Source: Math 126 Spring 2023 Final Exam, University of Washington
Tags: planes in 3D, parametric lines, normal vectors, cross product, line of intersection, line-surface intersection, paraboloid, MATH 126, Calculus III, multivariable calculus
Difficulty: Intermediate | Prerequisites: Vectors, dot product, cross product (Chapter 12.1–12.5)
This material sits at the foundation of multivariable calculus. Before you can do partial derivatives or multiple integrals, you need fluency with lines and planes in three dimensions. The core skill is translating geometric descriptions (two planes intersecting, a line piercing a surface) into algebraic equations you can solve. If you are comfortable with vectors and the cross product from earlier in Chapter 12, you are ready for this.
Planes are defined by a normal vector and a point. Lines in 3D use parametric equations. To find where geometric objects meet, set their equations equal and solve for the parameter(s).
Normal vector
A vector perpendicular to a plane. It determines the plane's orientation in space. Think of it as an arrow sticking straight out of a flat sheet of paper.
Parametric equations (of a line)
Equations that express x, y, and z each as functions of a single parameter t. In simple terms, t acts like a slider: as you move t, you trace out the line.
Direction vector
A vector parallel to a line, pointing along it. If you have two points on the line, subtracting their coordinates gives you a direction vector.
Cross product
An operation on two vectors that returns a third vector perpendicular to both. This is how you find the normal to a plane when you know two vectors lying in it.
Paraboloid
A bowl-shaped surface described by an equation like z = x² + y² (opens upward) or z = c − x² − y² (opens downward). In simple terms, it is the 3D version of a parabola, rotated around its axis.
A plane is fully determined by a normal vector n = ⟨a, b, c⟩ and a point P₀ = (x₀, y₀, z₀) on the plane.
The equation is: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0
When two lines intersect, both direction vectors lie in the plane. Take their cross product to get the normal vector.
Example: direction vectors ⟨1, 1, −2⟩ and ⟨0, 4, −3⟩ give normal ⟨1, 1, −2⟩ × ⟨0, 4, −3⟩ = ⟨5, 3, 4⟩. With point (1, 0, 1), the plane is 5x + 3y + 4z = 9.
Two non-parallel planes intersect along a line. To find it:
Subtract one equation from the other to eliminate a variable.
Find two specific points on the line by choosing convenient values for the free parameter.
Use the two points to get a direction vector, then write parametric equations.
Example: x + y + z = 1 and 2x + y − 3z = −3. Subtracting gives −x + 4z = 4. Setting z = 1 gives x = 0, y = 0, so one point is (0, 0, 1). Setting z = 0 gives x = −4, y = 5, so another is (−4, 5, 0). Direction: ⟨−4, 5, −1⟩. Parametric form: x = −4t, y = 5t, z = 1 − t.
Write the line parametrically, then substitute into the surface equation.
Solve the resulting equation in t. Each value of t gives an intersection point.
Example: line through (0, 0, 1) and (3, 4, 1) has parametric form x = 3t, y = 4t, z = 1. Substituting into 100z = x² + y² gives 100 = 9t² + 16t² = 25t², so t = ±2. This yields intersection points (6, 8, 1) and (−6, −8, 1).
\text{Plane: } a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\text{Cross product: } \vec{u} \times \vec{v} = \langle u_2 v_3 - u_3 v_2,\; u_3 v_1 - u_1 v_3,\; u_1 v_2 - u_2 v_1 \rangle\text{Parametric line: } x = x_0 + at,\; y = y_0 + bt,\; z = z_0 + ctStudents often confuse direction vectors with normal vectors. A direction vector lies along the line or in the plane. A normal vector is perpendicular to the plane. They serve different roles.
When finding the line of intersection of two planes, some students try to cross the two plane equations rather than working with elimination. The cross product of the two normal vectors does give the direction of the intersection line, but you still need a point on the line.
Students sometimes forget that a line can intersect a quadric surface at more than one point (or not at all). Always solve the full equation in t and check all solutions.
The cross product is not commutative. ⟨u⟩ × ⟨v⟩ = −(⟨v⟩ × ⟨u⟩). Swapping the order flips the sign of your normal vector, which changes which side of the plane is "positive."
⚠️ Plane equations from intersecting lines (cross product of direction vectors) are a staple exam question.
⚠️ Line-of-intersection problems test both algebraic manipulation and geometric understanding. Be ready to find a point and a direction.
⚠️ Line-surface intersections can yield 0, 1, or 2 solutions. The exam expects you to find all of them.
⚠️ These skills recur in later topics: tangent planes (Chapter 14), surface integrals, and Stokes' theorem all build on the language of planes and normals.
True or False: The cross product of two direction vectors lying in a plane gives a vector parallel to that plane. (False, it gives the normal.)
Fill in the blank: To write parametric equations for a line, you need a ______ and a ______ vector. (point, direction)
True or False: Two non-parallel planes in 3D always intersect in a line. (True.)
Fill in the blank: If a line has parametric equations x = 3t, y = 4t, z = 1, substituting into 100z = x² + y² gives 100 = ______ t². (25)
Q: Two lines with direction vectors ⟨1, 1, −2⟩ and ⟨0, 4, −3⟩ intersect at (1, 0, 1). Find the equation of the plane containing both lines.
A: Cross product gives normal ⟨5, 3, 4⟩. Using point (1, 0, 1): 5(x−1) + 3y + 4(z−1) = 0, which simplifies to 5x + 3y + 4z = 9.
Q: Find parametric equations for the line of intersection of x + y + z = 1 and 2x + y − 3z = −3.
A: Subtract to get −x + 4z = 4. Two points: (0, 0, 1) and (−4, 5, 0). Direction ⟨−4, 5, −1⟩. Parametric form: x = −4t, y = 5t, z = 1 − t.
Q: Where does the line through (0, 0, 1) and (3, 4, 1) intersect the paraboloid 100z = x² + y²?
A: Parametric line: x = 3t, y = 4t, z = 1. Substituting: 100 = 25t², so t = ±2. Points: (6, 8, 1) and (−6, −8, 1).
This connects to tangent planes (Chapter 14) because the tangent plane at a point on a surface is defined by the gradient vector, which serves as the normal. The parametric-line toolkit also comes back in line integrals and flux calculations in Chapters 16–17.
equation of a plane, normal vector, cross product, parametric equations 3D, line of intersection of two planes, direction vector, line-surface intersection, paraboloid intersection, MATH 126, Calculus III, UW Math 126, multivariable geometry, analytic geometry 3D