Source: MATH 163 Calculus III, University of Washington
Difficulty: Intermediate | Prerequisites: Cross product (Section 12.4), dot product (Section 12.3), parametric equations from Calculus II.
In two dimensions, a line is described by a single equation like y = mx + b. That approach breaks down in three dimensions, where a single linear equation describes a plane, not a line. This section introduces three equivalent ways to represent lines in R3 (parametric, symmetric, and vector-valued) and shows how to define planes using a point and a normal vector. These representations are the foundation for nearly everything that follows in Calculus III, from surface integrals to Stokes' theorem.
A line in R3 is specified by a point it passes through and a direction vector, and can be written in parametric, symmetric, or vector-valued form. A plane is specified by a point and a normal vector, and its equation is ax + by + cz = d. You find the normal vector to a plane by taking the cross product of two vectors that lie in it.
Parametric equations (for a line in R3)
A set of three equations x(t) = x0 + At, y(t) = y0 + Bt, z(t) = z0 + Ct, where (x0, y0, z0) is a point on the line and (A, B, C) is the direction vector. Think of t as a dial: turn it and you slide along the line.
Direction vector
A vector (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.
Symmetric equations
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, it is the parametric form with the parameter eliminated.
Vector-valued function (for a line)
The form r(t) = r0 + vt, where r0 is a position vector to a point on the line and v is the direction vector. This packages the three parametric equations into a single vector expression.
Normal vector (to a plane)
A vector n = (a, b, c) that is perpendicular to every vector lying in the plane. It defines the plane's orientation in space.
Standard form of a plane
The equation ax + by + cz = d, where (a, b, c) is the normal vector and d = ax0 + by0 + cz0 for any point (x0, y0, z0) on the plane.
Vector equation of a plane
The form (r - r0) . n = 0, stating that any vector from a known point on the plane to an arbitrary point on the plane must be perpendicular to the normal vector.
To describe a line in R3, you need two things: a point the line passes through, and a direction it travels in.
Pick any point (x0, y0, z0) on the line.
Find a direction vector (A, B, C) along the line.
Write: x(t) = x0 + At, y(t) = y0 + Bt, z(t) = z0 + Ct.
If you are given two points, subtract one from the other to get the direction vector.
Worked example: find parametric equations for the line through (1, 2, 3) and (-4, 7, 13).
Direction vector: (-4 - 1, 7 - 2, 13 - 3) = (-5, 5, 10).
Using (1, 2, 3) as the point: x(t) = 1 - 5t, y(t) = 2 + 5t, z(t) = 3 + 10t.
Note: the parametrisation is not unique. Any point on the line and any scalar multiple of the direction vector give a different but equally valid set of parametric equations.
Solve each parametric equation for t, then set the expressions equal:
(x - x0)/A = (y - y0)/B = (z - z0)/C
From the example above: (x - 1)/(-5) = (y - 2)/5 = (z - 3)/10.
This form is compact, but it cannot be used when any component of the direction vector is zero (because you would be dividing by zero). In that case, write the zero component as a constant equation instead.
Package the parametric equations into a single vector:
r(t) = r0 + vt = (x0 + At, y0 + Bt, z0 + Ct)
Here r0 = (x0, y0, z0) is the position vector of the chosen point, and v = (A, B, C) is the direction vector. This is the form used most often in later sections of the course.
Example: r(t) = (1, -2, 5) + (3, 1, 7)t describes a line through (1, -2, 5) in the direction (3, 1, 7).
A plane requires a point and a normal vector (a vector perpendicular to the plane).
Start with the vector equation: (r - r0) . n = 0.
Expand in components: a(x - x0) + b(y - y0) + c(z - z0) = 0.
Simplify to standard form: ax + by + cz = d, where d = ax0 + by0 + cz0.
The coefficients a, b, c in the standard form are the components of the normal vector. This is a quick way to read off the normal from any plane equation.
To find the normal vector when you are given three points in the plane: form two vectors from those points, then take their cross product. The result is perpendicular to both, and therefore perpendicular to the plane.
\text{Parametric:} \quad x = x_0 + At, \quad y = y_0 + Bt, \quad z = z_0 + Ct\text{Symmetric:} \quad \frac{x - x_0}{A} = \frac{y - y_0}{B} = \frac{z - z_0}{C}\text{Vector-valued:} \quad \vec{r}(t) = \vec{r}_0 + \vec{v}\,t\text{Plane (vector form):} \quad (\vec{r} - \vec{r}_0) \cdot \vec{n} = 0\text{Plane (standard form):} \quad ax + by + cz = dParametric lines in 3D are how ray tracing works in computer graphics: each pixel on your screen corresponds to a ray (a line with a starting point and direction), and the renderer checks which objects that ray intersects. Every reflection, shadow, and refraction in a 3D-rendered scene starts with a parametric line equation.
Plane equations are used in computer-aided design and manufacturing to define flat surfaces, cutting planes, and mirror planes of symmetry in molecular chemistry.
Students often try to write y = mx + b style equations for lines in R3. A single linear equation in three variables describes a plane, not a line. Lines in R3 require parametric equations (or the symmetric/vector forms).
The direction vector is not unique. Multiplying it by any non-zero scalar gives a different, equally valid direction vector for the same line. The parametric equations will look different but trace the same line.
When converting to symmetric form, students sometimes forget that if one component of the direction vector is zero, that variable is simply constant and does not appear in the chain of equal fractions.
In the plane equation ax + by + cz = d, the normal vector is (a, b, c), not (a, b, c, d). Students sometimes include d as a component of the normal vector.
Expect to be asked to write parametric equations for a line through two given points. This is a near-certain exam question.
Converting between parametric, symmetric, and vector-valued forms is commonly tested. Be fluent in all three.
Finding the equation of a plane given three points (form two vectors, cross product for the normal, plug into standard form) is a standard multi-step problem.
The distance from a point to a line (using the cross product) was flagged in the notes as an upcoming topic. It is likely to appear on the final.
True or false: a single linear equation in x, y, z describes a line in R3.
Fill in the blank: to write parametric equations for a line, you need a ___ and a ___.
True or false: the direction vector of a line is unique.
Fill in the blank: the coefficients of x, y, z in the standard plane equation are the components of the ___.
True or false: the vector equation of a plane is (r - r0) . n = 0.
Answers: 1. False (it describes a plane). 2. Point, direction vector. 3. False (any scalar multiple works). 4. Normal vector. 5. True.
Q: Write parametric equations for the line through (3, -1, 4) and (0, 2, -5).
A: Direction vector: (0 - 3, 2 - (-1), -5 - 4) = (-3, 3, -9). Parametric equations: x(t) = 3 - 3t, y(t) = -1 + 3t, z(t) = 4 - 9t.
Q: Convert the parametric equations x = 2 + t, y = -1 + 3t, z = 5 - 2t into symmetric form.
A: Solve for t: (x - 2)/1 = (y + 1)/3 = (z - 5)/(-2).
Q: Write the vector-valued function for the line through (1, -2, 5) with direction vector (3, 1, 7).
A: r(t) = (1, -2, 5) + (3, 1, 7)t = (1 + 3t, -2 + t, 5 + 7t).
Q: Find the equation of the plane with normal vector (2, -1, 3) passing through the point (1, 4, -2).
A: 2(x - 1) + (-1)(y - 4) + 3(z + 2) = 0. Expanding: 2x - y + 3z = -8.
Q: Find the equation of the plane containing the points (1, 0, 0), (0, 1, 0), and (0, 0, 1).
A: Form two vectors: (0 - 1, 1 - 0, 0 - 0) = (-1, 1, 0) and (0 - 1, 0 - 0, 1 - 0) = (-1, 0, 1). Cross product: (1, 1, 1). Using point (1, 0, 0): 1(x - 1) + 1(y) + 1(z) = 0, so x + y + z = 1.
Lines in R3 become the foundation for vector-valued functions and curves in space (Sections 13.1 onwards), where the parameter t represents time and the line becomes a trajectory. Plane equations connect directly to partial derivatives and tangent planes later in the course: the tangent plane to a surface at a point uses the same ax + by + cz = d structure, with the normal found from the gradient. Cross products from Section 12.4 are the tool you will use to find normals to planes whenever three points are given.
Parametric equations, symmetric equations, vector-valued function, direction vector, position vector, line in R3, equation of a plane, normal vector, standard form of a plane, point-normal form, three-dimensional lines, MATH 163, Calculus III, Section 12.5, parametric line 3D, plane equation from three points, ray tracing, tangent plane