Difficulty: Intermediate | Prerequisites: Vector notation, dot product, cross product basics
Once you can work with vectors, the next step is describing geometric objects in 3D space. Lines and planes are the building blocks: you need a line equation to describe the path of a particle or the edge where two surfaces meet, and you need a plane equation to describe a flat surface, a tangent plane, or a constraint. Calculus III gives you three ways to write a line (vector, parametric, symmetric) and one standard form for a plane. Knowing when and how to convert between them is a core exam skill.
A line in 3D needs a point and a direction vector, and can be written in vector, parametric, or symmetric form. A plane needs a point and a normal vector. Most problems give you the ingredients in disguise, and the work is extracting the point and the direction (or normal) before plugging in.
Vector equation of a line
A single equation r(t) = r₀ + tv that describes every point on a line using a position vector r₀ and a direction vector v, scaled by parameter t. Think of it as: pick a starting point, then walk in the direction v; the parameter t says how far.
Parametric equations of a line
Three separate equations, one per axis: x = x₀ + at, y = y₀ + bt, z = z₀ + ct. These are the component-by-component version of the vector equation. In simple terms, each coordinate gets its own formula.
Symmetric equations of a line
Equations with no parameter t, found by solving each parametric equation for t and setting them equal: (x – x₀)/a = (y – y₀)/b = (z – z₀)/c. Think of it as: the parameter-free version of the line.
Direction vector
The vector ⟨a, b, c⟩ that tells you which way the line runs. Any scalar multiple of a direction vector works equally well.
Normal vector (to a plane)
A vector ⟨a, b, c⟩ perpendicular to every vector lying in the plane. It defines the plane's orientation. In simple terms, it is the vector that sticks straight out of the plane.
Equation of a plane
The standard form a(x – x₀) + b(y – y₀) + c(z – z₀) = 0, where (x₀, y₀, z₀) is any point on the plane and ⟨a, b, c⟩ is the normal vector.
The vector equation packs the whole line into one expression: r(t) = r₀ + tv.
r₀ is the position vector of your known point P₀: ⟨x₀, y₀, z₀⟩
v is the direction vector: ⟨a, b, c⟩
The final form looks like: r(t) = ⟨x₀, y₀, z₀⟩ + t⟨a, b, c⟩
Example: if the point is (1, 2, 3) and v = ⟨4, 5, 6⟩, then r(t) = ⟨1, 2, 3⟩ + t⟨4, 5, 6⟩.
Break the vector equation into its three components:
x = x₀ + at
y = y₀ + bt
z = z₀ + ct
Using the same example: x = 1 + 4t, y = 2 + 5t, z = 3 + 6t. This is the most common form used in calculus.
Solve each parametric equation for t, then set them all equal:
(x – x₀)/a = (y – y₀)/b = (z – z₀)/c
How to build them:
Subtract the point coordinate from the variable (e.g., x – x₀)
Divide by the direction component (e.g., divide by a)
Set the three fractions equal
The trap: if one of the direction components is 0 (say b = 0), you cannot divide by it. Instead, write the other two as equal and add the zero-component coordinate as a separate equation. For example, if b = 0: (x – x₀)/a = (z – z₀)/c, y = y₀.
To write any line equation you need two ingredients:
A point (x₀, y₀, z₀) on the line
A direction vector (a, b, c)
Then plug into: x = x₀ + at, y = y₀ + bt, z = z₀ + ct.
To write a plane equation you need two ingredients:
A point (x₀, y₀, z₀) on the plane
A normal vector ⟨a, b, c⟩ perpendicular to the plane
Then plug into: a(x – x₀) + b(y – y₀) + c(z – z₀) = 0.
Note the difference: for a line, (a, b, c) is the direction the line travels. For a plane, (a, b, c) is the normal, the direction perpendicular to the plane.
\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v} = \langle x_0, y_0, z_0 \rangle + t\langle a, b, c \ranglex = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}a(x - x_0) + b(y - y_0) + c(z - z_0) = 0Students mix up the role of (a, b, c). For a line, it is the direction vector (parallel to the line). For a plane, it is the normal vector (perpendicular to the plane). These are not interchangeable.
When writing symmetric equations, students often try to divide by 0 when a direction component is zero. You cannot do this. Write the zero-component as a standalone equation (e.g. y = y₀) and set the remaining fractions equal.
Students sometimes forget that any scalar multiple of a direction vector describes the same line. The direction ⟨2, 4, 6⟩ is equivalent to ⟨1, 2, 3⟩ for line purposes.
The displacement vector between two points on a line is a valid direction vector. Students sometimes use the points themselves instead of the difference.
⚠️ Converting between equation forms is a staple exam question. You must be able to go from vector to parametric to symmetric and back.
⚠️ The b = 0 trap in symmetric equations appears regularly. If you see a direction vector with a zero component, handle it before you get stuck.
⚠️ Plane problems often give you three points instead of a normal. To find the normal: take two vectors in the plane (from the three points), then cross them.
⚠️ Finding the line of intersection of two planes: solve the two plane equations simultaneously and express the result in parametric form.
Line equations describe the trajectory of a projectile, a laser beam, or a robot arm moving through space. Plane equations describe flat surfaces, walls, or constraints in optimisation problems. In computer graphics, ray tracing uses parametric line equations to trace light paths through a 3D scene, and plane equations determine where those rays hit surfaces.
True or false: a line in 3D can be fully described by a single point.
False. You need a point and a direction vector.
Fill in the blank: the normal vector to a plane is ______ to the plane.
Perpendicular (orthogonal).
True or false: if the direction vector is ⟨0, 3, 1⟩, you can write symmetric equations with three fractions.
False. The zero component means you write y = y₀ separately and only set the other two fractions equal.
Q: Write parametric equations for the line through (2, –1, 4) with direction vector ⟨3, 0, –5⟩.
A: x = 2 + 3t, y = –1, z = 4 – 5t.
Q: Find the equation of the plane through (1, 2, 3) with normal vector ⟨4, –1, 2⟩.
A: 4(x – 1) + (–1)(y – 2) + 2(z – 3) = 0, which simplifies to 4x – y + 2z = 8.
Q: Convert the parametric equations x = 3 + 2t, y = 1 – t, z = 5 + 4t into symmetric form.
A: (x – 3)/2 = (y – 1)/(–1) = (z – 5)/4.
Q: Two planes are given by 2x + y – z = 1 and x – y + 3z = 5. Find a direction vector for their line of intersection.
A: Cross the normal vectors: ⟨2, 1, –1⟩ × ⟨1, –1, 3⟩ = ⟨(1)(3) – (–1)(–1), –((2)(3) – (–1)(1)), (2)(–1) – (1)(1)⟩ = ⟨2, –7, –3⟩.
Line equations feed directly into parametric curves and the derivative ladder (velocity, acceleration). Plane equations reappear in tangent planes to surfaces and in setting up double and triple integrals over regions bounded by planes.
vector equation of a line, parametric equations 3D, symmetric equations, direction vector, normal vector, equation of a plane, point-normal form, line of intersection of two planes, Calc III, Calculus III, lines and planes, 3D geometry, multivariable calculus