[ Coordinate method in space ] [ Parametric equations of a line ]

A line $l$ passes through the point $M 0\left(x 0 ; y_{0} ; z 0\right)$ parallel to a non-zero vector $\vec{m}=(a ; b ; c)$. Find the necessary and sufficient condition for the point $M(x ; y ; z)$ to lie on the line $l$.

#
Point $M$ lies on the line $l$ if and only if the vector $\overrightarrow{M_{0} M}$ is collinear with the vector $\vec{m}$, which means,

$$
\begin{aligned}
& M\left(x 0 ; y_{0} ; z 0\right) \in l \Leftrightarrow \overrightarrow{M_{0} M}=\vec{m} \Leftrightarrow \\
& \left\{\begin{array} { l } 
{ x - x _ { 0 } = a t } \\
{ y - y _ { 0 } = b t } \\
{ z - z _ { 0 } = c t }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=x_{0}+a t \\
y=y_{0}+b t \\
z=z_{0}+c t
\end{array}\right.\right.
\end{aligned}
$$

Thus, the parametric equations of the line are obtained. By eliminating $t$ from the system of three equations with four unknowns, we obtain the canonical equation of the line:

![](https://cdn.mathpix.com/cropped/2024_05_06_b3d98ec29b07c8345931g-24.jpg?height=91&width=337&top_left_y=543&top_left_x=855)

## Answer

$$
\left\{\begin{array}{l}
x=x_{0}+a t \\
y=y_{0}+b t \\
z=z_{0}+c t, \text { or } \frac{x-x_{0}}{a}=\frac{y-z_{0}}{b}=\frac{z-z_{0}}{a}
\end{array}\right.
$$