Difficulty: Intermediate | Prerequisites: Vectors in two dimensions, dot product basics, cross product basics (MATH 124/125 or equivalent).
Lines and planes are the fundamental geometric objects of three-dimensional space, and nearly every topic in Calculus III builds on them. You need lines to describe particle trajectories and tangent directions, and you need planes to describe tangent surfaces and constraints. Distance and projection formulas let you measure how far apart objects are and decompose vectors into useful components. If you are comfortable with vectors, dot products, and cross products from MATH 124/125, you have everything you need here.
Lines in 3D are defined by a point and a direction vector, expressed as parametric or symmetric equations. Planes are defined by a point and a normal vector. The distance formula measures point-to-plane separation, and projections break one vector into components along another.
Parametric equations (of a line)
Equations of the form x = x₀ + ta, y = y₀ + tb, z = z₀ + tc, where (x₀, y₀, z₀) is a point on the line, ⟨a, b, c⟩ is the direction vector, and t is a parameter. Think of it as: each value of t gives you a different point along the line, like a slider.
Symmetric equations (of a line)
The form (x - x₀)/a = (y - y₀)/b = (z - z₀)/c, obtained by solving each parametric equation for t and setting them equal. In simple terms, this is a way to describe a line without explicitly writing the parameter t.
Normal vector
A vector n = ⟨a, b, c⟩ that is perpendicular to a plane. The coefficients a, b, c appear directly in the plane equation a(x - x₀) + b(y - y₀) + c(z - z₀) = 0. Think of it as: the vector that "sticks straight out" from the plane's surface.
Scalar projection (component)
The signed length of the shadow of vector b onto the direction of vector a: comp_a b = (a · b) / |a|. In simple terms, this tells you how much of b points in the direction of a.
Vector projection
The actual vector that represents b's shadow along a: proj_a b = ((a · b) / |a|²) a. Think of it as: the piece of b that lies along a, as a vector with both magnitude and direction.
Distance from a point to a plane
The perpendicular distance D = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²), where the plane is ax + by + cz + d = 0 and the point is (x₁, y₁, z₁). In simple terms, the shortest straight-line distance from a point to the nearest spot on the plane.
To define a line in three dimensions, you need two pieces of information:
A point on the line: P₀ = (x₀, y₀, z₀)
A direction vector: v = ⟨a, b, c⟩
Parametric form
The position vector is r = r₀ + tv, which expands to:
x = x₀ + ta
y = y₀ + tb
z = z₀ + tc
As t varies over all real numbers, you trace out every point on the line.
Symmetric form
Solve each parametric equation for t and set them equal:
(x - x₀)/a = (y - y₀)/b = (z - z₀)/c = t
If one of a, b, or c is zero, that coordinate is constant, so you write it separately (for example, if c = 0, then z = z₀ and you only set the x and y fractions equal).
To define a plane, you need:
A point on the plane: A = (x₀, y₀, z₀)
A normal vector: n = ⟨a, b, c⟩
Equation of a plane
a(x - x₀) + b(y - y₀) + c(z - z₀) = 0
This expands to ax + by + cz = d, where d = ax₀ + by₀ + cz₀.
The normal vector's components become the coefficients in the plane equation directly. If you are given two vectors that lie in the plane, take their cross product to find the normal.
Given a plane ax + by + cz + d = 0 and a point P₁ = (x₁, y₁, z₁), the perpendicular distance is:
D = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²)
The numerator is the absolute value of plugging the point into the left side of the plane equation. The denominator is the magnitude of the normal vector. This formula works because you are projecting the vector from any point on the plane to P₁ onto the unit normal.
Note: make sure the plane equation is in the form ax + by + cz + d = 0 (everything on one side) before applying this formula.
Scalar projection (component of b onto a)
comp_a b = (a · b) / |a| = |b| cos θ
This gives you a signed number: positive if b has a component in the direction of a, negative if it points the other way.
Vector projection (projection of b onto a)
proj_a b = (a · b / |a|²) a = ((a · b) / (a · a)) a
This gives you the actual vector, pointing along a, whose length equals the scalar projection. Geometrically, it is the shadow of b dropped perpendicularly onto the line through a.
The relationship between the two: the vector projection is the scalar projection times the unit vector in the direction of a.
Line (parametric): r = r₀ + tv, i.e. x = x₀ + ta, y = y₀ + tb, z = z₀ + tc
Line (symmetric): (x - x₀)/a = (y - y₀)/b = (z - z₀)/c
Plane: a(x - x₀) + b(y - y₀) + c(z - z₀) = 0
Distance (point to plane): D = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²)
Scalar projection: comp_a b = (a · b) / |a|
Vector projection: proj_a b = ((a · b) / |a|²) a
Parametric line equations describe the trajectory of any object moving at constant velocity: plug in a time and you get the position. Computer graphics use these equations constantly for ray tracing, where each ray of light is a parametric line tested for intersection with surfaces.
The distance formula shows up in collision detection (how close is an object to a boundary?) and in least-squares fitting (minimising perpendicular distances from data points to a best-fit plane).
Students often confuse the direction vector of a line with the normal vector of a plane. The direction vector is parallel to the line; the normal vector is perpendicular to the plane. They play completely different roles.
Students sometimes forget the absolute value in the distance formula and end up with a negative distance, which is geometrically meaningless.
Scalar projection and vector projection are frequently mixed up. The scalar projection is a number (possibly negative). The vector projection is a vector.
When converting to symmetric form, students sometimes leave a zero in a denominator. If one component of the direction vector is zero, that coordinate is constant and must be written separately.
⚠️ Converting between parametric and symmetric forms of a line is a staple exam question. Practise going both directions.
⚠️ The point-to-plane distance formula appears frequently, often embedded in a larger problem (e.g. finding the distance between two parallel planes).
⚠️ You will almost certainly need to compute a vector projection as a step inside a curvature, acceleration, or work problem.
⚠️ Know how to find the equation of a plane given three points: form two vectors from the three points, take their cross product to get the normal, then plug into the plane equation.
True or false: The direction vector of a line is the same as the normal vector of the plane containing that line.
Fill in the blank: The symmetric equations of a line are obtained by solving each parametric equation for ______ and setting them equal.
True or false: The scalar projection of b onto a can be negative.
Fill in the blank: The denominator of the point-to-plane distance formula is the ______ of the normal vector.
True or false: proj_a b is always a scalar.
Q: Find the parametric equations of the line through (1, 2, 3) with direction vector ⟨4, 5, 6⟩.
A: x = 1 + 4t, y = 2 + 5t, z = 3 + 6t.
Q: Find the equation of the plane through (2, 1, -1) with normal vector ⟨3, -2, 5⟩.
A: 3(x - 2) - 2(y - 1) + 5(z + 1) = 0, which simplifies to 3x - 2y + 5z + 1 = 0.
Q: Find the distance from the point (1, 0, -3) to the plane 2x - y + 2z - 4 = 0.
A: D = |2(1) - 1(0) + 2(-3) - 4| / √(4 + 1 + 4) = |2 - 0 - 6 - 4| / 3 = |-8| / 3 = 8/3.
Q: Compute proj_a b where a = ⟨1, 2, 2⟩ and b = ⟨3, 0, 4⟩.
A: a · b = 3 + 0 + 8 = 11. |a|² = 1 + 4 + 4 = 9. So proj_a b = (11/9)⟨1, 2, 2⟩ = ⟨11/9, 22/9, 22/9⟩.
Q: A line passes through (0, 0, 0) and (1, 1, 1). Write its symmetric equations.
A: Direction vector = ⟨1, 1, 1⟩. Symmetric form: x/1 = y/1 = z/1, or simply x = y = z.
Lines connect directly to arc length and curvature: the tangent line at a point on a curve uses the same parametric machinery, with the direction vector replaced by r'(t). Planes connect to tangent planes for multivariable functions, where the normal vector comes from the gradient. Projections are the engine behind decomposing acceleration into tangential and normal components, which is covered in the curvature and acceleration notes.
Parametric equations, symmetric equations, direction vector, position vector, normal vector, equation of a plane, point-to-plane distance, perpendicular distance, scalar projection, component of a vector, vector projection, dot product, cross product, MATH 126, Calculus III, multivariable calculus, 3D geometry, lines in space, planes in space, UW MATH 126