A straight line in space is completely fixed by two pieces of information: one point on it and one direction along it. Both the vector and the Cartesian descriptions are just two ways of writing that single idea.
Vector form
Let $\vec a$ be the position vector of a known point $A$ on the line and let $\vec b$ be any vector along the line (a direction vector). A point with position vector $\vec r$ lies on the line exactly when $\vec r-\vec a$ is parallel to $\vec b$, i.e. $\vec r-\vec a=\lambda\vec b$ for some scalar $\lambda$. Rearranged:
$$\vec r=\vec a+\lambda\,\vec b,\qquad \lambda\in\mathbb{R}.$$
As the parameter $\lambda$ runs over all real numbers, $\vec r$ sweeps out the whole line. The same line has many valid equations — any point on it can play the role of $\vec a$, and any non-zero multiple of $\vec b$ can play the role of $\vec b$.
Cartesian (symmetric) form
Write $\vec a=x_1\hat i+y_1\hat j+z_1\hat k$ and $\vec b=a\hat i+b\hat j+c\hat k$, so $(a,b,c)$ are direction ratios. Equating components of $\vec r=\vec a+\lambda\vec b$ and eliminating $\lambda$ gives the symmetric form:
$$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}.$$
This says the same thing three times over: each ratio equals $\lambda$. Reading it backwards is a key skill — the numbers subtracted in the numerators give a point $(x_1,y_1,z_1)$ on the line, and the denominators are direction ratios $(a,b,c)$.
A warning about zero denominators: if, say, $c=0$ the symbol $\dfrac{z-z_1}{0}$ is shorthand for the condition $z=z_1$ (the line is parallel to the $xy$-plane). You never actually divide by zero — that fraction is read as "$z-z_1=0$".
Using direction cosines
If instead the direction cosines $(l,m,n)$ are used, the form becomes $\dfrac{x-x_1}{l}=\dfrac{y-y_1}{m}=\dfrac{z-z_1}{n}=r$, where the common value $r$ is then the actual signed distance of the point $(x,y,z)$ from $(x_1,y_1,z_1)$ along the line.
Line through two points
Given two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$, take $\vec a$ as $A$ and $\vec b=\vec{AB}=\vec b_2-\vec b_1$. In vector form $\vec r=\vec a+\lambda(\vec b_2-\vec b_1)$, and in Cartesian form:
$$\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}.$$
Converting between the forms
The two forms are interchangeable. Vector to Cartesian: read off the point from $\vec a$ and the DRs from $\vec b$, then slot them into the symmetric form. Cartesian to vector: read off the point and the DRs, then write $\vec r=(x_1\hat i+y_1\hat j+z_1\hat k)+\lambda(a\hat i+b\hat j+c\hat k)$.
Deeper Insight — "point plus direction" is the whole story. Both forms encode the same two ingredients, so most mistakes come from mixing up which number is a coordinate and which is a direction ratio. A reliable habit: whenever you see a line, immediately extract its $(\text{point},\ \text{DRs})$ pair, and whenever you must write a line, decide those two things first. Parallelism is then trivial — two lines are parallel iff they share direction ratios (one set is a multiple of the other) — and a point lies on a line iff substituting its coordinates makes all three ratios equal. Almost every line problem in the chapter is an application of this single pair.
Write the vector equation of the line through $(1,0,2)$ with direction $\hat i+\hat j-\hat k$.
Solution- Point: $\vec a=\hat i+0\hat j+2\hat k=\hat i+2\hat k$.
- Direction: $\vec b=\hat i+\hat j-\hat k$.
- Apply $\vec r=\vec a+\lambda\vec b$.
Answer: $\vec r=(\hat i+2\hat k)+\lambda(\hat i+\hat j-\hat k).$
Write the Cartesian equation of the line through $(2,-1,3)$ with DRs $1,2,2$.
Solution- Point $(x_1,y_1,z_1)=(2,-1,3)$, DRs $(a,b,c)=(1,2,2)$.
- Substitute into the symmetric form, taking care with the sign of $-1$.
Answer: $\dfrac{x-2}{1}=\dfrac{y+1}{2}=\dfrac{z-3}{2}.$
Find the Cartesian equation of the line through $(1,2,3)$ and $(2,4,5)$.
Solution- DRs $=(2-1,\ 4-2,\ 5-3)=(1,2,2)$.
- Use the point $(1,2,3)$ in the symmetric form.
Answer: $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{2}.$
From $\dfrac{x-1}{2}=\dfrac{y+3}{-1}=\dfrac{z}{4}$, read a point and the direction ratios.
Solution- Set each numerator to zero: $x=1,\ y=-3,\ z=0$ gives a point.
- The denominators are the DRs.
Answer: Point $(1,-3,0)$; direction ratios $(2,-1,4)$.
Find the vector and Cartesian equations of the line through $(2,-1,1)$ and $(4,1,3)$.
Solution- DRs $=(4-2,\ 1-(-1),\ 3-1)=(2,2,2)$, simplify to $(1,1,1)$.
- Vector form with point $\vec a=2\hat i-\hat j+\hat k$.
- Cartesian form with the same point and DRs.
Answer: $\vec r=(2\hat i-\hat j+\hat k)+\lambda(\hat i+\hat j+\hat k)$; $\dfrac{x-2}{1}=\dfrac{y+1}{1}=\dfrac{z-1}{1}.$
Convert $\vec r=(\hat i+\hat j)+\lambda(2\hat i-\hat j+\hat k)$ to Cartesian form.
Solution- Point from $\vec a$: $(1,1,0)$.
- DRs from $\vec b$: $(2,-1,1)$.
- Substitute.
Answer: $\dfrac{x-1}{2}=\dfrac{y-1}{-1}=\dfrac{z}{1}.$
Find the Cartesian equation of the line through the origin with direction ratios $3,-1,2$.
Solution- Point is $(0,0,0)$, so the subtracted constants vanish.
- Use DRs $(3,-1,2)$.
Answer: $\dfrac{x}{3}=\dfrac{y}{-1}=\dfrac{z}{2}.$
Find the vector and Cartesian equations of the line through $(1,2,3)$ parallel to $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}$.
Solution- Parallel lines share direction ratios, so $\vec b=(2,3,6)$.
- Use the given point $(1,2,3)$ for $\vec a$.
- Write both forms.
Answer: $\vec r=(\hat i+2\hat j+3\hat k)+\lambda(2\hat i+3\hat j+6\hat k)$; $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{6}.$
Does the point $(3,2,1)$ lie on the line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{1}$?
Solution- Substitute the coordinates and evaluate each ratio.
- $\dfrac{3-1}{2}=1$, $\dfrac{2+1}{3}=1$, $\dfrac{1}{1}=1$.
- All three are equal (to $\lambda=1$).
Answer: Yes — all three ratios equal $1$, so the point lies on the line.
Find the coordinates of the point on $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{1}$ at parameter $\lambda=2$.
Solution- Set each ratio equal to $\lambda=2$.
- $x-1=2(2)=4\Rightarrow x=5$.
- $y+1=3(2)=6\Rightarrow y=5$.
- $z=1(2)=2$.
Answer: The point is $(5,5,2)$.
Write the equation of the line through $(0,1,-2)$ parallel to the $z$-axis.
Solution- The $z$-axis has direction ratios $(0,0,1)$.
- Two denominators are zero, read as $x=0$ and $y=1$; only $z$ varies.
Answer: $\vec r=(\hat j-2\hat k)+\lambda\hat k$, i.e. $x=0,\ y=1$ with $z$ free.