Equations of lines in space: parametric and symmetric forms

2026-10-22 huynhvietkhanh Comments: 0

Write line equations from points and direction vectors, handle zero components, and find the intersection line of two planes.

High school3D Coordinate Geometry

Updated: October 22, 2026

A point and a nonzero direction vector determine a line. From these data, write parametric equations and then convert to symmetric equations when the denominators permit.

Learning goals: write and convert line equations; find points on a line; handle direction vectors with a component equal to $0$, and lines defined as intersections of two planes.

Prerequisites: point coordinates, direction vectors, linear systems, and the cross product.

On this page: Definitions · Method · Examples · Practice · Further study

1. Parametric and symmetric equations

Let $M_0(x_0;y_0;z_0)$ and $\vec u=(a;b;c)\ne\vec0$ be given. A point $M(x;y;z)$ lies on the line through $M_0$ with direction vector $\vec u$ if and only if $\overrightarrow{M_0M}=t\vec u$ for some real number $t$. Thus its parametric equations are

$$d:\begin{cases}x=x_0+at,\\y=y_0+bt,\\z=z_0+ct,\end{cases}\quad t\in\mathbb R.$$

All three coordinates must use the same value of $t$. If $a,b,c$ are all different from $0$, eliminating the parameter yields the symmetric equations

$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}.$$

If a direction-vector component is $0$, keep the corresponding point coordinate constant. For example, through $M_0(1;2;3)$ with direction vector $(2;0;-1)$, write $y=2$ and $(x-1)/2=(z-3)/(-1)$. Never write a fraction with denominator $0$. Parametric equations remain valid whenever $\vec u\ne\vec0$.

2. Choosing a point and direction vector

Through two points: if $A\ne B$, choose the point $A$ and direction vector $B-A$. Choosing another point on the line, or multiplying the direction vector by a number different from $0$, may change the equations but leaves their point set unchanged.

Intersection of two planes: for $(P):A_1x+B_1y+C_1z+D_1=0$ and $(Q):A_2x+B_2y+C_2z+D_2=0$, first check that their normal vectors are not parallel. The planes then meet along a line; choose a common solution as a point on it, and take the cross product of the normals as its direction vector.

If the normal vectors are parallel, the two equations may describe parallel or coincident planes; do not conclude that they define a line. When seeking a common solution, a chosen coordinate value may make the remaining system inconsistent. Try another method instead of concluding that there is no intersection line.

3. Four worked examples

Example 1. A point and direction vector

Through $M_0(-5;4;6)$, with direction vector $(-1;2;-3)$, the line has equations

$$\begin{cases}x=-5-t,\\y=4+2t,\\z=6-3t,\end{cases}\quad t\in\mathbb R.$$

$$\frac{x+5}{-1}=\frac{y-4}{2}=\frac{z-6}{-3}.$$

Setting $t=0$ recovers the point $M_0$; the coefficients of $t$ are precisely the direction-vector components.

Example 2. Two distinct points

Let $A(1;2;4)$ and $B(3;-1;1)$. Since $B-A=(2;-3;-3)\ne\vec0$, the line $AB$ has equations

$$\begin{cases}x=1+2t,\\y=2-3t,\\z=4-3t,\end{cases}\quad t\in\mathbb R.$$

$$\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z-4}{-3}.$$

Setting $t=0$ gives $A$; setting $t=1$ gives $B$. These substitutions check both a point on the line and its direction.

Example 3. Reading points and converting equations

$$d_1:\frac{x-1}{1}=\frac{y+2}{-4}=\frac{z-3}{2},$$

$$d_2:\begin{cases}x=2-t,\\y=t,\\z=3-3t,\end{cases}\quad t\in\mathbb R.$$

Set the common value in the equations of $d_1$ equal to $s$. Then $x=1+s$, $y=-2-4s$, and $z=3+2s$. The choices $s=0,1$ yield $(1;-2;3)$ and $(2;-6;5)$, respectively; a direction vector is $(1;-4;2)$.

For $d_2$, the choices $t=0,1$ yield $(2;0;3)$ and $(1;1;0)$, respectively. The direction vector is $(-1;1;-3)$, so the symmetric equations are

$$\frac{x-2}{-1}=y=\frac{z-3}{-3}.$$

Example 4. Intersection of two planes

$$d:\begin{cases}2x+3y+z-6=0,\\x-y+2z-2=0.\end{cases}$$

The normal vectors $(2;3;1)$ and $(1;-1;2)$ have cross product $(7;-3;-5)\ne\vec0$. The point $A(1;1;1)$ satisfies both equations. Hence

$$d:\begin{cases}x=1+7t,\\y=1-3t,\\z=1-5t,\end{cases}\quad t\in\mathbb R,$$

$$\frac{x-1}{7}=\frac{y-1}{-3}=\frac{z-1}{-5}.$$

Substitution of the parametric expressions into the plane equations gives coefficients of $t$ equal to $14-9-5=0$ and $7+3-10=0$, respectively; the constant terms are also $0$. Thus every point of the line lies in both planes.

4. Practice

Find the line through $A(1;0;0)$ and $B(1;2;0)$.
For $x=1+2t$, $y=-1+t$, and $z=3-t$, determine whether $M(5;1;1)$ lies on the line.
Write equations for the intersection of the planes $x+y=1$ and $z=2$.

Hints and answers

Choose direction vector $(0;1;0)$: $x=1$, $y=t$, $z=0$, $t\in\mathbb R$. Do not use denominator $0$.
Yes: the same value $t=2$ satisfies all three coordinate equations.
Choose the point $(1;0;2)$ and direction vector $(1;-1;0)$: $x=1+t$, $y=-t$, $z=2$.

Common mistakes: replacing $y$ by $z$ in a fraction; omitting a constant coordinate when the corresponding direction-vector component is $0$; using three independent parameters; checking only two coordinates; using the cross product $\vec0$ as a direction vector.

5. Further study and resources

Direction vectors (English lesson) · Equations of planes (English lesson) · Intersecting lines (English lesson) · Parallel lines (English lesson).

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