Date: Winter 2026 | Source: Stewart, Calculus: Early Transcendentals, 9th ed., Section 12.5
Difficulty: Intermediate | Prerequisites: Vectors in 3D (Section 12.2), dot product (12.3), cross product (12.4)
This section introduces the equations that describe lines in three-dimensional space. In two dimensions you can write a line as y = mx + b, but that form breaks down once you add a third axis. Instead, you describe 3D lines using a point and a direction vector, which leads to three equivalent representations: vector equations, parametric equations, and symmetric equations. You will also learn to determine whether two lines in space are parallel, intersecting, or skew (a relationship that does not exist in 2D).
A line in 3D is pinned down by one point on it and a direction vector. From those two ingredients you can write vector, parametric, or symmetric equations. Two lines in space can be parallel, intersecting, or skew, and you tell them apart by comparing direction vectors and then checking whether a shared point exists.
Direction vector
A vector v = ⟨a, b, c⟩ that points along the line. Any scalar multiple of a direction vector is also a valid direction vector for the same line. In simple terms, it tells you which way the line goes and how steeply it climbs or falls in each coordinate direction.
Position vector
The vector r₀ = ⟨x₀, y₀, z₀⟩ from the origin to a known point on the line. Think of it as the starting location you build the line from.
Vector equation of a line
The equation r(t) = r₀ + tv, where t is a real-valued parameter. It says: start at r₀ and walk in the direction of v, with t controlling how far and which way you walk.
Parametric equations
The component form of the vector equation: x = x₀ + at, y = y₀ + bt, z = z₀ + ct. Each coordinate is written as a separate function of the parameter t.
Symmetric equations
The parameter-free form obtained by solving each parametric equation for t and setting them equal: (x − x₀)/a = (y − y₀)/b = (z − z₀)/c. If one direction component is zero, that coordinate is stated separately as a constant.
Parallel lines
Two lines whose direction vectors are scalar multiples of each other. They point the same (or exactly opposite) way and never meet.
Skew lines
Lines that are not parallel and do not intersect. This can only happen in three or more dimensions. Think of two roads on different levels of a motorway interchange: they are not parallel, yet they never cross.
Intersecting lines
Lines that share exactly one point. Their direction vectors are not parallel, and you can find parameter values that give the same (x, y, z) on both lines.
Given a point P₀ = (x₀, y₀, z₀) on the line and a direction vector v = ⟨a, b, c⟩:
r(t) = ⟨x₀, y₀, z₀⟩ + t⟨a, b, c⟩
This is the most compact form. Every value of t gives a different point on the line.
Break the vector equation into components:
x(t) = x₀ + at
y(t) = y₀ + bt
z(t) = z₀ + ct
These are what WebAssign usually asks you to enter.
Solve each parametric equation for t and set them equal:
(x − x₀) / a = (y − y₀) / b = (z − z₀) / c
If one of a, b, c is zero, you cannot divide by it. Instead, state that coordinate as a constant. For example, if b = 0, write (x − x₀) / a = (z − z₀) / c, y = y₀.
Two common setups appear in homework problems:
Line parallel to a given line. Read the direction vector straight from the given line's parametric equations. If the given line is x = 4 − 4t, y = −1 + 3t, z = 6 + 9t, the direction vector is ⟨−4, 3, 9⟩ (the coefficients of t).
Line through two points. If the line passes through P₁ = (x₁, y₁, z₁) and P₂ = (x₂, y₂, z₂), the direction vector is v = P₂ − P₁ = ⟨x₂ − x₁, y₂ − y₁, z₂ − z₁⟩. Use either point as the base point.
When a line must be perpendicular to two given vectors u and w, compute v = u × w. The cross product is, by definition, perpendicular to both input vectors, so it serves as the direction vector of the line.
Worked example (from the homework): find the line through (5, 5, 0) perpendicular to both i + j = ⟨1, 1, 0⟩ and j + k = ⟨0, 1, 1⟩.
v = ⟨1, 1, 0⟩ × ⟨0, 1, 1⟩ = ⟨(1)(1) − (0)(1), (0)(0) − (1)(1), (1)(1) − (1)(0)⟩ = ⟨1, −1, 1⟩
Parametric equations: x = 5 + t, y = 5 − t, z = t.
Symmetric equations: x − 5 = −(y − 5) = z.
To find where a line crosses a coordinate plane, set the appropriate coordinate to zero in the parametric equations and solve for t:
xy-plane: set z = 0, solve for t, then compute x and y
xz-plane: set y = 0, solve for t, then compute x and z
yz-plane: set x = 0, solve for t, then compute y and z
Worked example: the line through (5, −3, 7) with direction ⟨−1, 3, −2⟩ has parametric equations x = 5 − t, y = −3 + 3t, z = 7 − 2t.
xy-plane (z = 0): 7 − 2t = 0, so t = 7/2. Point: (3/2, 15/2, 0).
yz-plane (x = 0): 5 − t = 0, so t = 5. Point: (0, 12, −3).
xz-plane (y = 0): −3 + 3t = 0, so t = 1. Point: (4, 0, 5).
Given two lines L₁ (with direction v₁) and L₂ (with direction v₂):
Check whether v₁ and v₂ are parallel (one is a scalar multiple of the other). If yes, the lines are parallel. Done.
If not parallel, set the parametric equations equal and solve the resulting system. Use different parameter names (t for L₁, s for L₂) because the lines may reach the same point at different parameter values.
Solve two of the three equations for t and s. Then check the third equation.
If the third equation is satisfied, the lines intersect. Plug the values back in to get the intersection point.
If the third equation fails, the lines are skew.
L₁: x = 6 + 4t, y = 8 − 2t, z = 2 + 6t (direction ⟨4, −2, 6⟩)
L₂: x = 1 + 4s, y = 3 − 2s, z = 4 + 5s (direction ⟨4, −2, 5⟩)
Step 1: ⟨4, −2, 6⟩ and ⟨4, −2, 5⟩ are not scalar multiples (the first two ratios are 1, but 6/5 ≠ 1). Not parallel.
Step 2: Set equations equal.
6 + 4t = 1 + 4s → 4t − 4s = −5
8 − 2t = 3 − 2s → −2t + 2s = −5
From the second equation: t − s = 5/2. From the first: t − s = −5/4. These contradict each other, so there is no solution.
The lines are skew.
L₁: x = 7 − 3t, y = 8 + 12t, z = 5 − 6t (direction ⟨−3, 12, −6⟩)
L₂: x = 8 + 2s, y = −8s, z = 5 + 4s (direction ⟨2, −8, 4⟩)
Step 1: Check ratios: −3/2 = 12/(−8) = −6/4 = −3/2. All equal. The direction vectors are parallel.
The lines are parallel (or the same line, but for this problem they are distinct because plugging a point from one into the other does not satisfy the equations).
\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v} = \langle x_0 + at,\; y_0 + bt,\; z_0 + ct \rangle\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \quad \text{(symmetric equations)}\mathbf{v} = \mathbf{u} \times \mathbf{w} = \langle u_2 w_3 - u_3 w_2,\; u_3 w_1 - u_1 w_3,\; u_1 w_2 - u_2 w_1 \rangleThe first formula is the vector/parametric equation. The second is the symmetric form. The third is the cross product, used when you need a direction vector perpendicular to two given vectors.
Parametric line equations describe the trajectory of any object moving in a straight line through space: a laser beam in optics, a projectile in the absence of gravity, or the path a 3D printer head follows between waypoints. In computer graphics, ray tracing fires parametric lines from a virtual camera and tests where they intersect surfaces to render images.
Students often use the same parameter letter (t) for both lines when checking intersection. You must use different parameters (t and s) because the two objects may reach the same point at different parameter values.
Students sometimes assume non-parallel lines must intersect. In 3D, two non-parallel lines can be skew, meaning they pass each other without ever meeting. This is the most common mistake on exam problems about line relationships.
When a direction-vector component is zero, students try to write (y − y₀)/0 in the symmetric equations. Instead, pull that coordinate out as a separate equation y = y₀.
Students sometimes confuse the base point and the direction vector when reading parametric equations. The constant terms are the base point. The coefficients of t are the direction vector.
⚠️ "Find parametric and symmetric equations" is a near-certain exam question. Know the conversion cold.
⚠️ The parallel/skew/intersecting classification question appears regularly. Practise the three-step procedure until it is automatic.
⚠️ Cross-product problems (line perpendicular to two vectors) combine Sections 12.4 and 12.5. Expect at least one on the final.
⚠️ Coordinate-plane intersection is quick to test and quick to mark. Set the right coordinate to zero, solve for t, compute the other two.
True or false: Two non-parallel lines in 3D must intersect.
Fill in the blank: The direction vector of the line x = 3 + 7t, y = −2t, z = 1 + 5t is ⟨___, ___, ___⟩.
True or false: If a direction-vector component is zero, the corresponding coordinate is constant along the line.
Fill in the blank: To find where a line crosses the xz-plane, set ___ = 0.
True or false: Symmetric equations can always be written as a single chain of equal fractions with no side condition.
Answers: 1. False (they can be skew). 2. ⟨7, −2, 5⟩. 3. True. 4. y. 5. False (a zero direction component forces a separate equation).
Q: Find parametric equations for the line through (2, −1, 4) parallel to the line x = 1 + 3t, y = 5 − t, z = 2t.
A: Direction vector is ⟨3, −1, 2⟩. Parametric equations: x = 2 + 3t, y = −1 − t, z = 4 + 2t.
Q: Write symmetric equations for the line through (0, 0, 0) and (6, 7, −1).
A: Direction vector is ⟨6, 7, −1⟩. Symmetric equations: x/6 = y/7 = z/(−1), which simplifies to x/6 = y/7 = −z.
Q: Find the direction vector for a line that is perpendicular to both ⟨1, 0, 2⟩ and ⟨0, 3, 1⟩.
A: Compute the cross product: ⟨(0)(1) − (2)(3), (2)(0) − (1)(1), (1)(3) − (0)(0)⟩ = ⟨−6, −1, 3⟩.
Q: The line x = 3 − 2t, y = 1 + t, z = −4 + 3t crosses the xy-plane at what point?
A: Set z = 0: −4 + 3t = 0, so t = 4/3. Then x = 3 − 8/3 = 1/3, y = 1 + 4/3 = 7/3. The point is (1/3, 7/3, 0).
Q: Are the lines L₁: x = 1 + 2t, y = 3 − t, z = 5 + 4t and L₂: x = 3 + 4s, y = 1 − 2s, z = 7 + 8s parallel, skew, or intersecting?
A: Direction vectors are ⟨2, −1, 4⟩ and ⟨4, −2, 8⟩. Since ⟨4, −2, 8⟩ = 2⟨2, −1, 4⟩, the lines are parallel.
This material connects directly to Section 12.6 (planes), where you will use many of the same tools: a point plus a normal vector defines a plane the way a point plus a direction vector defines a line. The cross product from Section 12.4 reappears here for perpendicularity problems and will keep appearing throughout the rest of the course.
Parametric equations for lines are a special case of the vector-valued functions you will study in Chapter 13. There, the direction vector can change from point to point, giving curves instead of straight lines.
The parallel/skew/intersecting classification is also the foundation for later work on distances between lines (Section 12.5 Part 2) and angles between lines and planes.
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