Difficulty: Intermediate | Prerequisites: Vectors, dot product, cross product (MATH 124/125 material)
This topic sits at the core of the multivariable portion of MATH 126. You are working in three dimensions now, and everything builds on being able to describe lines and planes with equations. If you can find a normal vector and write a parametric line, you can handle intersections, angles between planes, and projections onto surfaces. You should already be comfortable with basic vector operations (dot product, cross product, magnitude) before diving in.
Lines in 3D are described by parametric equations using a point and a direction vector. Planes are described by a normal vector and a point on the plane. To find where things intersect (two planes, a line and a surface), you substitute parametric expressions into the relevant equations and solve. The cross product is your main tool for producing normal vectors.
Normal vector
A vector perpendicular to a plane (or surface). For a plane, the normal vector appears directly in the plane equation: if the normal is ⟨a, b, c⟩, the plane equation is a(x − x₀) + b(y − y₀) + c(z − z₀) = 0. In simple terms, the normal tells you which way the plane "faces."
Direction vector
A vector that points along a line. In parametric form, the direction vector gives the coefficients of the parameter t. Think of it as the velocity of a point travelling along the line.
Parametric equations (of a line)
Equations of the form x = x₀ + at, y = y₀ + bt, z = z₀ + ct, where (x₀, y₀, z₀) is a known point on the line and ⟨a, b, c⟩ is the direction vector. In simple terms, you pick a starting point and walk along the direction vector as t changes.
Cross product
An operation on two vectors in 3D that produces a third vector perpendicular to both. Computed as a 3×3 determinant with i, j, k in the top row. The cross product is the standard way to find a normal vector when you have two direction vectors lying in a plane.
Paraboloid
A surface of the form z = x² + y² (circular) or z = x²/a² + y²/b² (elliptic). It looks like an upward-opening bowl. In this exam, it appears as 100z = x² + y².
Extract the direction vector from each line's parametric equations (the coefficients of t).
Compute the cross product of the two direction vectors. This gives the normal vector to the plane.
Use the normal vector and a known point on the plane (e.g. the intersection point) to write the plane equation: n · (r − r₀) = 0.
Worked example: Lines x = 1 + t, y = t, z = 1 − 2t and x = 1, y = 4t, z = 1 − 3t intersect at (1, 0, 1). Direction vectors are ⟨1, 1, −2⟩ and ⟨0, 4, −3⟩. Cross product: ⟨(1)(−3) − (−2)(4), (−2)(0) − (1)(−3), (1)(4) − (1)(0)⟩ = ⟨−3 + 8, 0 + 3, 4 − 0⟩ = ⟨5, 3, 4⟩. Plane equation: 5(x − 1) + 3(y − 0) + 4(z − 1) = 0, which simplifies to 5x + 3y + 4z = 9.
The direction of the line is perpendicular to both plane normals, so take the cross product of the two normal vectors.
Find one specific point on the line by setting one variable (say z = t) and solving the two plane equations simultaneously.
Write parametric equations using the point and the direction vector.
Worked example: Planes x + y + z = 1 and 2x + y − 3z = −3. Normals: ⟨1, 1, 1⟩ and ⟨2, 1, −3⟩. Cross product: ⟨(1)(−3) − (1)(1), (1)(2) − (1)(−3), (1)(1) − (1)(2)⟩ = ⟨−4, 5, −1⟩. To find a point, set z = 0: x + y = 1 and 2x + y = −3, giving x = −4, y = 5. So parametric equations: x = −4 − 4t, y = 5 + 5t, z = −t.
Write the line in parametric form using the two given points.
Substitute the parametric expressions for x, y, z into the surface equation.
Solve for t, then plug back in to get the intersection point(s).
Worked example: Line through (0, 0, 1) and (3, 4, 1): direction ⟨3, 4, 0⟩, so x = 3t, y = 4t, z = 1. Substitute into 100z = x² + y²: 100(1) = 9t² + 16t², so 25t² = 100, t² = 4, t = ±2. Points: (6, 8, 1) and (−6, −8, 1).
Use one equation to eliminate a variable, then parametrise what remains.
Substitute the plane equation into the surface equation and look for a recognisable curve (often an ellipse or circle).
Worked example: Surface 5x² + y² − z² = 4 and plane z = x. Substituting z = x: 5x² + y² − x² = 4, so 4x² + y² = 4, i.e. x²/1 + y²/4 = 1. This is an ellipse. Parametrise as x = cos t, y = 2 sin t, z = cos t. Vector function: r(t) = ⟨cos t, 2 sin t, cos t⟩.
Cross product formula:
a × b = ⟨a₂b₃ − a₃b₂, a₃b₁ − a₁b₃, a₁b₂ − a₂b₁⟩
Plane equation (point-normal form):
a(x − x₀) + b(y − y₀) + c(z − z₀) = 0, where ⟨a, b, c⟩ is the normal and (x₀, y₀, z₀) is a point on the plane.
Parametric line through point P in direction d:
r(t) = P + td, i.e. x = x₀ + at, y = y₀ + bt, z = z₀ + ct
Plane equations describe physical surfaces in engineering and computer graphics. Every polygon in a 3D game or CAD model is defined by a plane equation, and intersection calculations are how ray tracing determines what you see on screen.
Students often forget to use the cross product (not the dot product) to find the normal to a plane from two direction vectors. The dot product gives a scalar, not a vector.
When finding the line of intersection of two planes, students sometimes try to "solve" the two plane equations for x, y, z directly. You cannot solve two equations for three unknowns without introducing a parameter.
A direction vector of ⟨0, 4, −3⟩ does not mean the line passes through the origin. The direction vector only tells you the line's orientation, not its position.
When substituting a parametric line into a surface equation, students sometimes forget there may be two (or zero) solutions. Always solve the resulting equation completely.
⚠️ Cross product problems appear on nearly every MATH 126 final. Know the determinant formula cold.
⚠️ "Find parametric equations" means your answer must include a parameter (usually t). Giving a single point or a Cartesian equation loses marks.
⚠️ The exam instructions say "exact form." Leave answers with square roots and fractions rather than decimal approximations.
True or false: The cross product of two vectors always lies in the plane spanned by those two vectors. (False. It is perpendicular to that plane.)
Fill in the blank: To find the normal vector to a plane containing two known direction vectors, compute their ______. (cross product)
True or false: A line in 3D can be described by a single equation. (False. A line in 3D requires two equations or a set of parametric equations.)
Fill in the blank: Substituting z = x into 5x² + y² − z² = 4 gives ______ = 4. (4x² + y²)
True or false: If a line is parallel to a plane, the line and the plane have zero points of intersection. (Not always true. The line could lie entirely in the plane, giving infinitely many intersection points.)
Q: Two lines x = 1 + t, y = t, z = 1 − 2t and x = 1, y = 4t, z = 1 − 3t intersect at (1, 0, 1). Find the equation of the plane containing both lines.
A: Direction vectors are ⟨1, 1, −2⟩ and ⟨0, 4, −3⟩. Normal = ⟨1, 1, −2⟩ × ⟨0, 4, −3⟩ = ⟨5, 3, 4⟩. Using point (1, 0, 1): 5(x − 1) + 3y + 4(z − 1) = 0, i.e. 5x + 3y + 4z = 9.
Q: Find parametric equations for the line of intersection of the planes x + y + z = 1 and 2x + y − 3z = −3.
A: Cross product of normals ⟨1, 1, 1⟩ × ⟨2, 1, −3⟩ = ⟨−4, 5, −1⟩ gives the direction. Setting z = 0 and solving: x = −4, y = 5. Parametric equations: x = −4 − 4t, y = 5 + 5t, z = −t.
Q: Find all intersection points of the line through (0, 0, 1) and (3, 4, 1) with the paraboloid 100z = x² + y².
A: Line: x = 3t, y = 4t, z = 1. Substituting: 100 = 9t² + 16t² = 25t², so t = ±2. Points: (6, 8, 1) and (−6, −8, 1).
Q: Find a vector function for the curve of intersection of 5x² + y² − z² = 4 and z = x.
A: Substitute z = x: 4x² + y² = 4. Parametrise the ellipse as x = cos t, y = 2 sin t, then z = cos t. r(t) = ⟨cos t, 2 sin t, cos t⟩.
This connects to curvature (next topic), because once you can parametrise a curve of intersection, you can compute its curvature and arc length. It also feeds into double integrals: the region of integration is often bounded by surfaces described in exactly this way. Understanding plane equations is a prerequisite for tangent planes to surfaces (Problem 3 of this exam).
plane equation, normal vector, cross product, parametric equations, direction vector, line of intersection, paraboloid, elliptic paraboloid, surface intersection, curve of intersection, vector function, 3D geometry, MATH 126, Calculus III, UW Math 126, planes and lines in space