A Point on the Line and the Direction of the Line are Given

Theorem 6.11 (parametric vector equation). The line through the fixed point with position vector $\vec a$, parallel to a given vector $\vec b$, has vector equation

Proof. If $P$ (position vector $\vec r$) is any point on the line, $\vec{AP}=\vec r-\vec a$ is parallel to $\vec b$, so $\vec r-\vec a=t\vec b$ for some scalar $t$; rearranging gives $\vec r=\vec a+t\vec b$. As $t$ ranges over $\mathbb R$, $\vec a+t\vec b$ sweeps out every point of the line — this is the parametric form.

(b) Non-parametric form. Since $\vec{AP}\parallel\vec b$ we equally have $\vec{AP}\times\vec b=\vec 0$, i.e.

a single vector equation with no free parameter.

(c) Cartesian equations. Write $P=(x,y,z)$, $A=(x_1,y_1,z_1)$ and $\vec b=b_1\hat i+b_2\hat j+b_3\hat k$. Substituting into $\vec r=\vec a+t\vec b$ and comparing $\hat i,\hat j,\hat k$-coefficients gives $x-x_1=tb_1,\ y-y_1=tb_2,\ z-z_1=tb_3$, conventionally written as the Cartesian (symmetric) equations

If $l,m,n$ are the direction cosines of the line (proportional to $b_1,b_2,b_3$), the equations may equally be written with $l,m,n$ in place of $b_1,b_2,b_3$. …