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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• 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:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(1-2) = -1}
\\ {~\color{magenta}    2    }    &{{}}    &{b_1}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(-2-3) = -5}
\\ {~\color{magenta}    3    }    &{{}}    &{c_1}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(3-(-4)) = 7}
\\ \end{array}}$  

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)

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:

Miscellaneous Exercise

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

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27.2 - Equation of A Line in Space

In the previous section, we saw the effect when direction ratios of two lines are proportional. In this section, we will see equation of line.

Equation of a line in space

In Class XI, we have seen the equation of line in two dimensional plane. Now we will see the equation in three dimensional space.
It can be explained in 10 steps:
1. Suppose that, we want a line to satisfy a condition:
The line must pass through a particular point $\small{U(x_1,y_1,z_1)}$.
• But here we encounter a problem. Infinite lines can pass through U. Because there are infinite directions possible through U. We cannot think about a unique line which satisfies the condition.

2. So we add one more condition. The two 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 will be one and only one line which can satisfy the two conditions simultaneously. Our aim is to write the equation of such a line.

3. In fig.27.4 below, the magenta line $\small{L}$ satisfies two conditions.
(i) It passes through the given point $\small{U(x_1,y_1,z_1)}$
(ii) It is parallel to the given vector $\small{\vec{v}}$
• Our aim is to write the equation of this magenta line $\small{L}$.

Fig.27.4

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

5. The position vector of U is $\small{\vec{u}}$

6. Imagine that, there is a vector $\small{\vec{UP}}$ between U and P
• $\small{\vec{UP}}$ is parallel to $\small{\vec{v}}$
• So we can write: $\small{\vec{UP}~=~\lambda \vec{v}}$
Where $\small{\lambda}$ is some real number.

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

8. 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 $\small{L}$

9. Therefore, the vector equation of the line passing through U and parallel to $\small{\vec{v}}$ is: $\small{\vec{r}~=~\vec{u}+\lambda\vec{v}}$

10. Let us see an example:
• The vector equation of the line through (5,2,−4) and which is parallel to the vector $\small{3\hat{i}+2\hat{j}-8\hat{k}}$ can be obtained in 2 steps:
(i) $\small{\vec{u}}$ is the position vector of the given point (5,2,−4). So $\small{\vec{u} = 5\hat{i}+2\hat{j}-4\hat{k}}$
(ii) Then the required vector equation is:
$\small{5\hat{i}+2\hat{j}-4\hat{k}~+~\lambda \left(3\hat{i}+2\hat{j}-8\hat{k} \right)}$

Derivation of Cartesian form

This can be done in 7 steps:
1. In the fig.27.4 above, $\small{\vec{v}}$ is parallel to the magenta line $\small{L}$. Let the component form of the vector be: $\small{\vec{v}=v_1 \hat{i}+v_2\hat{j}+v_3\hat{k}}$.
• Based on the component form of $\small{\vec{v}}$, we can write:
Vector $\small{\vec{v}}$ has a set of direction ratios: $\small{v_1,~v_2,~v_3}$. This we proved in the previous section.

2. Suppose another line $\small{L_1}$ is also parallel to $\small{\vec{v}}$. Then $\small{L_1}$ will also have a set of direction ratios $\small{a_{L1},~b_{L1},~c_{L1}~~\text{as:}~~v_1,~v_2,~v_3}$.

3. $\small{L_1}$ will be parallel to $\small{L}$.
• Then the direction ratios of $\small{L~\text{and}~L_1}$ are proportional. This we proved in the previous section. We can write:
$\small{\frac{a_L}{a_{L1}}~=~\frac{b_L}{b_{L1}}~=~\frac{b_L}{b_{L1}}}$

4. So our next task is to find $\small{a_{L},~b_{L},~c_{L}}$
• We have two points $\small{U(x_1,y_1,z_1)~\text{and}~P(x,y,z)}$ on the line $\small{L}$
• Then a set of direction ratios of $\small{L}$ can be taken as: $\small{(x – x_1),~(y – y_1),~(z – z_1)}$

5. So from (3), we get:
$\small{\frac{x – x_1}{v_1}~=~\frac{y – y_1}{v_2}~=~\frac{z – z_1}{v_3}}$
• This is the equation of $\small{L}$ in Cartesian form.

6. Let us see an example:
The Cartesian equation of the line through (5,2,−4) and which is parallel to the vector $\small{3\hat{i}+2\hat{j}-8\hat{k}}$ can be obtained in 3 steps:
(i) The required line $\small{L}$, passes through (5,2,−4).
• Let (x,y,z) be any point on the line.
• Then (x−5), (y−2), (z+4) is a set of direction ratios of the required line.
(ii) Any line $\small{L_1}$ parallel to $\small{3\hat{i}+2\hat{j}-8\hat{k}}$, will have a set of direction ratios 3, 2, −8
(iii) The lines $\small{L ~\text{and}~ L_1}$ are parallel.
• So the direction ratios of the two lines will be proportional. Therefore, we can write:
$\small{\frac{x – 5}{3}~=~\frac{y – 2}{2}~=~\frac{z + 4}{-8}}$
• This is the equation of $\small{L}$ in Cartesian form.

7. Now we will see an interesting fact. It can be written in (ii) steps:
(i) Consider the Cartesian equation obtained in (6) above. The three fractions are equal. So we can write:
$\small{\frac{x – 5}{3}~=~\frac{y – 2}{2}~=~\frac{z + 4}{-8} = \lambda}$
(ii) $\small{\lambda}$ can be any real number.
• Let us put $\small{\lambda = -1}$. Then we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x – 5}{3}}    & {~=~}    &{-1}    \\
{~\color{magenta}     {}   }    &{\Rightarrow}    &{x}    & {~=~}    &{2}    \\
{~\color{magenta}    2    }    &{{}}    &{\frac{y – 2}{2}}    & {~=~}    &{-1}    \\
{~\color{magenta}     {}   }    &{\Rightarrow}    &{y}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{}}    &{\frac{z + 4}{-8}}    & {~=~}    &{-1}    \\
{~\color{magenta}     {}   }    &{\Rightarrow}    &{z}    & {~=~}    &{4}    \\
\end{array}}$

• So (2,0,4) is a point on the line. This is shown in the actual plot of $\small{L}$ in the fig.27.5 below:

Fig.27.5

• Note that in the fig. above, $\small{L}$ is parallel to $\small{\vec{v}}$, and is passing through U(5,2,−4)

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Alternate method to derive the Cartesian form

This can be written in 4 steps:
1. Based on fig.27.4 above, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}}    & {~=~}    &{(x_1)\hat{i}+(y_1)\hat{j}+(z_1)\hat{k}}    \\
{~\color{magenta}    2    }    &{{}}    &{\vec{v}}    & {~=~}    &{(v_1)\hat{i}+(v_2)\hat{j}+(v_3)\hat{k}}    \\
{~\color{magenta}    3    }    &{{}}    &{\vec{r}}    & {~=~}    &{x\hat{i}+y\hat{j}+z\hat{k}}    \\
\end{array}}$

2. We have the vector form:
$\small{\vec{r}~=~\vec{u}+\lambda\vec{v}}$
• Substituting the vectors, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}}    & {~=~}    &{\vec{u}+\lambda\vec{v}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x\hat{i}+y\hat{j}+z\hat{k}}    & {~=~}    &{(x_1)\hat{i}+(y_1)\hat{j}+(z_1)\hat{k}+\lambda\left[(v_1)\hat{i}+(v_2)\hat{j}+(v_3)\hat{k} \right]}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x\hat{i}+y\hat{j}+z\hat{k}}    & {~=~}    &{\left[x_1+ \lambda(v_1)\right]\hat{i}+\left[y_1+ \lambda(v_2)\right]\hat{j}+\left[z_1+ \lambda(v_3)\right]\hat{k}}    \\
\end{array}}$

3. Equating the coefficients of $\small{\hat{i},~\hat{j}~\text{and}~\hat{k}}$, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x}    & {~=~}    &{x_1+ \lambda(v_1)}    \\
{~\color{magenta}    2    }    &{{}}    &{y}    & {~=~}    &{y_1+ \lambda(v_2)}    \\
{~\color{magenta}    3    }    &{{}}    &{z}    & {~=~}    &{z_1+ \lambda(v_3)}    \\
\end{array}}$

4. Eliminating $\small{\lambda}$, we get:
$\small{\frac{x - x_1}{v_1}~=~\frac{y - y_1}{v_2}~=~\frac{z - z_1}{v_3}}$

---

Now we will see some solved examples

Solved example 27.8
Find the equation of the line which passes through the point (1,2,3) and is parallel to the vector $\small{3\hat{i}+2\hat{j}-2\hat{k}}$.
Solution:
Part (i): Derivation of vector form
1. $\small{\vec{u}}$ is the position vector of the given point U(1,2,3). So $\small{\vec{u} = \hat{i}+2\hat{j}+3\hat{k}}$
2. Then the required vector equation is:
$\small{\vec{r}=\hat{i}+2\hat{j}+3\hat{k}~+~\lambda \left(3\hat{i}+2\hat{j}-2\hat{k} \right)}$

Part (ii): Derivation of Cartesian form
1. The required line $\small{L}$, passes through (1,2,3).
• Let (x,y,z) be any point on the line.
• Then (x−1), (y−2), (z−3) is a set of direction ratios of the required line.
(ii) Any line $\small{L_1}$ parallel to $\small{3\hat{i}+2\hat{j}-2\hat{k}}$, will have a set of direction ratios 3, 2, −2
(iii) The lines $\small{L ~\text{and}~ L_1}$ are parallel.
• So the direction ratios of the two lines will be proportional. Therefore, we can write:
$\small{\frac{x – 1}{3}~=~\frac{y – 2}{2}~=~\frac{z - 3}{-2}}$
• This is the equation of $\small{L}$ in Cartesian form. 

Solved example 27.9
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $\small{2\hat{i}-\hat{j}+4\hat{k}}$ and is in the direction $\small{\hat{i}+2\hat{j}-\hat{k}}$.
Solution:
Part (i): Derivation of vector form
1. $\small{\vec{u}}$ is the position vector of the given point. This vector is already given to us. So $\small{\vec{u} = 2\hat{i}-\hat{j}+4\hat{k}}$
2. Then the required vector equation is:
$\small{\vec{r}=2\hat{i}-\hat{j}+4\hat{k}~+~\lambda \left(\hat{i}+2\hat{j}-\hat{k} \right)}$

Part (ii): Derivation of Cartesian form
1. Based on the given position vector, we can write:
The required line $\small{L}$, passes through (2,−1,4).
• Let (x,y,z) be any point on the line.
• Then (x−2), (y+1), (z−4) is a set of direction ratios of the required line.
(ii) Any line $\small{L_1}$ parallel to $\small{\hat{i}+2\hat{j}-\hat{k}}$, will have a set of direction ratios 1, 2, −1
(iii) The lines $\small{L ~\text{and}~ L_1}$ are parallel.
• So the direction ratios of the two lines will be proportional. Therefore, we can write:
$\small{\frac{x – 2}{1}~=~\frac{y +1}{2}~=~\frac{z - 4}{-1}}$
• This is the equation of $\small{L}$ in Cartesian form. 

Solved example 27.10
Find the Cartesian equation of the line that passes through the point (−2,4,−5) and parallel to the line given by $\small{\frac{x+3}{3}~=~\frac{y-4}{5}~=~\frac{x+8}{6}}$.
Solution:
1. The Cartesian form of a line passing through $\small{U\left(x_1,y_1,z_1 \right)}$, and is parallel to the vector $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$ is:
$\small{\frac{x-x_1}{v_1}~=~\frac{y-y_1}{v_2}~=~\frac{z-z_1}{v_3}}$

2. $\small{v_2,~v_3,~v_4}$ can be taken as the direction ratios of any line parallel to $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$

3. In our present problem, the required line $\small{L}$ must be parallel to the line $\small{\frac{x+3}{3}~=~\frac{y-4}{5}~=~\frac{x+8}{6}}$
• This line has direction ratios 3, 5, 6
• So the required line $\small{L}$ must be parallel to the vector $\small{\vec{v}=3 \hat{i}+5 \hat{j}+6 \hat{k}}$  

4. The required line $\small{L}$ is said to pass through (−2,4,−5).
• So this line has the Cartesian equation:
$\small{\frac{x+2}{3}~=~\frac{y-4}{5}~=~\frac{z+5}{6}}$

Solved example 27.11
The Cartesian equation of a line is
$\small{\frac{x+3}{2}~=~\frac{y-5}{4}~=~\frac{x+6}{2}}$
Write its vector form
Solution:
1. The Cartesian form of a line passing through $\small{U\left(x_1,y_1,z_1 \right)}$, and is parallel to the vector $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$ is:
$\small{\frac{x-x_1}{v_1}~=~\frac{y-y_1}{v_2}~=~\frac{z-z_1}{v_3}}$

2. $\small{v_2,~v_3,~v_4}$ can be taken as the direction ratios of any line parallel to $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$

3. So in our present problem, the required line $\small{L}$ must be parallel to the vector
$\small{\vec{v}=2 \hat{i}+4 \hat{j}+2 \hat{k}}$

4. The given Cartesian form is:
$\small{\frac{x+3}{2}~=~\frac{y-5}{4}~=~\frac{x+6}{2}}$
• So the given line passes through $\small{U(-3,5,-6)}$
• So the position vector of $\small{U}$ is:
$\small{\vec{u}=-3\hat{i}+5\hat{j}-6\hat{k}}$

5. The general vector form is: $\small{\vec{r}=\vec{u}+\lambda \vec{v}}$
• So for the present problem, we get:
$\small{\vec{r}=-3\hat{i}+5\hat{j}-6\hat{k}+\lambda \left(2 \hat{i}+4 \hat{j}+2 \hat{k} \right)}$

Easy method:
After becoming familiar with the above five steps, the reader may use the easy method. It can be written in three steps:
1. From the numerators of the given Cartesian form, we get the coordinates of U. These coordinates are the scalar components of $\small{\vec{u}}$.
• So we get: $\small{\vec{u}=-3\hat{i}+5\hat{j}-6\hat{k}}$
2. From the denominators of the given Cartesian form, we get the scalar components of $\small{\vec{v}}$.
• So we get: $\small{\vec{v}=2 \hat{i}+4 \hat{j}+2 \hat{k}}$
3. Now we can easily write $\small{\vec{r}=\vec{u}+\lambda \vec{v}}$

Solved example 27.12
The Cartesian equation of a line is
$\small{\frac{x-5}{2}~=~\frac{y+4}{4}~=~\frac{x-6}{2}}$
Write its vector form
Solution:
1. The Cartesian form of a line passing through $\small{U\left(x_1,y_1,z_1 \right)}$, and is parallel to the vector $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$ is:
$\small{\frac{x-x_1}{v_1}~=~\frac{y-y_1}{v_2}~=~\frac{z-z_1}{v_3}}$

2. $\small{v_2,~v_3,~v_4}$ can be taken as the direction ratios of any line parallel to $\small{\vec{v}=v_1 \hat{i}+v_2 \hat{j}+v_3 \hat{k}}$

3. So in our present problem, the required line $\small{L}$ must be parallel to the vector
$\small{\vec{v}=3 \hat{i}+7 \hat{j}+2 \hat{k}}$

4. The given Cartesian form is:
$\small{\frac{x-5}{2}~=~\frac{y+4}{4}~=~\frac{x-6}{2}}$
• So the given line passes through $\small{U(5,-4,6)}$
• So the position vector of $\small{U}$ is:
$\small{\vec{u}=5\hat{i}-4\hat{j}+6\hat{k}}$

5. The general vector form is: $\small{\vec{r}=\vec{u}+\lambda \vec{v}}$
• So for the present problem, we get:
$\small{\vec{r}=5\hat{i}-4\hat{j}+6\hat{k}+\lambda \left(3 \hat{i}+7 \hat{j}+2 \hat{k} \right)}$

Easy method:
After becoming familiar with the above five steps, the reader may use the easy method. It can be written in three steps:
1. From the numerators of the given Cartesian form, we get the coordinates of U. These coordinates are the scalar components of $\small{\vec{u}}$.
• So we get: $\small{\vec{u}=5\hat{i}-4\hat{j}+6\hat{k}}$
2. From the denominators of the given Cartesian form, we get the scalar components of $\small{\vec{v}}$.
• So we get: $\small{\vec{v}=3 \hat{i}+7 \hat{j}+2 \hat{k}}$
3. Now we can easily write $\small{\vec{r}=\vec{u}+\lambda \vec{v}}$

---

In the next section, we will see equation of a line passing through two given points.

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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 ratios 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 ratios 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 complete the result in (7). We get:

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

◼ 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.

Direction ratios of a line, when the line is parallel to a vector

This can be explained in 5 steps:
1. Consider a line $\small{L}$, which is parallel to a vector:
$\small{\vec{v}=v_1 \hat{i}+v_2\hat{j}+v_3\hat{k}}$

2. We know that the direction cosines of $\small{\vec{v}}$ are:
$\small{\frac{v_1}{\left|\vec{v} \right|},~\frac{v_2}{\left|\vec{v} \right|},~\frac{v_3}{\left|\vec{v} \right|}}$

3. Let $\small{a_L,~b_L,~c_L}$ be the direction ratios of $\small{L}$.
• The direction cosines of $\small{L}$ will be same as in (2) above. This is because, $\small{L~\text{and}~\vec{v}}$ are parallel.
• Then we can write: $\small{\frac{v_1}{\left|\vec{v} \right|\,a_L}~=~\frac{v_2}{\left|\vec{v} \right|\,b_L}~=~\frac{v_3}{\left|\vec{v} \right|\,c_L}}$
• If we multiply all three fractions by $\small{\left|\vec{v} \right|}$, the equality will not change. So we get:
$\small{\frac{v_1}{a_L}~=~\frac{v_2}{b_L}~=~\frac{v_3}{c_L}}$   
• Since the three fractions are equal, we can write:
$\small{\frac{v_1}{a_L}~=~\frac{v_2}{b_L}~=~\frac{v_3}{c_L}~=~\lambda}$

4. Here $\small{\lambda}$ can be any real number.
• When $\small{\lambda = 1}$, we get:
$\small{a_L = v_1,~b_L = v_2,~c_L = v_3}$

5. So we can write:
When the line $\small{L}$ is parallel to the vector $\small{\vec{v}=v_1 \hat{i}+v_2\hat{j}+v_3\hat{k}}$, the numbers $\small{v_1,~v_2,~v_3}$ can be considered as a set of direction ratios of $\small{L}$
• This fact will become useful in the next section where we see lines parallel to a given vector.

6. Always remember that, for a vector (or a directed line),
    ♦ There is only one set of direction cosines
    ♦ There are infinite sets of direction ratios
That is why in step (5) above, we wrote: "a set of direction ratios". We cannot write: "the set of direction ratios" 

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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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