Lesson 01

Equations of Lines in R² and R³

A line is the simplest object you can build from a point and a direction — but in three dimensions the familiar slope–intercept form breaks down. This lesson rebuilds the equation of a line from scratch using vectors, then shows how the vector, parametric and symmetric forms are really the same statement written three different ways.

By the end of this lesson

  • Write the vector equation of a line given a point and a direction vector, or given two points on the line.
  • Convert freely between vector, parametric and symmetric equations of a line in R³.
  • Explain why a linear equation like \(4x - 3y + 2 = 0\) describes a line in R² but a plane in R³.
  • Identify when a line’s symmetric form cannot be written directly, and express that case correctly.

Section 1

From slope to direction vector

In earlier grades a line in R² was written using its slope \(m\) and a point: \(y - y_1 = m(x - x_1)\). That slope is really encoding a direction — for every \(1\) unit moved in \(x\), the line moves \(m\) units in \(y\). We can package the same information as a direction vector \(\vv{m} = (1,\,m)\), or more generally as any scalar multiple of it.

Once direction is written as a vector, the idea extends immediately to three dimensions, where “slope” no longer makes sense but a direction vector still does. This is the shift the whole lesson is built around: stop thinking in slopes, start thinking in points and directions.

The vector equation of a line in two-space A coordinate grid with a straight line through the point P zero at (1, 1). The position vector r zero runs from the origin to P zero, and the direction vector m is drawn from P zero to the point (3, 2), one step along the line. x y P0(1, 1) r0 m (3, 2) L
fig. 1.1The line \(L\) through \(P_0(1,1)\) in the direction of \(\vv{m}\), passing through \((3,2)\). The direction vector can be read straight off the diagram as \(\vv{m} = (3,2) - (1,1) = (2,1)\).

Deriving the vector equation

\[\vv{r} = \vv{r}_0 + t\vv{m}\]
From a known point, every multiple of the direction
\[(x,y,z) = (x_0,y_0,z_0) + t(m_1,m_2,m_3)\]
Component by component
\[x = x_0 + tm_1,\qquad y = y_0 + tm_2,\qquad z = z_0 + tm_3\]
Parametric form — one equation per coordinate, one shared \(t\)
Live form converter Interactive

Change the point and the direction vector, and all four forms rewrite themselves. Try setting a component of \(\vv{m}\) to zero and watch what happens to the symmetric form.

Vector
\(\vv{r} = (1,\,2,\,3) + t(2,\,-1,\,4)\)
Parametric
\(x = 1 + 2t,\quad y = 2 - t,\quad z = 3 + 4t\)
Symmetric
\(\dfrac{x-1}{2} = \dfrac{y-2}{-1} = \dfrac{z-3}{4}\)
At t = 1
\(\vv{r} = (3,\,1,\,7)\)

Section 2

Direction vectors and normal vectors in R²

A line in R² can be described by a direction vector, pointing along the line, or by a normal vector, pointing perpendicular to it — and the two are always at right angles to each other. If a line has equation \(Ax + By + C = 0\), the vector \(\vv{n} = (A,B)\) is automatically normal to it, and swapping and negating the components gives a valid direction vector: \(\vv{m} = (B,-A)\) or \(\vv{m} = (-B,A)\).

This relationship is the reason the normal form of a line (\(Ax + By + C = 0\)) and the vector form (\(\vv{r} = \vv{r}_0 + t\vv{m}\)) always agree. They describe the same line from two different geometric perspectives, and you will use both throughout this unit, depending on which is more convenient for a given question.

A line in two-space with its direction and normal vectors A line through the point (1, 2) with direction vector m equal to (3, 4) drawn along it, and normal vector n equal to (4, negative 3) drawn perpendicular to it and marked with a right-angle symbol. x y m = (3, 4) n = (4, −3) (1, 2) 4x − 3y + 2 = 0
fig. 1.2The line \(4x - 3y + 2 = 0\) through \((1,2)\), with its direction vector \(\vv{m} = (3,4)\) and a corresponding normal vector \(\vv{n} = (4,-3)\), shown perpendicular to one another.

Section 3

One equation, two meanings

Here is a subtlety that trips many students up: the single linear equation \(x + 2y = 4\) is not automatically “a line”. In R², where there are only two variables to satisfy, it describes a line. That exact same equation in R³ — where \(z\) is simply unrestricted — describes an entire plane, because every value of \(z\) is allowed as long as \(x\) and \(y\) satisfy the relationship.

This is why a single equation is never enough to pin down a line in three dimensions. You need either the vector or parametric form, built from a point and a direction, or two intersecting planes, which is Lesson 04. Keeping the distinction clear now makes the jump into Lesson 02 on planes much more intuitive.

The same equation as a line in two-space and a plane in three-space Two panels side by side. On the left, x plus 2y equals 4 drawn on a two-dimensional grid as a straight line. On the right, the same equation drawn against three-dimensional axes as a vertical plane running parallel to the z-axis. x y x + 2y = 4 in ℝ² — a line x y z x + 2y = 4 in ℝ³ — a plane
fig. 1.3The same equation, \(x + 2y = 4\), plotted in R² (left), where it is a line, and in R³ (right), where the unrestricted \(z\)-axis turns it into a plane.

Practice

Worked examples

Example 1 · routine

Find the vector, parametric and symmetric equations of the line through \(A(2,-1,3)\) and \(B(5,1,-2)\).

  1. Find the direction vector: \(\vv{m} = B - A = \bigl(5-2,\ 1-(-1),\ -2-3\bigr) = (3,\,2,\,-5)\).
  2. Choose either point as \(\vv{r}_0\) — taking \(A(2,-1,3)\) — and write the vector equation: \(\vv{r} = (2,-1,3) + t(3,2,-5)\).
  3. Split that into parametric form, \(x = 2+3t,\quad y = -1+2t,\quad z = 3-5t\), then solve each for \(t\) and set them equal.
Vector \(\vv{r} = (2,-1,3) + t(3,2,-5)\); parametric \(x = 2+3t,\ y = -1+2t,\ z = 3-5t\); symmetric \(\dfrac{x-2}{3} = \dfrac{y+1}{2} = \dfrac{z-3}{-5}\).
Example 2 · demanding

A line passes through \(P(4,-2,1)\) and is parallel to \(\dfrac{x-1}{2} = \dfrac{y+3}{0} = \dfrac{z}{-4}\). Write its symmetric equation, and explain what the zero in the denominator means.

  1. The direction vector of the given line is \(\vv{m} = (2,\,0,\,-4)\), read straight off the denominators. The new line is parallel, so it shares that direction vector.
  2. Write the vector equation through \(P\): \(\vv{r} = (4,-2,1) + t(2,0,-4)\), giving \(x = 4+2t,\quad y = -2,\quad z = 1-4t\).
  3. Since \(y\) never changes — the middle component of \(\vv{m}\) is zero — the symmetric form cannot carry a fraction for \(y\). Instead \(y = -2\) is stated separately, alongside the symmetric ratio for \(x\) and \(z\).
\(\dfrac{x-4}{2} = \dfrac{z-1}{-4},\qquad y = -2\) — the zero denominator signals that the line lies entirely in the plane \(y = -2\), moving only in the \(x\)- and \(z\)-directions.

Watch out

Common mistakes

  • Leaving a zero in a denominator of the symmetric form, such as writing \(\dfrac{y+3}{0}\) in a final answer. A zero denominator is never valid. Pull that variable out as a separately stated equation instead, as in Example 2.
  • Expecting your vector equation to match the answer key character for character. A different choice of \(\vv{r}_0\), or a different sign or scale on the direction vector, still describes the same line. Check by testing whether a given point satisfies your equation, not by comparing its exact form to someone else’s.