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 such a line.

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)

---

• To summarize, we can write:
    ♦ $\small{\vec{r}~=~\vec{u}+\lambda\vec{v}}$ is the vector equation
    ♦ $\small{\frac{x – x_1}{v_1}~=~\frac{y – y_1}{v_2}~=~\frac{z – z_1}{v_3}}$ is the Cartesian equation
• In the next section, we will see equation of a line passing through two given points.

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