Difficulty: Intermediate | Prerequisites: Vectors, dot product, cross product (§12.3 and §12.4)
This material sits at the heart of multivariable geometry. Sections 12.5 and 12.6 move from vector algebra into the equations that describe lines, planes and surfaces in three-dimensional space. If you can compute dot products and cross products confidently, you have what you need. If not, revisit §12.3 and §12.4 before going further. Nearly everything in later chapters (partial derivatives, surface integrals, Stokes' theorem) assumes fluency with these representations.
Lines in 3D are described by a point and a direction vector; planes by a point and a normal vector. Parallel planes share the same normal, and the distance between them (or from a point to a plane) follows a clean formula. Many problems reduce to setting up the right vector equation and reading off the geometry.
Normal vector
A vector perpendicular to a plane (or surface). Every plane in 3D can be specified by one point on it and a normal vector. Think of it as the arrow that sticks straight out of the plane's surface.
Direction vector
A vector parallel to a line, giving the line its orientation in space. Any nonzero scalar multiple of a direction vector describes the same line. In simple terms, it tells you which way the line points.
Parametric equations of a line
A way to express a line using a parameter t: x = x₀ + at, y = y₀ + bt, z = z₀ + ct, where (x₀, y₀, z₀) is a point on the line and ⟨a, b, c⟩ is the direction vector. Think of t as time: as t varies, you trace every point on the line.
Symmetric equations of a line
(x - x₀)/a = (y - y₀)/b = (z - z₀)/c, obtained by solving each parametric equation for t and setting them equal. Only valid when a, b and c are all nonzero.
Scalar (linear) equation of a plane
a(x - x₀) + b(y - y₀) + c(z - z₀) = 0, or equivalently ax + by + cz = d. The coefficients ⟨a, b, c⟩ form the normal vector. In simple terms, this is the single equation that every point on the plane satisfies.
Parallel planes
Two planes are parallel when their normal vectors are scalar multiples of each other. Parallel planes never intersect. The distance between them is constant everywhere.
Distance from a point to a plane
For the plane ax + by + cz = d and the point (x₁, y₁, z₁), the distance is |ax₁ + by₁ + cz₁ - d| / √(a² + b² + c²). Think of it as projecting the displacement onto the normal direction.
A plane is determined by a point P₀ = (x₀, y₀, z₀) and a normal vector n = ⟨a, b, c⟩.
The scalar equation is a(x - x₀) + b(y - y₀) + c(z - z₀) = 0.
Expanding gives the general form ax + by + cz = d, where d = ax₀ + by₀ + cz₀.
The plane x = 5 has normal vector ⟨1, 0, 0⟩. Any plane parallel to it has the form x = k for some constant k.
Two planes are parallel when they share the same normal vector (up to a scalar multiple).
To find all planes parallel to ax + by + cz = d₁ and a distance D away: the second plane has equation ax + by + cz = d₂, and the distance between them is |d₂ - d₁| / √(a² + b² + c²) = D.
This gives two solutions (one on each side of the original plane).
Example: planes parallel to x = 5 and 2 units away are x = 3 and x = 7.
Example: planes parallel to x - y + z = 0 and √3 units away. The normal is ⟨1, -1, 1⟩ with magnitude √3, so |d| / √3 = √3 gives d = ±3, yielding x - y + z = 3 and x - y + z = -3.
A line through P₀ = (x₀, y₀, z₀) with direction v = ⟨a, b, c⟩ has the vector equation r(t) = r₀ + tv.
Parametric form: x = x₀ + at, y = y₀ + bt, z = z₀ + ct.
If you are given two points, the direction vector is their difference.
Example: the line through (2, 0, 1) and (2, 0, -5) has direction ⟨0, 0, -6⟩, which simplifies to ⟨0, 0, 1⟩.
The set of points equidistant from two points A and B lies on the plane that is the perpendicular bisector of segment AB.
To find the locus equidistant from three points A, B, C: intersect the perpendicular bisector planes of AB and AC. The result is typically a line.
Method: set |P - A|² = |P - B|² and |P - A|² = |P - C|², expand, and solve the resulting linear system.
Between two points on moving objects: if positions are r₁(t) and r₂(t), the squared distance is a quadratic in t. Minimise it by differentiating and setting the derivative to zero.
Between two skew lines: the shortest distance is |(P₂ - P₁) · (v₁ × v₂)| / |v₁ × v₂|, where P₁, P₂ are points on each line and v₁, v₂ are their direction vectors.
If v₁ × v₂ = 0, the lines are parallel, and the distance reduces to a point-to-line calculation.
To check whether a line segment intersects a sphere: parametrise the segment, substitute into the sphere equation, and check whether the resulting quadratic has a solution with 0 ≤ t ≤ 1.
To check whether a line blocks the view: determine whether the line passes between the two observers. This requires finding the closest point of approach and checking geometry.
\text{Plane: } a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\text{Line (vector): } \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}\text{Line (parametric): } x = x_0 + at,\; y = y_0 + bt,\; z = z_0 + ct\text{Distance, point to plane: } D = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}\text{Distance between parallel planes } ax+by+cz=d_1 \text{ and } ax+by+cz=d_2\text{: } D = \frac{|d_1 - d_2|}{\sqrt{a^2+b^2+c^2}}\text{Distance between skew lines: } D = \frac{|(\mathbf{P}_2 - \mathbf{P}_1) \cdot (\mathbf{v}_1 \times \mathbf{v}_2)|}{|\mathbf{v}_1 \times \mathbf{v}_2|}Line and plane equations underpin computer graphics: every ray-traced scene tests millions of ray-plane and ray-surface intersections per frame. The distance-between-lines formula is used in robotics to compute clearance between moving parts. GPS trilateration is, at its core, finding the intersection of surfaces defined by distance equations.
Students often think the coefficients in ax + by + cz = d give a direction vector of the plane. They do not. The vector ⟨a, b, c⟩ is the normal (perpendicular to the plane), not a direction lying in it.
A common error is forgetting the absolute value in the distance formula, which can produce negative distances. Distance is always non-negative.
Students sometimes assume two lines in 3D that do not intersect must be parallel. In three dimensions, non-parallel lines can be skew (they simply miss each other in different planes).
When finding the line equidistant from three points, students often try to find a single point equidistant from all three rather than recognising the answer is a line (the intersection of two planes).
⚠️ The point-to-plane distance formula appears repeatedly on exams. Be sure you can write it from memory and apply it quickly.
⚠️ Finding a plane through three points (compute two in-plane vectors, take their cross product for the normal) is a standard exam question.
⚠️ Skew-line distance problems are a favourite. Know the cross-product formula and practise setting it up from word problems.
⚠️ "Equidistant from" problems test whether you can translate a geometric condition into algebra. Expect at least one on the exam.
True or false: the normal vector to the plane 2x - 3y + z = 7 is ⟨2, -3, 1⟩.
Fill in the blank: two planes are parallel if and only if their normal vectors are ______ of each other.
True or false: in 3D, two lines that do not intersect must be parallel.
Fill in the blank: the distance from the point (1, 0, 0) to the plane x + y + z = 1 is ______.
True or false: the symmetric equations of a line can always be written regardless of the direction vector's components.
Answers: 1. True. 2. Scalar multiples. 3. False (they may be skew). 4. 0 (the point lies on the plane). 5. False (if any component of the direction vector is zero, the symmetric form is undefined for that coordinate).
Q: Find the equation of all planes parallel to x = 5 that are exactly 2 units away from it.
A: The plane x = 5 has normal ⟨1, 0, 0⟩. Parallel planes have the form x = k. The distance |k - 5| / 1 = 2 gives k = 3 or k = 7. The planes are x = 3 and x = 7.
Q: Find the equation of the line equidistant from A(2, 1, -3), B(3, 4, -1) and C(-2, 1, -1).
A: Set |P - A|² = |P - B|² and expand to get a plane equation. Do the same with |P - A|² = |P - C|². The intersection of these two planes is the required line. Expand the squared distances, cancel the x² + y² + z² terms, and solve the resulting two linear equations simultaneously.
Q: Two spaceships travel as r₁(t) = ⟨-1, 3, 2⟩ + t⟨-3, 1, -1⟩ and r₂(t) = ⟨4, -2, 1⟩ + t⟨2, 8, 4⟩. Find the shortest distance between the ships.
A: Compute |r₁(t) - r₂(t)|² as a function of t, which is a quadratic. Differentiate with respect to t, set the derivative to zero, and solve for t. Substitute back to get the minimum distance. (Note: both ships share the same parameter t, since they travel simultaneously.)
Q: Find the shortest distance between the two trajectory lines of the ships above.
A: The lines may be evaluated at different parameter values (say s and t). Use the skew-line distance formula: D = |(P₂ - P₁) · (v₁ × v₂)| / |v₁ × v₂|. Compute v₁ × v₂ = ⟨-3, 1, -1⟩ × ⟨2, 8, 4⟩, then dot (P₂ - P₁) with the result, and divide by the magnitude.
Q: Determine whether the line segment from (-1, 0, 2) to (1, 0, 2) intersects the unit sphere centred at the origin.
A: Parametrise the segment as (1 - 2t)(-1, 0, 2) + 2t(1, 0, 2) or equivalently (-1 + 2t, 0, 2) for 0 ≤ t ≤ 1. Substitute into x² + y² + z² = 1: (-1 + 2t)² + 0 + 4 = 1, so (-1 + 2t)² = -3, which has no real solution. The segment does not intersect the sphere.
This connects to §12.6 (quadric surfaces) because planes are the simplest surfaces, and many surface problems start with the plane case. The cross product (§12.4) is the primary tool for computing normals and distances here. Later, in §14.6 and §16.6, you will compute tangent planes to surfaces and flux through surfaces, both of which rely on normal vectors. The parametric line equation reappears in §13.1 as the simplest vector-valued function.
lines in 3D, planes in 3D, parametric equations of a line, symmetric equations, vector equation of a line, scalar equation of a plane, normal vector, direction vector, parallel planes, distance from point to plane, distance between parallel planes, skew lines, distance between skew lines, perpendicular bisector plane, equidistant locus, line-plane intersection, line-sphere intersection, Math 215, Calculus III, Stewart §12.5, multivariable calculus