Let L be a line in space passing through point $\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0, \mathrm{z}_0\right)$. Let $\overrightarrow{\mathbf{b}}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$ be a vector parallel to L . Then, for any point on line $\mathrm{Q}(\mathrm{x}, \mathrm{y}, \mathrm{z})$, we know that vector PQ is parallel to vector b . Thus, there is a scalar, $\lambda$, such that $\overrightarrow{P Q}=\lambda \overrightarrow{\mathbf{b}}$, which gives,

$
\begin{gathered}
\overrightarrow{P Q}=\lambda \tilde{\mathbf{b}} \\
\left(x-x_0\right) \hat{i}+\left(y-y_0\right) \hat{j}+\left(z-z_0\right) \hat{k}=\lambda(a \hat{i}+b \hat{j}+c \hat{k})
\end{gathered}
$

Using vector operations, we can rewrite,
$
\begin{aligned}
(x \hat{i}+y \hat{j}+z \hat{k})-\left(x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}\right) & =\lambda(a \hat{i}+b \hat{j}+c \hat{k}) \\
(x \hat{i}+y \hat{j}+z \hat{k}) & =\left(x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}\right)+\lambda(a \hat{i}+b \hat{j}+c \hat{k})
\end{aligned}
$

Setting $\overrightarrow{\mathbf{r}}=x \hat{i}+y \hat{j}+z \hat{k}$ and $\overrightarrow{\mathbf{r}}_0=x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}$ we now have the vector equation of a line:
$
\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}}
$

Cartesian Form

Vector equation of a line $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}}$ shows that the following equations are simultaneously true: $x=x_0+\lambda a ; y=y_0+\lambda b ; z=z_0+\lambda c$
If we solve each of these equations for the component variables $\mathbf{x}, \mathbf{y}$, and $\mathbf{z}$, we get a set of equations in which each variable is defined in terms of the parameter $\lambda$ and that, together, describe the line. This set of three equations forms a set of parametric equations of a line:

If we solve each of the equations for $\boldsymbol{\lambda}$ assuming $\mathbf{a}, \mathbf{b}$, and $\mathbf{c}$ are non-zero, we get a different description of the same line:
$
\frac{x-x_0}{a}=\lambda, \quad \frac{y-y_0}{b}=\lambda, \quad \frac{z-z_0}{c}=\lambda
$

These are parametric equations of the line. Eliminating the parameter $\lambda$ from the above equation, we get,
$
\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}
$

This is the Cartesian equation of the line.

NOTE:

If $\mathbf{I}, \mathbf{m}, \mathbf{n}$ are the direction cosines of the line, the equation of the line is
$
\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}
$

Equation of a line passing through two given points
Let $\mathbf{P}\left(\mathbf{x}_0, \mathbf{y}_0, \mathbf{z}_0\right)$ and $\mathbf{Q}\left(\mathbf{x}_1, \mathbf{y}_1, \mathbf{z}_1\right)$ be the points on a line, and let $\overrightarrow{\mathbf{p}}=x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}$ and $\overrightarrow{\mathbf{q}}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}$.
Let $\overrightarrow{\mathbf{r}}$ be the position vector of an arbitrary point $\mathbf{R}(\mathbf{x}, \mathbf{y}, \mathbf{z})$ lying on this line

We want to find a vector equation for the line PQ.

Using $\mathbf{P}$ as our known point on the line, and $\overrightarrow{P Q}=\left(x_1-x_0\right) \hat{i}+\left(y_1-y_0\right) \hat{j}+\left(z_1-z_0\right) \hat{k}$ as the direction vector, the equation $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}}$ gives
$
\begin{aligned}
\overrightarrow{\mathbf{r}} & =\overrightarrow{\mathbf{p}}+\lambda(\overrightarrow{P Q}) \\
\overrightarrow{\mathbf{r}} & =\overrightarrow{\mathbf{p}}+\lambda(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}})
\end{aligned}
$

Cartesian Form

We can also find parametric equations for the line segment $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{p}}+\lambda(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}})$
As position vector of point $R(x, y, z)$ is
$
\overrightarrow{\mathbf{r}}=(x \hat{i}+y \hat{j}+z \hat{k})
$
$
\begin{aligned}
\overrightarrow{\mathbf{r}} & =\overrightarrow{\mathbf{p}}+\lambda(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}}) \\
\hat{x} \hat{i}+y \hat{j}+z \hat{k} & =x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}+\lambda\left[\left(x_1-x_0\right) \hat{i}+\left(y_1-y_0\right) \hat{j}+\left(z_1-z_0\right) \hat{k}\right]
\end{aligned}
$

Equating the like coefficients of $\hat{i}, \hat{j}, \hat{k}$, we get
$
x=x_0+\lambda\left(x_1-x_0\right) ; y=y_0+\lambda\left(y_1-y_0\right) ; z=z_0+\lambda\left(z_1-z_0\right)
$

On eliminating $\lambda$, we obtain
$
\frac{x-x_0}{x_1-x_0}=\frac{y-y_0}{y_1-y_0}=\frac{z-z_0}{z_1-z_0}
$
which is the equation of the line in Cartesian form.