    &{\frac{y-y_1}{b}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\frac{y-y_1}{b}}    & {~=~}    &{\frac{x-x_1}{a}}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{y-y_1}    & {~=~}    &{\frac{b}{a}\left(x - x_1 \right)}    \\
{~\color{magenta}    4    }    &{\Rightarrow}    &{y-y_1}    & {~=~}    &{\frac{b}{a}\left(x \right)~-~\frac{b}{a}\left(x_1 \right)}    \\
{~\color{magenta}    5    }    &{\Rightarrow}    &{y}    & {~=~}    &{\frac{b}{a}\left(x \right)~-~\frac{b}{a}\left(x_1 \right)+y_1}    \\
{~\color{magenta}    6    }    &{\Rightarrow}    &{y}    & {~=~}    &{\frac{b}{a}\left(x \right)~+~\left[y_1 - \frac{b}{a}\left(x_1 \right) \right]}    \\
\end{array}}$

3. Let us examine the terms in the above result:
• There is one term in the L.H.S. It is a variable term.
• In the R.H.S, there are two terms. The first one is a variable term and the second one is a constant term.

4. We are familiar with this type of equations. It is of the form $\small{y = mx + c}$ It is the equation of the line with slope ‘m’ and y-intercept ‘c’.
• So we can write:
    ♦ In 3D, the real numbers $\small{a,~b~\text{and}~c}$ are the scalars of the parallel vector.
    ♦ In 2D, the real number $\small{\frac{b}{a}}$ is the slope of the line.

---

Let us see an example. It can be written in 3 steps:
1. Consider the equation of a line in 2D:
$\small{\frac{x+4}{3}~=~\frac{y-5}{2}}$
• It is clear that, the line passes through (−4,5).
• Also, it has a slope of $\small{\frac{b}{a} = \frac{2}{3}}$
• Then the line will make an angle of $\small{\tan^{-1}\left(\frac{2}{3} \right)~=~33.69^ \circ}$ with the x-axis
• The y-intercept of this line is:
$\small{y_1 - \frac{b}{a}\left(x_1 \right)~=~5 - \frac{2}{3}\left(-4 \right)~=~5 + \frac{8}{3} = 7.67}$
• This line is shown in yellow color in fig.27.8 below:

Fig.27.8

2. Consider another line with the same values of a and b:
$\small{\frac{x-1}{3}~=~\frac{y-4}{2}}$
• It is clear that, the line passes through (1,4).
• Also, it has a slope of $\small{\frac{b}{a} = \frac{2}{3}}$
• Then this line also will make an angle of $\small{\tan^{-1}\left(\frac{2}{3} \right)~=~33.69^ \circ}$ with the x-axis
• The y-intercept of this line is:
$\small{y_1 - \frac{b}{a}\left(x_1 \right)~=~4 - \frac{2}{3}\left(1 \right)~=~4 - \frac{2}{3} = 3.33}$
• This line is shown in red color in fig.27.8 above.

3. Let us form a 2D vector using the above values of a and b. We get:
$\small{\vec{a}=a\hat{i}+b\hat{j}=3\hat{i}+2\hat{j}}$
• It is shown in magenta color in fig.27.8 above. We see that, $\small{\vec{a}}$ is parallel to both the red and yellow lines.

---

The information obtained from the above discussion can be extended to 3D also. It can be written in 3 steps:
1. We are given a line in 3D:
$\small{\frac{x-x_1}{a}~=~\frac{y-y_1}{b}~=~\frac{z-z_1}{c}}$
• Then we can write a 3D vector: $\small{a\hat{i}+b\hat{j}+c\hat{k}}$   
This vector will be parallel to the given line.
2. The converse can also be written:
We are given a 3D vector: $\small{a\hat{i}+b\hat{j}+c\hat{k}}$   
Then we can write a 3D line:
$\small{\frac{x-x_1}{a}~=~\frac{y-y_1}{b}~=~\frac{z-z_1}{c}}$
• This line will be parallel to the given vector
3. The numbers $\small{x_1,~y_1,~z_1}$ can be calculated by equating all three fractions to any convenient real number.

---

The above discussion is applicable to the other type of the line also. It can be explained in 3 steps:
1. We have seen the other type of equation of a line:
$\small{\frac{x-x_1}{x_2 - x_1}~=~\frac{y-y_1}{y_2 - y_1}~=~\frac{z-z_1}{z_2 - z_1}}$
• Here, the line passes through two points:
$\small{U\left(x_1,y_1,z_1 \right)~\text{and}~V\left(x_2,y_2,z_2 \right)}$ 

2. Suppose that, we are considering the XY-plane only. Then all the points will lie on that plane. There is no z-axis.
• In such a situation, the equation becomes:
$\small{\frac{x-x_1}{x_2 - x_1}~=~\frac{y-y_1}{y_2 - y_1}}$
• This can be rearranged as shown below:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x-x_1}{x_2 - x_1}}    & {~=~}    &{\frac{y-y_1}{y_2 - y_1}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\frac{y-y_1}{y_2 - y_1}}    & {~=~}    &{\frac{x-x_1}{x_2 - x_1}}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{y-y_1}    & {~=~}    &{\frac{y_2 - y_1}{x_2 - x_1}\left(x - x_1 \right)}    \\
\end{array}}$

3. We are familiar with this type of equations. It is used in 2D when a line passes through two points $\small{U\left(x_1,y_1 \right)~\text{and}~V\left(x_2,y_2 \right)}$ .
• So we can write:
    ♦ In 3D, the real numbers $\small{\left(x_2 - x_1 \right)~\text{and}~\left(y_2 - y_1 \right)}$ are the scalars of the parallel vector.
    ♦ In 2D, the real number $\small{\frac{y_2 - y_1}{x_2 - x_1}}$ is the slope of the line.

---

Now we will see angle between two lines. It can be explained in steps:

1. In fig.27.9 below, two lines $\small{L_1~\text{and}~L_2}$ pass through the origin O.

Fig.27.9

2. The direction ratios of the two lines are given:
    ♦ $\small{L_1}$ has the direction ratios $\small{a_1,~b_1~\text{and}~c_1}$
    ♦ $\small{L_2}$ has the direction ratios $\small{a_2,~b_2~\text{and}~c_2}$

3. We want to find the acute angle $\small{\theta}$ between the two lines.

4. First we mark any two convenient points:
    ♦ Point $\small{P}$ on $\small{L_1}$
    ♦ Point $\small{Q}$ on $\small{L_2}$

5. Now we can imagine two vectors:
    ♦ $\small{\vec{OP}}$ aligned with $\small{L_1}$
    ♦ $\small{\vec{OQ}}$ aligned with $\small{L_2}$

6. Now we can think about parallel vectors
    ♦ $\small{\vec{p}=a_1 \hat{i}+b_1 \hat{j}+c_1\hat{k}}$ will be parallel to $\small{\vec{OP}}$
    ♦ $\small{\vec{q}=a_2 \hat{i}+b_2 \hat{j}+c_2\hat{k}}$ will be parallel to $\small{\vec{OQ}}$

7. So we can write about the angles:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\theta}    & {~=~}    &{\text{Angle between}~L_1~\text{and}~L_2}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\text{Angle between}~\vec{OP}~\text{and}~\vec{OQ}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\text{Angle between}~\vec{p}~\text{and}~\vec{q}}    \\
\end{array}}$

8. We can easily find the angle between $\small{\vec{p}~\text{and}~\vec{q}}$ because, both are in the component form. We can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\theta}    & {~=~}    &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{a_1 a_2 ~+~ b_1 b_2~+~c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\,\sqrt{a_2^2 + b_2^2 + c_2^2}}}    \\
\end{array}}$

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Suppose that, the two lines $\small{L_1~\text{and}~L_2}$ do not pass through the origin. In such a situation, we can use another method. It can be written in 4 steps:
1. Imagine a third line $\small{L_1’}$ such that:
    ♦ $\small{L_1’}$ is parallel to $\small{L_1}$
    ♦ $\small{L_1’}$ passes through the origin
• Since $\small{L_1’}$ is parallel to $\small{L_1}$, both will have the same direction ratios.

2. Imagine a fourth line $\small{L_2’}$ such that:
    ♦ $\small{L_2’}$ is parallel to $\small{L_2}$
    ♦ $\small{L_2’}$ passes through the origin
• Since $\small{L_2’}$ is parallel to $\small{L_2}$, both will have the same direction ratios.

3. Now we can find the angle between $\small{L_1'~\text{and}~L_2'}$.

4. The required angle between $\small{L_1~\text{and}~L_2}$ will be same as the angle between $\small{L_1'~\text{and}~L_2'}$ 

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Now we will see some solved examples

Solved example 27.15
Find the angle between the pair of lines given by
$\small{3\hat{i}+2\hat{j}-4\hat{k}~+~\lambda\left(\hat{i}+2\hat{j}+2\hat{k} \right)}$
and $\small{5\hat{i}-2\hat{j}~+~\mu\left(3\hat{i}+2\hat{j}+6\hat{k} \right)}$.
Solution:
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~\hat{i}+2\hat{j}+2\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~3\hat{i}+2\hat{j}+6\hat{k}}$
• Reason: In vector form, the vector inside braces is a vector parallel to the line.

2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\theta}    & {~=~}    &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{(1)(3)~+~(2)(2)~+~(2)(6)}{\sqrt{1^2 + 2^2 + 2^2}\,\sqrt{3^2 + 2^2 + 6^2}}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{19}{\sqrt{9}\,\sqrt{49}}~=~\frac{19}{21}}    \\
{~\color{magenta}    4    }    &{\Rightarrow}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{19}{21} \right)}    \\
\end{array}}$

Solved example 27.16
Find the angle between the following pairs of lines
(i) $\small{2\hat{i}-5\hat{j}+\hat{k}~+~\lambda\left(3\hat{i}+2\hat{j}+6\hat{k} \right)}$ and
$\small{7\hat{i}-6\hat{j}~+~\mu\left(\hat{i}+2\hat{j}+2\hat{k} \right)}$
(ii) $\small{3\hat{i}+\hat{j}-2\hat{k}~+~\lambda\left(\hat{i}-\hat{j}-2\hat{k} \right)}$ and
$\small{2\hat{i}-\hat{j}-56\hat{k}~+~\mu\left(3\hat{i}-5\hat{j}-4\hat{k} \right)}$
Solution:
Part (i):
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~3\hat{i}+2\hat{j}+6\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~\hat{i}+2\hat{j}+2\hat{k}}$
• Reason: In vector form, the vector inside braces is a vector parallel to the line.

2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\theta}    & {~=~}    &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{(3)(1)~+~(2)(2)~+~(6)(2)}{\sqrt{3^2 + 2^2 + 6^2}\,\sqrt{1^2 + 2^2 + 2^2}}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{19}{\sqrt{49}\,\sqrt{9}}~=~\frac{19}{21}}    \\
{~\color{magenta}    4    }    &{\Rightarrow}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{19}{21} \right)}    \\
\end{array}}$

Part (ii):
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~\hat{i}-\hat{j}-2\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~3\hat{i}-5\hat{j}-4\hat{k}}$
• Reason: In vector form, the vector inside braces is a vector parallel to the line.

2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\theta}    & {~=~}    &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{(1)(3)~+~(-1)(-5)~+~(-2)(-4)}{\sqrt{1^2 + (-1)^2 + (-2)^2}\,\sqrt{3^2 + (-5)^2 + (-4)^2}}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{16}{\sqrt{6}\,\sqrt{50}}~=~\frac{16}{\sqrt{6}\,(5)\sqrt{2}}~=~\frac{8\sqrt{2} \sqrt{2}}{\sqrt{3}\sqrt{2}\,(5)\sqrt{2}}~=~\frac{8}{5\sqrt{3}}}    \\
{~\color{magenta}    4    }    &{\Rightarrow}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{8}{5\sqrt{3}} \right)}    \\
\end{array}}$

Solved example 27.17
Find the angle between the pair of lines given by
$\small{\frac{x+3}{3}~=~\frac{y-1}{5}~=~\frac{z+3}{4}}$
and $\small{\frac{x+1}{1}~=~\frac{y-4}{1}~=~\frac{z-5}{2}}$.
Solution:
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~3\hat{i}+5\hat{j}+4\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~\hat{i}+\hat{j}+2\hat{k}}$
• Reason: The direction ratios in Cartesian form, can be used as the scalars of a vector parallel to the line.

2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\theta}    & {~=~}    &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{(3)(1)~+~(5)(1)~+~(4)(2)}{\sqrt{3^2 + (5)^2 + (4)^2}\,\sqrt{1^2 + (1)^2 + (2)^2}}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{16}{\sqrt{50}\,\sqrt{6}}~=~\frac{16}{(5)\sqrt{2}\,\sqrt{6}}~=~\frac{8\sqrt{2} \sqrt{2}}{(5)\sqrt{2}\,\sqrt{3}\sqrt{2}}~=~\frac{8}{5\sqrt{3}}}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\frac{8\sqrt{3}}{5\sqrt{3}\,\sqrt{3}}~=~\frac{8\sqrt{3}}{15}}    \\
{~\color{magenta}    5    }    &{\Rightarrow}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{8}{5\sqrt{3}} \right)}    \\
\end{array}}$

Solved example 27.18
Find the angle between the following pairs of lines
(i) $\small{\frac{x-2}{2}~=~\frac{y-1}{5}~=~\frac{z+3}{-3}}$ and
$\small{\frac{x+2}{-1}~=~\frac{y-4}{8}~=~\frac{z-5}{4}}$
(ii) $\small{\frac{x}{2}~=~\frac{y}{2}~=~\frac{z}{1}}$ and
$\small{\frac{x-5}{4}~=~\frac{y-2}{1}~=~\frac{z-3}{8}}$
Solution:
Part (i):
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~2\hat{i}+5\hat{j}-3\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~-\hat{i}+8\hat{j}+4\hat{k}}$
• Reason: The direction ratios in Cartesian form, can be used as the scalars of a vector parallel to the line.

2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\theta}    & {~=~}    &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{(2)(-1)~+~(5)(8)~+~(-3)(4)}{\sqrt{2^2 + (5)^2 + (-3)^2}\,\sqrt{(-1)^2 + (8)^2 + (4)^2}}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{26}{\sqrt{38}\,\sqrt{81}}~=~\frac{26}{9\sqrt{38}}}    \\
{~\color{magenta}    4    }    &{\Rightarrow}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{26}{9\sqrt{38}} \right)}    \\
\end{array}}$

Part (ii):
1. Let us write $\small{\vec{p}~\text{and}~\vec{q}}$:
• For the first line we can write:
$\small{\vec{p}~=~2\hat{i}+2\hat{j}+\hat{k}}$
• For the second line we can write:
$\small{\vec{q}~=~4\hat{i}+\hat{j}+8\hat{k}}$
• Reason: The direction ratios in Cartesian form, can be used as the scalars of a vector parallel to the line.

2. Now we can write the angle:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\theta}    & {~=~}    &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{(2)(4)~+~(2)(1)~+~(1)(8)}{\sqrt{2^2 + (2)^2 + (1)^2}\,\sqrt{(4)^2 + (1)^2 + (8)^2}}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{18}{\sqrt{9}\,\sqrt{81}}~=~\frac{18}{(3)\,(9)}~=~\frac{2}{3}}    \\
{~\color{magenta}    4    }    &{\Rightarrow}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{2}{3} \right)}    \\
\end{array}}$

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In the next section, we will see condition for two lines to be parallel or perpendicular.

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