Lesson 01

Equations of Lines in R² and R³

A line is the simplest object in this topic and the one every later lesson is built on. Get comfortable moving between its four forms now — the work in Lessons 04 and 05 is almost entirely a matter of choosing the form that makes the question easy.

By the end of this lesson

  • Write a line in vector, parametric, symmetric and Cartesian form from a point and a direction, or from two points.
  • Convert freely between the forms, including when a direction component is zero.
  • Explain why one equation gives a line in R² but a plane in R³.
  • Decide whether two differently-written equations describe the same line.

A point and a direction

A line is determined by a point on it and a direction vector \(\vv{m}\) parallel to it. If \(\vv{r}_0\) is the position vector of a known point \(P_0\) and \(\vv{r}\) is the position vector of a general point on the line, then the same line can be written four ways.

The idea behind all four is the same one picture: start at \(P_0\) and walk some multiple of \(\vv{m}\). The multiple is the parameter \(t\). Every point of the line corresponds to exactly one value of \(t\), and every value of \(t\) lands on the line — which is what makes the parametric description so useful later, when we substitute it into the equation of a plane.

If you are given two points rather than a point and a direction, subtract them: \(\vv{m} = \overrightarrow{AB} = B - A\). That single step turns almost every “line through two points” question into the routine case.

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. 1.1Starting at \(P_0(1,1)\) and stepping once along \(\vv{m}=(2,1)\) reaches \((3,2)\). Every other point of the line is another value of \(t\).

The four forms

Vector
\(\vv{r} = \vv{r}_0 + t\vv{m},\quad t \in \R\)
Parametric
\(x = x_0 + tm_1,\quad y = y_0 + tm_2,\quad z = z_0 + tm_3\)
Symmetric
\(\dfrac{x-x_0}{m_1} = \dfrac{y-y_0}{m_2} = \dfrac{z-z_0}{m_3},\quad m_1,m_2,m_3 \neq 0\)
Cartesian (R² only)
\(Ax + By + C = 0,\ \text{with normal } \vv{n} = (A,B)\)

Read them from left to right and they are the same sentence three times. The vector form says “position equals start plus \(t\) steps”; the parametric form is that statement written out one coordinate at a time; and the symmetric form is what you get by solving each parametric equation for \(t\) and setting the results equal. That last derivation is worth doing once by hand — it explains both why the denominators are the components of \(\vv{m}\) and why a zero component breaks the form.

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
\(\dfrac{x-1}{2} = \dfrac{y-2}{-1} = \dfrac{z-3}{4}\)

R²: the Cartesian form and the normal

In two-space a line has a second personality. As well as a direction \(\vv{m}\) along it, it has a normal \(\vv{n}\) perpendicular to it, and that normal is what makes the single scalar equation \(Ax+By+C=0\) possible: the coefficients \((A,B)\) are a normal vector.

Converting between the two views is a single move. Rotating \(\vv{m}=(m_1,m_2)\) by a quarter turn gives \(\vv{n}=(m_2,-m_1)\), and you can check it in one line: \(\vv{m} \cdot \vv{n} = m_1m_2 - m_2m_1 = 0\). Then substitute the known point to find \(C\).

Going the other way is just as quick: read \(\vv{n}=(A,B)\) straight off the equation, take \(\vv{m}=(-B,A)\), and find any point on the line by choosing a convenient value of one variable.

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 direction lies along the line; the normal stands across it. In R² either one determines the line, given a point.

R³: why one equation is not enough

In R² a single scalar equation describes a line because one normal direction pins it down. In R³ a single linear equation in \(x,y,z\) describes a plane, so a line in three-space needs either a vector form or a pair of equations — a point worth making clearly in your solutions, since it is a favourite exam question.

The counting argument is the one to write down. Three coordinates give three degrees of freedom. One linear equation removes one of them, leaving a two-dimensional solution set: a plane. A line has only one degree of freedom, so it needs two equations — which is exactly what the symmetric form is, once you notice that \(a = b = c\) is two equations wearing one pair of equals signs.

This is also why there is no “Cartesian form of a line in R³” row in the table above. Asking for one is asking for a single equation to do a job that takes two.

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, \(x+2y=4\), in two dimensions and in three. Nothing about the equation changed; the number of variables it is allowed to constrain did.

Zero components in the symmetric form

The symmetric form divides by each component of \(\vv{m}\), so it collapses the moment one of them is zero. Suppose \(\vv{m} = (3,0,-2)\). The parametric equations are \(x = x_0 + 3t,\ y = y_0,\ z = z_0 - 2t\), and the middle one contains no \(t\) at all: \(y\) never changes.

So do not write \(\dfrac{y-y_0}{0}\). Split that coordinate off as its own equation and set the remaining two ratios equal:

\[\dfrac{x-x_0}{3} = \dfrac{z-z_0}{-2},\qquad y = y_0\]

Geometrically the line is horizontal in the \(y\) direction: it lives entirely inside the plane \(y = y_0\). Two zero components would be even simpler — the line would then be parallel to a coordinate axis, and two of the three coordinates would be constant.

Same line, different equations

Two lines are parallel when their direction vectors are scalar multiples; three or more points are collinear when the vectors joining them are scalar multiples. Any nonzero scalar multiple of \(\vv{m}\) is an equally valid direction vector, so equations of the same line may look different yet be equivalent.

That means a marking scheme cannot simply compare your answer to one string of symbols, and neither should you. To decide whether two equations describe the same line, ask two questions in this order:

  1. Are the direction vectors scalar multiples of one another? If not, the lines are certainly different.
  2. Does the point from one equation satisfy the other? If it does, the lines coincide; if it does not, they are parallel and distinct.

Both checks are needed. Two parallel lines pass the first and fail the second, and that is precisely the difference between a system with infinitely many solutions and one with none — the distinction Lesson 04 is built around.

Practice

Worked examples

Example 1 · routine

All four forms from a point and a direction

Write the line through \(P(1,2,3)\) with direction \(\vv{m} = (2,-1,4)\) in vector, parametric and symmetric form.

Show solution
  1. Vector form is a direct substitution: \(\vv{r} = (1,\,2,\,3) + t(2,\,-1,\,4),\ t\in\R\).
  2. Read off one coordinate at a time: \(x = 1+2t,\quad y = 2-t,\quad z = 3+4t\).
  3. Solve each for \(t\) and set them equal: \(\dfrac{x-1}{2} = \dfrac{y-2}{-1} = \dfrac{z-3}{4}\).
  4. Check: \(t=1\) gives \((3,1,7)\), and \(\tfrac{3-1}{2} = \tfrac{1-2}{-1} = \tfrac{7-3}{4} = 1\ \checkmark\)
Example 2 · routine

A line through two points

Find a vector equation of the line through \(A(4,0,-1)\) and \(B(1,3,5)\).

Show solution
  1. The direction is the vector joining them: \(\vv{m} = \overrightarrow{AB} = B - A = (-3,\,3,\,6)\).
  2. Divide by the common factor \(-3\) to keep the numbers small: \(\vv{m} = (1,\,-1,\,-2)\). Any nonzero multiple is equally correct.
  3. Use either given point: \(\vv{r} = (4,\,0,\,-1) + t(1,\,-1,\,-2),\ t\in\R\).
  4. Check that \(B\) is on it: \(t=-3 \Rightarrow (1,\,3,\,5)\ \checkmark\)
Example 3 · standard

The R² round trip: direction to normal to Cartesian

A line in R² passes through \((1,2)\) with direction \(\vv{m} = (3,4)\). Find its Cartesian equation.

Show solution
  1. Rotate the direction a quarter turn to get a normal: \(\vv{n} = (4,\,-3)\). Check: \((3)(4) + (4)(-3) = 0\ \checkmark\)
  2. The coefficients of the Cartesian equation are the components of the normal, so the equation is \(4x - 3y + C = 0\).
  3. Substitute the given point: \(4(1) - 3(2) + C = 0 \Rightarrow C = 2\).
  4. Therefore \(4x - 3y + 2 = 0\). Check a second point of the line, \((4,6)\): \(16 - 18 + 2 = 0\ \checkmark\)
Example 4 · demanding

A symmetric form with a zero component

Write the symmetric equations of the line through \((2,-1,5)\) with direction \(\vv{m} = (3,0,-2)\), and describe the line geometrically.

Show solution
  1. Parametric first: \(x = 2 + 3t,\quad y = -1,\quad z = 5 - 2t\).
  2. The \(y\)-equation has no \(t\) in it, so it cannot be solved for \(t\). It becomes a standalone condition.
  3. Solve the other two for \(t\) and set them equal: \(\dfrac{x-2}{3} = \dfrac{z-5}{-2},\qquad y = -1\).
  4. Geometrically: the line lies entirely in the plane \(y=-1\), running parallel to the \(xz\)-plane. Writing \(\tfrac{y+1}{0}\) would have been meaningless, and would have cost a mark.

Watch out

Three ways to lose a mark here
  • Assuming your answer is wrong because it looks different. Any nonzero multiple of \(\vv{m}\) is a valid direction, and any point of the line is a valid starting point, so two correct answers can share no digits at all. Settle it by testing a point, not by comparing strings.
  • Putting a zero in a denominator. \(\tfrac{y-y_0}{0}\) is not a slip of notation, it is undefined. Split that coordinate off as its own equation.
  • Sign errors when a coordinate is negative. With \(y_0 = -1\) the numerator is \(y+1\), not \(y-1\). Write \(y-(-1)\) in rough and simplify on the next line.