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:
$\small{l^2 + m^2 +n^2 = 1}$

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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:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l_1 l_2~+~m_1 m_2~+~n_1 n_2}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left(\lambda_1 a_1 \right)\left(\lambda_2 a_2 \right)~+~\left(\lambda_1 b_1 \right)\left(\lambda_2 b_2 \right)~+~\left(\lambda_1 c_1 \right)\left(\lambda_2 c_2 \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\lambda_1 \lambda_2\left(a_1 a_2~+~b_1 b_2~+~c_1 c_2 \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{a_1 a_2~+~b_1 b_2~+~c_1 c_2}    & {~=~}    &{0}
\\ \end{array}}$

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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:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$

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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.

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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.

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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.

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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:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\theta}    & {~=~}    &{\frac{\vec{p}.\vec{q}}{\left|\vec{p} \right|\,\left|\vec{q} \right|}}    \\
{~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{a_1 a_2 ~+~ b_1 b_2~+~c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\,\sqrt{a_2^2 + b_2^2 + c_2^2}}}    \\
\end{array}}$

---

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 $\small{L_1'~\text{and}~L_2'}$ 

---

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

---

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

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27.3 - Line Through Two Given Points

In the previous section, we saw the equation of a line in vector form and Cartesian form. We saw that, if two conditions are satisfied, we will get a unique line. The conditions are:
(i) The line must pass through a particular point $\small{U(x_1,y_1,z_1)}$.
(ii) The line must have a particular direction.

• There is another possibility for obtaining a unique line. This is when the line passes through two given points U and V.

◼ So we can write:
Two cases give unique line:
Case I: When the following two conditions are satisfied:
(i) The line must pass through a particular point $\small{U(x_1,y_1,z_1)}$.
(ii) The line must have a particular direction.
Case II: When the following single condition is satisfied:
The line must pass through two given points $\small{U(x_1,y_1,z_1)~\text{and}~V(x_2,y_2,z_2)}$.

We saw case I in the previous section. In this section, we will see case II. It can be explained in 9 steps:

1. In fig.27.6 below, the magenta line $\small{L}$ satisfies the condition because it passes through two given points $\small{U(x_1,y_1,z_1)~\text{and}~V(x_2,y_2,z_2)}$.
• Our aim is to write the equation of this magenta line $\small{L}$.

Fig.27.6

2. A random point $\small{P(x,y,z)}$ is marked on the line $\small{L}$
• The position vector of P is $\small{\vec{r}}$

3. We have the position vectors of the given points also:
• The position vector of U is $\small{\vec{u}}$
• The position vector of V is $\small{\vec{v}}$

4. Imagine that, there are two vectors between U, V and P:
• $\small{\vec{UP}}$ between U and P
• $\small{\vec{UV}}$ between U and V

5. Consider the triangle OUP. Applying the triangle rule of vector addition, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}}    & {~=~}    &{\vec{u}+\vec{UP}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{UP}}    & {~=~}    &{\vec{r} - \vec{u}}    \\
\end{array}}$

6. Consider the triangle OUV. Applying the triangle rule of vector addition, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{v}}    & {~=~}    &{\vec{u}+\vec{UV}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{UV}}    & {~=~}    &{\vec{v} - \vec{u}}    \\
\end{array}}$

7. We obtained two vectors: $\small{\vec{UP}~\text{and}~\vec{UV}}$
• Those two vectors are collinear. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{UP}}    & {~=~}    &{\lambda\,\vec{UV}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{r} - \vec{u}}    & {~=~}    &{\lambda\left(\vec{v} - \vec{u} \right)}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{r}}    & {~=~}    &{\vec{u} + \lambda\left(\vec{v} - \vec{u} \right)}    \\
\end{array}}$
• In the above equation, we can input infinite values (real numbers) for $\small{\lambda}$.
• For each value of $\small{\lambda}$, we get a unique $\small{\vec{r}}$
• Each unique $\small{\vec{r}}$ gives a unique point on the line $\small{L}$
• So infinite values of $\small{\lambda}$ will give infinite points on $\small{L}$. Those infinite points together will give us the line $\small{L}$

8. Therefore, the vector equation of the line passing through two points U and V is:
$\small{\vec{r}~=~\vec{u} + \lambda\left(\vec{v} - \vec{u} \right)}$

9. Let us see an example:
• The vector equation of the line through (−1,0,2) and (3,4,6) can be obtained in 2 steps:
(i) Write the position vectors of the given points:
• $\small{\vec{u}}$ is the position vector of the given point U(−1,0,2).
So $\small{\vec{u} = -\hat{i}+2\hat{k}}$
• $\small{\vec{v}}$ is the position vector of the given point V(3,4,6).
So $\small{\vec{v} = 3\hat{i}+4\hat{j}+6\hat{k}}$
(ii) Then the required vector equation can be obtained as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}}    & {~=~}    &{\vec{u} + \lambda\left(\vec{v} - \vec{u} \right)}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{r}}    & {~=~}    &{-\hat{i}+2\hat{k}~+~\lambda\left[3\hat{i}+4\hat{j}+6\hat{k} - \left(-\hat{i}+2\hat{k} \right) \right]}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{r}}    & {~=~}    &{-\hat{i}+2\hat{k}~+~\lambda\left(4\hat{i}+4\hat{j}+4\hat{k} \right)}    \\
\end{array}}$

Derivation of Cartesian form

This can be done in 5 steps:
1. In the fig.27.6 above, the magenta line $\small{L}$ passes through U and V.
• So the set of direction cosines of $\small{L}$ is:
$\small{\frac{x_2 - x_1}{UV},~\frac{y_2 - y_1}{UV},~\frac{z_2 - z_1}{UV}}$
    ♦ Here UV is the length of the line segment UV
(See Solved example 27.4)

2. Also in the same fig.27.6 above, the magenta line $\small{L}$ passes through U and P.
• So one of the many sets of direction ratios of $\small{L}$ is:
$\small{\left(x - x_1 \right),~\left(x - x_1 \right),~\left(x - x_1 \right)}$
(See section 27.1)

3. For any line, its direction cosines and direction ratios are proportional. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x_2 - x_1}{(x - x_1)UV }}& {~=~}    &{\frac{y_2 - y_1}{(y - y_1)UV }}    & {~=~}    &{\frac{z_2 - z_1}{(z - z_1)UV }}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{x - x_1}{x_2 - x_1}}& {~=~}    &{\frac{y - y_1}{y_2 - y_1}}    & {~=~}    &{\frac{z - z_1}{z_2 - z_1}}
\\ \end{array}}$ 
• This is the equation of $\small{L}$ in Cartesian form.

4. Let us see an example:
The Cartesian equation of the line through (−1,0,2) and (3,4,6) is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x + 1}{3 + 1}}& {~=~}    &{\frac{y - 0}{4 - 0}}    & {~=~}    &{\frac{z - 2}{6 - 2}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{x + 1}{4}}& {~=~}    &{\frac{y}{4}}    & {~=~}    &{\frac{z - 2}{4}}
\\ \end{array}}$

5. Now we will see an interesting fact. It can be written in (ii) steps:
(i) Consider the Cartesian equation obtained in (4) above. The three fractions are equal. So we can write:
$\small{\frac{x + 1}{4}~=~\frac{y}{4}~=~\frac{z - 2}{4} = \lambda}$
(ii) $\small{\lambda}$ can be any real number.
• Let us put $\small{\lambda = -1}$. Then we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x + 1}{4}}    & {~=~}    &{-1}    \\
{~\color{magenta}    {}    }    &{\Rightarrow}    &{x}    & {~=~}    &{-5}    \\
{~\color{magenta}    2    }    &{{}}    &{\frac{y}{4}}    & {~=~}    &{-1}    \\
{~\color{magenta}    {}    }    &{\Rightarrow}    &{y}    & {~=~}    &{-4}    \\
{~\color{magenta}    3    }    &{{}}    &{\frac{z - 2}{4}}    & {~=~}    &{-1}    \\
{~\color{magenta}   {}     }    &{\Rightarrow}    &{z}    & {~=~}    &{-2}    \\
\end{array}}$

• So (−5,−4,−2) is a point on the line. This is shown in the actual plot of $\small{L}$ in the fig.27.7 below:

Fig.27.7

Alternate method to derive the Cartesian form

This can be written in 4 steps:
1. Based on fig.27.6 above, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}}    & {~=~}    &{(x_1)\hat{i}+(y_1)\hat{j}+(z_1)\hat{k}}    \\
{~\color{magenta}    2    }    &{{}}    &{\vec{v}}    & {~=~}    &{(x_2)\hat{i}+(y_2)\hat{j}+(z_2)\hat{k}}    \\
{~\color{magenta}    3    }    &{{}}    &{\vec{r}}    & {~=~}    &{x\hat{i}+y\hat{j}+z\hat{k}}    \\
\end{array}}$

2. We have the vector form:
$\small{\vec{r}~=~\vec{u} + \lambda\left(\vec{v} - \vec{u} \right)}$
• Substituting the vectors, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}}    & {~=~}    &{\vec{u} + \lambda\left(\vec{v} - \vec{u} \right)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x\hat{i}+y\hat{j}+z\hat{k}}    & {~=~}    &{(x_1)\hat{i}+(y_1)\hat{j}+(z_1)\hat{k}}    \\
{~\color{magenta}    {}    }    &{{}}    &{}    & {{}}    &{+~\lambda\Big[(x_2)\hat{i}+(y_2)\hat{j}+(z_2)\hat{k}~-~\left[(x_1)\hat{i}+(y_1)\hat{j}+(z_1)\hat{k} \right] \Big]}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x\hat{i}+y\hat{j}+z\hat{k}}    & {~=~}    &{(x_1)\hat{i}+(y_1)\hat{j}+(z_1)\hat{k}}    \\
{~\color{magenta}    {}    }    &{{}}    &{}    & {{}}    &{+~\lambda\Big[(x_2 - x_1)\hat{i}+(y_2 - y_1)\hat{j}+(z_2 - z_1)\hat{k} \Big]}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x\hat{i}+y\hat{j}+z\hat{k}}    & {~=~}    &{(x_1)\hat{i}+(y_1)\hat{j}+(z_1)\hat{k}}    \\
{~\color{magenta}    {}    }    &{{}}    &{}    & {{}}    &{+~\lambda(x_2 - x_1)\hat{i}+\lambda(y_2 - y_1)\hat{j}+\lambda(z_2 - z_1)\hat{k} }    \\
\end{array}}$

3. Equating the coefficients of $\small{\hat{i},~\hat{j}~\text{and}~\hat{k}}$, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x}    & {~=~}    &{x_1+ \lambda(x_2 - x_1)}    \\
{~\color{magenta}    2    }    &{{}}    &{y}    & {~=~}    &{y_1+ \lambda(y_2 - y_1)}    \\
{~\color{magenta}    3    }    &{{}}    &{z}    & {~=~}    &{z_1+ \lambda(z_2 - z_1)}    \\
\end{array}}$

4. Eliminating $\small{\lambda}$, we get:
$\small{\frac{x - x_1}{x_2 - x_1}~=~\frac{y - y_1}{y_2 - y_1}~=~\frac{z - z_1}{z_2 - z_1}}$

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Now we will see some solved examples

Solved example 27.13
Find the vector and Cartesian equations of the line that passes through the origin and (5,−2,3).
Solution:
Part (i): Vector form
1. Write the position vectors of the given points:
• $\small{\vec{u}}$ is the position vector of the given point U(0,0,0).
So $\small{\vec{u} = 0\hat{i}+0\hat{j}+0\hat{k}}$
• $\small{\vec{v}}$ is the position vector of the given point V(5,−2,3).
So $\small{\vec{v} = 5\hat{i}-2\hat{j}+3\hat{k}}$

2. Then the required vector equation can be obtained as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}}    & {~=~}    &{\vec{u} + \lambda\left(\vec{v} - \vec{u} \right)}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{r}}    & {~=~}    &{0\hat{i}+0\hat{j}+0\hat{k}~+~\lambda\left[5\hat{i}-2\hat{j}+3\hat{k} - \left(0\hat{i}+0\hat{j}+0\hat{k} \right) \right]}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{r}}    & {~=~}    &{\lambda\left(5\hat{i}-2\hat{j}+3\hat{k} \right)}    \\
\end{array}}$

Part (ii): Cartesian form
The Cartesian equation of the line through (0,0,0) and (5,−2,3) is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x - x_1}{x_2 - x_1}}& {~=~}    &{\frac{y - y_1}{y_2 - y_1}}    & {~=~}    &{\frac{z - z_1}{z_2 - z_1}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{x -0}{5 - 0}}& {~=~}    &{\frac{y - 0}{-2 - 0}}    & {~=~}    &{\frac{z - 0}{3 - 0}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x}{5}}& {~=~}    &{\frac{y}{-2}}    & {~=~}    &{\frac{z}{3}}
\\ \end{array}}$

Solved example 27.14
Find the vector and Cartesian equations of the line that passes through the points (3,−2,−5) and (3,−2,6).
Solution:
Part (i): Vector form
1. Write the position vectors of the given points:
• $\small{\vec{u}}$ is the position vector of the given point U(3,−2,−5).
So $\small{\vec{u} = 3\hat{i}-2\hat{j}-5\hat{k}}$
• $\small{\vec{v}}$ is the position vector of the given point V(3,−2,6).
So $\small{\vec{v} = 3\hat{i}-2\hat{j}+6\hat{k}}$

2. Then the required vector equation can be obtained as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}}    & {~=~}    &{\vec{u} + \lambda\left(\vec{v} - \vec{u} \right)}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{r}}    & {~=~}    &{3\hat{i}-2\hat{j}-5\hat{k}~+~\lambda\left[3\hat{i}-2\hat{j}+6\hat{k} - \left(3\hat{i}-2\hat{j}-5\hat{k} \right) \right]}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{r}}    & {~=~}    &{3\hat{i}-2\hat{j}-5\hat{k}~+~\lambda\left(11\hat{k} \right)}    \\
\end{array}}$

Part (ii): Cartesian form
The Cartesian equation of the line through (3,−2,−5) and (3,−2,6) is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x - x_1}{x_2 - x_1}}& {~=~}    &{\frac{y - y_1}{y_2 - y_1}}    & {~=~}    &{\frac{z - z_1}{z_2 - z_1}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{x -3}{3 - 3}}& {~=~}    &{\frac{y +2}{-2 + 2}}    & {~=~}    &{\frac{z +5}{6 + 5}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x-3}{0}}& {~=~}    &{\frac{y+2}{0}}    & {~=~}    &{\frac{z+5}{11}}
\\ \end{array}}$

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

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