Three Dimensional Geometry · Lesson 2 of 5

Representing a Line in Space

“Build line equations from a point and a direction, then move confidently between vector and coordinate forms.”

Learning Objectives

• Explain why one point and a nonzero direction determine a line. • Derive vector, parametric, and Cartesian equations of a line. • Construct a line through two distinct points. • Convert between forms and test whether a point lies on a line. • Handle zero direction components without dividing by zero.

A location and a direction

A direction alone describes many parallel lines. To choose one of them, fix a point that the line must pass through. You can then reach every point on the line by moving forwards or backwards from this anchor in the chosen direction. This is the geometric meaning of a line equation.

Let $A$ be the fixed point with position vector $a = O A$ , and let $b \neq = 0$ be a vector parallel to the line. If $P$ is any point on that line, then $A P$ is parallel to $b$ . Therefore $A P = λ b$ for some real number $λ$ . The number $λ$ is called a parameter: changing it selects a different point.

An anchor and a variable displacement— The position vector of P is the anchor vector plus a displacement parallel to the line.

Vector addition gives $O P = O A + A P$ . Calling $O P$ the variable vector $r$ , we obtain the equation below. Conversely, every real choice of $λ$ gives a point along the same direction from $A$ , so the equation describes the whole infinite line.

Vector equation of a lineLaTeX

r = a + λ b, λ \in R, b \neq = 0

The vector

a

fixes a point, while

b

fixes a direction. At

λ = 0

we are at the anchor; positive and negative values move in opposite directions.

The parameter is not automatically a distance. The displacement has length $∣ λ ∣ ∣ b ∣$ . It equals $∣ λ ∣$ only if the direction vector has unit length. Likewise, the anchor vector and the direction vector do different jobs: replacing one by the other usually changes the line.

The line extends in both directions

The full line requires $λ \in R$ . Restricting $λ \geq 0$ gives a ray from the anchor. Restricting $0 \leq λ \leq 1$ gives only one segment when the direction is the displacement between its endpoints. Do not add such restrictions unless the question asks for them.

Separate the three coordinates

A vector equation packages three coordinate equations into one statement. Expanding it makes each coordinate change visible. This form is particularly useful for finding points and for cases in which a direction component is zero.

Write $a = x_{1} \hat{i} + y_{1} \hat{j} + z_{1} \hat{k}$ and $b = a \hat{i} + b \hat{j} + c \hat{k}$ . A general point has $r = x \hat{i} + y \hat{j} + z \hat{k}$ . Equate the coefficients of the three unit vectors; equality of vectors requires equality of every component.

Parametric equationsLaTeX

x = x_{1} + λa, y = y_{1} + λb, z = z_{1} + λ c

The same parameter

λ

must be used in all three equations. The numbers

a, b, c

are the direction ratios, while

x_{1}, y_{1}, z_{1}

locate the anchor.

If all three direction ratios are nonzero, solve each equation for $λ$ and equate the results. This eliminates the parameter and gives the symmetric Cartesian form. Its three fractions do not have to equal one; they equal the same freely varying parameter.

Cartesian form with nonzero denominatorsLaTeX

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}

This displayed form assumes

a, b, c \neq = 0

. Direction cosines

l, m, n

may replace the direction ratios in the denominators, with the same care about zero components.

Example — Build both forms from a parallel vector

Problem
Find the line through $(5, 2, - 4)$ parallel to $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$ .

1.Use the given point as the anchor and the given parallel vector as the direction: $r = (5, 2, - 4) + λ (3, 2, - 8)$ . In unit-vector notation this is $r = 5 \hat{i} + 2 \hat{j} - 4 \hat{k} + λ (3 \hat{i} + 2 \hat{j} - 8 \hat{k})$ .
2.The coordinate equations are $x = 5 + 3 λ$ , $y = 2 + 2 λ$ , and $z = - 4 - 8 λ$ .
3.Eliminate the parameter: $(x - 5) /3 = (y - 2) /2 = (z + 4) / (- 8)$ . Setting the common value to zero recovers the given anchor.

Example — A new anchor with the same direction

Problem
Write an equation of the line through $(1, 2, 3)$ parallel to $3 \hat{i} + 2 \hat{j} - 2 \hat{k}$ .

1.The direction ratios are $(3, 2, - 2)$ ; they are not the coordinates of the anchor.
2.The vector equation is $r = (1, 2, 3) + t (3, 2, - 2)$ , where $t$ is real.
3.Equivalently, $x = 1 + 3 t$ , $y = 2 + 2 t$ , $z = 3 - 2 t$ , or $(x - 1) /3 = (y - 2) /2 = (z - 3) / (- 2)$ .

Example — Read a position vector carefully

Problem
A line passes through the point with position vector $2 \hat{i} - \hat{j} + 4 \hat{k}$ and has direction $\hat{i} + 2 \hat{j} - \hat{k}$ . Find vector and Cartesian equations.

1.The point is $(2, - 1, 4)$ and a direction vector is $(1, 2, - 1)$ .
2.Thus $r = (2, - 1, 4) + s (1, 2, - 1)$ . Expanding gives $x = 2 + s$ , $y = - 1 + 2 s$ , $z = 4 - s$ .
3.Eliminating $s$ gives $(x - 2) /1 = (y + 1) /2 = (z - 4) / (- 1)$ . The plus sign in $y + 1$ comes from subtracting the negative anchor coordinate.

Read a line equation in reverse

When a line is already written in Cartesian form, recover its anchor and direction rather than trying to memorise separate conversion rules. Give the common fraction a name, then solve for the coordinates. This also exposes signs that can otherwise be misread.

Example — Convert Cartesian form to vector form

Problem
Convert $(x - 5) /3 = (y + 4) /7 = (z - 6) /2$ to a vector equation.

1.Set all three fractions equal to $t$ . Then $x = 5 + 3 t$ , $y = - 4 + 7 t$ , and $z = 6 + 2 t$ .
2.At $t = 0$ the anchor is $(5, - 4, 6)$ , and the parameter coefficients give direction $(3, 7, 2)$ .
3.The equation is $r = (5, - 4, 6) + t (3, 7, 2)$ . In particular, $y + 4$ corresponds to an anchor coordinate of $- 4$ , not $4$ .

Equivalent equations can look different. For instance, starting at $a + 2 b$ instead of $a$ produces the same line because this new anchor already lies on it. Replacing $b$ by $k b$ , where $k \neq = 0$ , also keeps the line: a change of parameter absorbs the scale. To show two equations represent the same line, check both a shared direction and a shared point.

Use two distinct points

Two distinct points determine a direction as well as a location. Choose the first as the anchor and use the displacement to the second as the direction vector. This connects the two-point direction formula from the previous lesson to the vector line equation.

Line through two pointsLaTeX

r = a + λ (b - a), λ \in R, a \neq = b

Here

a

and

b

are the position vectors of two points, so

b - a

is their displacement. This use of

b

as a position vector differs from its role as a direction vector in the earlier formula.

Example — A line determined by two points

Problem
Find a line through $A = (1, - 2, 3)$ and $B = (3, 2, 1)$ , and check that it contains both points.

1.Subtract: $A B = (2, 4, - 2)$ . A vector equation is $r = (1, - 2, 3) + t (2, 4, - 2)$ .
2.At $t = 0$ this gives $A$ ; at $t = 1$ it gives $B$ . At $t = 2$ it gives $(5, 6, - 1)$ , showing that the unrestricted equation continues beyond the segment.
3.The symmetric form is $(x - 1) /2 = (y + 2) /4 = (z - 3) / (- 2)$ . We could divide the direction vector by two to use $(1, 2, - 1)$ instead; the point parameter would then be rescaled.

Zero components and point checks

A zero direction component means that one coordinate stays fixed. It does not make the line invalid. Return to the parametric equations to describe this directly, and remember that a point belongs to a line only when all coordinate conditions can hold at the same parameter value.

Example — Avoid a zero denominator

Problem
Write a line through $(2, - 1, 4)$ with direction $(0, 3, - 2)$ .

1.Its vector equation is $r = (2, - 1, 4) + t (0, 3, - 2)$ .
2.The coordinate equations are $x = 2$ , $y = - 1 + 3 t$ , and $z = 4 - 2 t$ . Thus a Cartesian description is $x = 2$ together with $(y + 1) /3 = (z - 4) / (- 2)$ .
3.There is no need to write $(x - 2) /0$ . The fixed coordinate $x = 2$ states the missing condition accurately.

Example — One parameter must satisfy every coordinate

Problem
Does $P = (5, 7, 2)$ lie on $r = (1, - 1, 4) + t (2, 4, - 1)$ ? What about $Q = (5, 7, 1)$ ?

1.The $x$ equation gives $5 = 1 + 2 t$ , so $t = 2$ . The $y$ equation then gives $- 1 + 4 (2) = 7$ , as required.
2.At $t = 2$ , the $z$ coordinate must be $4 - 2 = 2$ . Hence $P$ lies on the line.
3.The point $Q$ has the same $x$ and $y$ coordinates but the wrong $z$ coordinate. No single value of $t$ produces it, so it is not on the line.

Do not divide by a hidden coefficient

The expression $(1 - x) /3 = t$ gives $x = 1 - 3 t$ , so its direction component is $- 3$ . Similarly, $(7 y - 14) / (2 p) = t$ gives $y = 2 + (2 p /7) t$ . Rewrite unusual numerators into coordinate equations before reading a direction. If a denominator vanishes for a parameter value, use the parametric description to discuss that limiting direction instead of an undefined fraction.

Check your understanding

Read anchors, directions, and parameters as separate pieces of information. The following questions test whether you can move between the geometric picture and an equation without relying on its appearance alone.

Quick check

Which vector fixes an anchor in $r = (2, 3, - 1) + t (4, 0, 2)$ ?

Quick check

Which is a correct Cartesian description of $r = (2, 3, - 1) + t (4, 0, 2)$ ?

Quick check

Which direction vector joins $A = (0, 1, 2)$ to $B = (2, - 1, 3)$ ?

Quick check

For $r = a + t d$ with $d \neq = 0$ , which change always preserves the whole line?

Quick check

What anchor and direction follow from $(x + 2) /3 = (y - 1) / (- 4) = (z + 5) /2$ ?

Quick check

On $r = (0, 0, 1) + t (1, 2, 3)$ , which point corresponds to $t = - 1$ ?

Quick check

Why can two identical anchor points not be used alone to determine a unique line?

Practice Problems

Find a line through $(- 2, 4, - 5)$ parallel to $(x + 3) /3 = (y - 4) /5 = (z + 8) /6$ , in vector and Cartesian forms.
Find vector and Cartesian equations through $(1, 0, - 2)$ with direction $(2, - 3, 4)$ .
Convert $(x - 4) / (- 2) = (y + 1) /3 = (z - 2) /5$ to a vector equation and identify its anchor.
Find a line through $A = (0, 1, 2)$ and $B = (2, - 1, 3)$ .
Write a Cartesian description of the line through $(3, 2, - 1)$ parallel to $(0, 1, 0)$ .
Test whether $(3, - 3, 4)$ and $(3, - 3, 5)$ lie on $r = (1, 1, 2) + t (1, - 2, 1)$ .
Explain why $r = (1, 2, 3) + t (2, - 1, 4)$ and $r = (3, 1, 7) + s (- 4, 2, - 8)$ represent the same line.
If a direction vector is $(3, 0, 4)$ , what distance is travelled when the parameter increases from $- 1$ to $2$ ?

The parallel line supplies direction $(3, 5, 6)$ . Use $r = (- 2, 4, - 5) + t (3, 5, 6)$ and $(x + 2) /3 = (y - 4) /5 = (z + 5) /6$ . Its anchor need not match the anchor of the given parallel line.

Key Takeaways

• A fixed anchor and a nonzero direction determine $r = a + λ b$ . • Vector, parametric, and Cartesian equations describe the same line in different forms. • Two distinct points supply the direction $b - a$ . • All three coordinates must correspond to the same parameter value. • A zero direction component gives a constant coordinate; it does not permit division by zero.

Previous · Lesson 1

Directions and Coordinates in Space

Next · Lesson 3

Angles and Relationships between Lines

A location and a direction

Separate the three coordinates

Read a line equation in reverse

Use two distinct points

Zero components and point checks

Check your understanding

Quiz

Practice Problems

Key Takeaways