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:
$\small{\frac{x - x_1}{x_2 - x_1}~=~\frac{y - y_1}{y_2 - y_1}~=~\frac{z - z_1}{z_2 - z_1}}$
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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{x - x_1}{x_2 - x_1}}& {~=~} &{\frac{y - y_1}{y_2 - y_1}} & {~=~} &{\frac{z - z_1}{z_2 - z_1}}
\\ {~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{x -3}{3 - 3}}& {~=~} &{\frac{y +2}{-2 + 2}} & {~=~} &{\frac{z +5}{6 + 5}}
\\ {~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{x-3}{0}}& {~=~} &{\frac{y+2}{0}} & {~=~} &{\frac{z+5}{11}}
\\ \end{array}}$
pppq
xxx
• 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.
The link below gives a few more miscellaneous examples:
Miscellaneous Exercise
In the next section, we will see equation of a line in space.
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