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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{a_1 a_2~+~b_1 b_2~+~c_1 c_2} & {~=~} &{0}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\left(-3 \right)\left((-3/7)p\right)+\left((2/7)p \right)\left(1 \right)+\left(2 \right)\left(-5 \right)} & {~=~} &{0}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\frac{9p}{7} + \frac{2p}{7} - 10} & {~=~} &{0}
\\ {~\color{magenta} 4 } &{\Rightarrow} &{\frac{11p - 70}{7}} & {~=~} &{0}
\\ {~\color{magenta} 5 } &{\Rightarrow} &{11p - 70} & {~=~} &{0}
\\ {~\color{magenta} 6 } &{\Rightarrow} &{p} & {~=~} &{\frac{70}{11}}
\\ \end{array}}$
In the next section, we will see the shortest distance between two lines.
Previous
Contents
Next
Copyright©2026 Higher secondary mathematics.blogspot.com
