27.1 - Direction Ratios of Two Parallel Lines

In the previous section, we saw the basic details about direction cosines and direction ratios. We saw the direction cosines of the line through two points P and Q. In this section, we will see more details about direction ratios.

Direction ratios of a line passing through two points

This can be explained in 6 steps:
1. Let $\small{P(x_1,x_2,x_3)~\text{and}~Q(x_1,x_2,x_3)}$ be two points in space.
2. We obtained the direction cosines of the line passing through P and Q:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l~=~\cos \alpha}    & {~=~}    &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{}    &{m~=~\cos \beta}    & {~=~}    &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{}    &{n~=~\cos \gamma}    & {~=~}    &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$
3. If we can find any three numbers a, b and c which satisfies the condition
$\small{\frac{l}{a}~=~\frac{m}{b}~=~\frac{n}{c}}$, then we can say that a, b and c are the direction cosines of the line through P and Q
4. Let:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a}    & {~=~}    &{x_2 - x_1}
\\ {~\color{magenta}    2    }    &{}    &{b}    & {~=~}    &{y_2 - y_1}
\\ {~\color{magenta}    3    }    &{}    &{c}    & {~=~}    &{z_2 - z_1}
\\ \end{array}}$
5. We will check whether this a, b and c, satisfy the condition written in (3):
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{l}{a}}& {~=~}    &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|\left(x_2 - x_1 \right)}}    & {~=~}    &{\frac{1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{m}{b}}& {~=~}    &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|\left(y2 - y_1 \right)}}    & {~=~}    &{\frac{1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{n}{c}}& {~=~}    &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|\left(z_2 - z_1 \right)}}    & {~=~}    &{\frac{1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$
• The condition is satisfied.
• So $\small{\left(x_2 - x_1 \right),~\left(y_2 - y_1 \right),~\left(z_2 - z_1 \right)}$ are indeed direction ratios of the line through $\small{(x_1,x_2,x_3)~\text{and}~(x_1,x_2,x_3)}$
6. Note that, $\small{\left(x_1 - x_2 \right),~\left(y_1 - y_2 \right),~\left(z_1 - z_2 \right)}$ are also direction ratios of the line through $\small{(x_1,x_2,x_3)~\text{and}~(x_1,x_2,x_3)}$.
• This is shown below:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{l}{a}}& {~=~}    &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|\left(x_1 - x_2 \right)}}    & {~=~}    &{\frac{-1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{m}{b}}& {~=~}    &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|\left(y1 - y_2 \right)}}    & {~=~}    &{\frac{-1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{n}{c}}& {~=~}    &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|\left(z_1 - z_2 \right)}}    & {~=~}    &{\frac{-1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$

Lines having proportional direction ratios

This can be explained in 10 steps:
1. Consider two directed lines L₁ and L₂
L₁ has:
direction cosines $\small{l_1,~m_1,~n_1}$
direction ratios $\small{a_1,~b_1,~c_1}$
L₂ has:
direction cosines $\small{l_2,~m_2,~n_2}$
direction ratios $\small{a_2,~b_2,~c_2}$

2. For L₁, we can write: $\small{\frac{l_1}{a_1}~=~\frac{m_1}{b_1}~=~\frac{n_1}{c_1}}$

Since the three fractions are equal, we can write:
$\small{\frac{l_1}{a_1}~=~\frac{m_1}{b_1}~=~\frac{n_1}{c_1}~=~\lambda_1}$

3. For L₂, we can write: $\small{\frac{l_2}{a_2}~=~\frac{m_2}{b_2}~=~\frac{n_2}{c_2}}$

Since the three fractions are equal, we can write:
$\small{\frac{l_2}{a_2}~=~\frac{m_2}{b_2}~=~\frac{n_2}{c_2}~=~\lambda_2}$

4. Suppose that, the direction ratios of the two lines satisfy the condition:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$

Since the three fractions are equal, we can write:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}~=~\lambda_3}$

5. Substituting from (4) into (2), we get:
$\small{\frac{l_1}{a_2 \lambda_3}~=~\frac{m_1}{b_2 \lambda_3}~=~\frac{n_1}{c_2 \lambda_3}~=~\lambda_1}$

6. Substituting from (3) into (5), we get:
$\small{\frac{l_1}{\left(\frac{l_2}{\lambda_2} \right) \lambda_3}~=~\frac{m_1}{\left(\frac{m_2}{\lambda_2} \right) \lambda_3}~=~\frac{n_1}{\left(\frac{n_2}{\lambda_2} \right) \lambda_3}~=~\lambda_1}$

$\small{\Rightarrow \frac{l_1 \lambda_2}{l_2 \lambda_3}~=~\frac{m_1 \lambda_2}{m_2 \lambda_3}~=~\frac{n_1 \lambda_2}{n_2 \lambda_3}~=~\lambda_1}$

• Multiplying throughout by $\small{\left(\frac{\lambda_3}{\lambda_2} \right)}$, we get:

$\small{\frac{l_1}{l_2}~=~\frac{m_1}{m_2}~=~\frac{n_1}{n_2}~=~\lambda_1\left(\frac{\lambda_3}{\lambda_2} \right)~=~\lambda_4}$

• Note that $\small{\lambda_1,~\lambda_2}$ . . . etc., are just some real numbers. Multiplying them will give new real numbers.

7. So we get an important result:
When direction ratios of two lines satisfy the condition $\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}~=~\lambda_3}$,
their direction raios will satisfy the condition
$\small{\frac{l_1}{l_2}~=~\frac{m_1}{m_2}~=~\frac{n_1}{n_2}~=~\lambda_4}$  

We can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{b_1}{b_2}}    & {~=~}    &{\frac{c_1}{c_2}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{l_1}{l_2}}& {~=~}    &{\frac{m_1}{m_2}}    & {~=~}    &{\frac{n_1}{n_2}}
\\ \end{array}}$  

8. We have the direction cosines of L₁ and L₂. So we can write unit vectors parallel to them. We get:

$\small{\hat{L_1}= l_1\hat{i}+m_1\hat{j}+n_1\hat{k}}$

$\small{\hat{L_2}= l_2\hat{i}+m_2\hat{j}+n_2\hat{k}}$

9. Substituting from (7) into (8), we get:
$\small{\hat{L_1}= \left(l_2 \lambda_4 \right)\hat{i}+\left(m_2 \lambda_4 \right)\hat{j}+\left(n_2 \lambda_4 \right)\hat{k}}$

$\small{\Rightarrow \hat{L_1}= \lambda_4\left(l_2\hat{i}+m_2\hat{j}+n_2\hat{k} \right)}$

$\small{\Rightarrow \text{Unit vectors}~\hat{L_1}~\text{and}~\hat{L_2}~\text{are parallel}}$

$\small{\Rightarrow \text{Lines}~L_1~\text{and}~L_2~\text{are parallel}}$

10. So we can write the final result:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{b_1}{b_2}}    & {~=~}    &{\frac{c_1}{c_2}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{l_1}{l_2}}& {~=~}    &{\frac{m_1}{m_2}}    & {~=~}    &{\frac{n_1}{n_2}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{L_1}& {~~\parallel~}    &{L_2}    & {{}}    &{{}}
\\ \end{array}}$  

◼ Remarks:
• 1 (magenta color): This line gives the condition. It indicates that, the direction ratios of the two lines are proportional
• 2 (magenta color): If the condition is satisfied, the direction cosines of the two lines will be proportional.
• 3 (magenta color): If the condition is satisfied, the two lines will be parallel to each other.

   

Solved example 27.6
Show that the points A(2,3,−4), B(1,−2,3) and C(3,8,−11) are collinear.
Solution:
1. First we find the direction ratios of AB:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(1-2) = -1}
\\ {~\color{magenta}    2    }    &{{}}    &{b_1}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(-2-3) = -5}
\\ {~\color{magenta}    3    }    &{{}}    &{c_1}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(3-(-4)) = 7}
\\ \end{array}}$  

2. Next we find the direction ratios of BC:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_2}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(3-1) = 2}
\\ {~\color{magenta}    2    }    &{{}}    &{b_2}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(8-(-2)) = 10}
\\ {~\color{magenta}    3    }    &{{}}    &{c_2}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(-11-3) = -14}
\\ \end{array}}$ 

3. Next we calculate the three ratios:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{-1}{2}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{b_1}{b_2}}& {~=~}    &{\frac{-5}{10}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{c_1}{c_2}}& {~=~}    &{\frac{7}{-14}}    & {~=~}    &{-\frac{1}{2}}
\\ \end{array}}$  

4. We see that:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$
• That means, the direction ratios of the two lines AB and BC, are proportional.
• That means, AB and BC are parallel.

5. The two parallel lines AB and BC have one point B in common. So the points A, B and C are collinear.

Solved example 27.7
Show that the points (2,3,4), (−1,−2,1) and (5,8,7) are collinear.
Solution:
1. Let the three points be: A(2,3,4), B(−1,−2,1) and C(5,8,7)
• First we find the direction ratios of AB:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(-1-2) = -3}
\\ {~\color{magenta}    2    }    &{{}}    &{b_1}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(-2-3) = -5}
\\ {~\color{magenta}    3    }    &{{}}    &{c_1}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(1-4) = -3}
\\ \end{array}}$  

2. Next we find the direction ratios of BC:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_2}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(5-(-1)) = 6}
\\ {~\color{magenta}    2    }    &{{}}    &{b_2}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(8-(-2)) = 10}
\\ {~\color{magenta}    3    }    &{{}}    &{c_2}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(7-1) = 6}
\\ \end{array}}$ 

3. Next we calculate the three ratios:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{-3}{6}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{b_1}{b_2}}& {~=~}    &{\frac{-5}{10}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{c_1}{c_2}}& {~=~}    &{\frac{-3}{6}}    & {~=~}    &{-\frac{1}{2}}
\\ \end{array}}$  

4. We see that:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$
• That means, the direction ratios of the two lines AB and BC, are proportional.
• That means, AB and BC are parallel.

5. The two parallel lines AB and BC have one point B in common. So the points A, B and C are collinear.

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The link below gives a few more miscellaneous examples:

Exercise 27.1

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In the next section, we will see equation of a line in space.

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