This material covers how to describe lines and planes in three-dimensional space using vectors, and how to determine geometric relationships between them (parallel, perpendicular, skew, intersecting). You need the dot product, cross product, and projection formulas to solve most problems. If you can set up parametric equations for a line and write a normal-vector equation for a plane, you can handle the bulk of Midterm 1.
Difficulty: Intermediate | Prerequisites: MATH 124/125 (single-variable calculus), basic vector algebra (addition, scalar multiplication, magnitude).
MATH 126 (Calculus III) at the University of Washington opens with the geometry of three-dimensional space. This is where single-variable calculus hands off to multivariable work: instead of curves on a flat plane, you are now placing lines, planes, and curves in 3D and describing how they relate to one another. The tools you build here (parametric equations, normal vectors, dot and cross products) reappear constantly in later topics such as partial derivatives, multiple integrals, and vector fields. If you skipped MATH 125 or are rusty on vectors, review vector addition, magnitudes, and basic trigonometry before sitting this exam.
Parametric equations of a line
A way of expressing a line in 3D by writing each coordinate as a function of a single parameter t: x = x₀ + at, y = y₀ + bt, z = z₀ + ct. Think of it as: you pick a starting point and a direction, then t tells you how far along the line you have walked.
Direction vector
The vector ⟨a, b, c⟩ that points along a line. In simple terms, this is the vector that tells you which way the line goes.
Normal vector (to a plane)
A vector perpendicular to every vector lying in the plane. If the plane equation is ax + by + cz = d, the normal vector is ⟨a, b, c⟩. Think of it as: the arrow sticking straight out of a flat surface.
Dot product (scalar product)
a · b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ. In simple terms, this measures how much two vectors point in the same direction. Result is a number, not a vector.
Cross product (vector product)
a × b = ⟨a₂b₃ - a₃b₂, a₃b₁ - a₁b₃, a₁b₂ - a₂b₁⟩. The result is a new vector perpendicular to both a and b, with magnitude |a||b| sin θ. Think of it as: the vector that sticks out of the plane formed by a and b.
Scalar projection
comp_b a = (a · b) / |b|. The signed length of a's shadow on b.
Vector projection
proj_b a = ((a · b) / |b|²) b. The actual vector shadow of a onto b. Think of it as: if you shone a torch straight down onto the line of b, the shadow of a is the projection.
Skew lines
Two lines in 3D that are not parallel and do not intersect. They pass each other in different planes. This cannot happen in 2D, only in 3D.
Elliptic cylinder
A surface defined by an equation like 4y² + z² = 16 (no x term). It extends infinitely along the missing axis (here x). In simple terms, take an ellipse in the yz-plane and stretch it along the x-axis like a tube.
A line in 3D is specified by a point P₀ = (x₀, y₀, z₀) and a direction vector d = ⟨a, b, c⟩.
Parametric form: x = x₀ + at, y = y₀ + bt, z = z₀ + ct
Vector form: r(t) = r₀ + td = ⟨x₀ + at, y₀ + bt, z₀ + ct⟩
Symmetric form: (x - x₀)/a = (y - y₀)/b = (z - z₀)/c (only when a, b, c are all non-zero)
To read off the direction vector from parametric equations, collect the coefficients of t. For example, L: x = t, y = 1 + 2t, z = 2 + 3t has direction vector d = ⟨1, 2, 3⟩ and passes through (0, 1, 2) when t = 0.
A plane is specified by a point P₀ = (x₀, y₀, z₀) and a normal vector n = ⟨a, b, c⟩.
Standard form: a(x - x₀) + b(y - y₀) + c(z - z₀) = 0, which simplifies to ax + by + cz = d.
The normal vector is read directly from the coefficients: for 3x + y + 2z = 6, the normal is n = ⟨3, 1, 2⟩.
To find a plane that contains a given line and satisfies some condition (e.g. perpendicular to the xy-plane), you need to find the normal vector of the desired plane. Two approaches:
Cross-product method: if you know two vectors that lie in the plane, their cross product gives the normal.
Condition method: translate the geometric condition into a constraint on the normal. "Perpendicular to the xy-plane" means the plane's normal has no z-component (the normal lies in the xy-plane), so n = ⟨a, b, 0⟩.
For a plane containing a line, the normal must be perpendicular to the line's direction vector: n · d = 0.
Line perpendicular to a plane: the line's direction vector is parallel to (a scalar multiple of) the plane's normal vector. For plane P: 3x + y + 2z = 6, any line perpendicular to P has direction vector d = ⟨3, 1, 2⟩ (or any scalar multiple).
Line parallel to a plane: the line's direction vector is perpendicular to the plane's normal, i.e. d · n = 0.
Plane perpendicular to another plane: their normal vectors are perpendicular, i.e. n₁ · n₂ = 0. A plane perpendicular to the xy-plane has a normal with z-component equal to zero.
Skew lines: two lines in 3D that never meet and are not parallel. To construct two skew lines that both intersect the y-axis, give each line a point on the y-axis and a direction that moves off the y-axis in different planes (e.g. one goes in the xz-direction, the other in a different combination). The key check is that the system of equations from setting the two parametric forms equal has no solution and the direction vectors are not proportional.
Intersection of surfaces: to parametrise the curve where two surfaces meet (e.g. an elliptic cylinder and a plane), parametrise the simpler surface first (the cylinder, using trigonometric substitution), then solve for the remaining variable from the other equation.
Dot product uses:
Finding the angle between two vectors: cos θ = (a · b) / (|a| |b|)
Checking perpendicularity: a ⊥ b if and only if a · b = 0
Computing projections (see below)
Determining whether a particle speeds up or slows down: if the dot product of velocity and acceleration is positive, speed increases; if negative, speed decreases
Cross product uses:
Finding a vector perpendicular to two given vectors (the normal to a plane containing both)
Computing the area of a parallelogram spanned by a and b: Area = |a × b|
Computing the area of a triangle: Area = ½|a × b|
Determining orientation (right-hand rule)
Vector projection formula:
proj_b a = ((a · b) / (b · b)) b
The condition proj_v k = v/2 (from the exam) means the projection of k onto v equals half of v. Working through the formula: ((v · k) / (v · v)) v = v/2, so (v · k) / (v · v) = 1/2. If v = ⟨a, b, c⟩, then c / (a² + b² + c²) = 1/2, giving a² + b² + c² = 2c, or a² + b² = 2c - c² = c(2 - c). Pick any a, b, c satisfying this with a or b non-zero (so v is not parallel to k). For example, v = ⟨1, 0, 1⟩.
Triangle area via cross product:
For triangle OAB where O is the origin, the area = ½ |OA × OB|. To find points A and B on x + y + z = 1 such that the area equals 1, choose A and B on the plane and solve |OA × OB| = 2. A workable approach: pick A = (1, 0, 0) and B = (0, b₂, b₃) with b₂ + b₃ = 1, then compute the cross product and set its magnitude to 2.
Line: r(t) = ⟨x₀ + at, y₀ + bt, z₀ + ct⟩
Plane: a(x - x₀) + b(y - y₀) + c(z - z₀) = 0, or ax + by + cz = d
Dot product: a · b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ
Cross product: a × b = ⟨a₂b₃ - a₃b₂, a₃b₁ - a₁b₃, a₁b₂ - a₂b₁⟩
Projection: proj_b a = ((a · b) / |b|²) b
Area of parallelogram: |a × b|
Area of triangle: ½|a × b|
Angle between vectors: θ = arccos((a · b) / (|a||b|))
Plane equations describe surfaces in engineering: wing surfaces, walls, cutting planes in manufacturing. The cross product is used in computer graphics to find surface normals for lighting calculations. Projections underpin everything from GPS positioning (projecting satellite signals) to structural engineering (decomposing forces into components along and perpendicular to a beam).
Students often confuse the normal vector of a plane with a direction vector lying in the plane. The normal is perpendicular to the plane, not parallel to it.
"Perpendicular to the xy-plane" does not mean the plane's normal is ⟨0, 0, 1⟩. It means the plane contains the z-direction, so the normal has no z-component: n = ⟨a, b, 0⟩.
The cross product is not commutative: a × b = -(b × a). The sign flips, which matters when finding normals.
The dot product gives a scalar, the cross product gives a vector. Mixing these up in a projection formula is a common error.
Students sometimes forget that skew lines require both conditions: non-parallel and non-intersecting. Two non-parallel lines in 3D might still intersect at a point.
⚠️ "Find the equation of a plane containing a line and perpendicular to [something]" is a staple exam question. Always identify two constraints on the normal vector.
⚠️ Projection problems often require you to work backwards from a given projection to find the original vector. Set up the formula and solve for unknowns.
⚠️ "Give an example of..." questions (skew lines, a vector with a certain projection) test whether you understand the definitions well enough to construct objects, not just verify them.
⚠️ Parametrising the intersection of two surfaces appears regularly. Practise the trig-substitution approach for cylinders.
True or false: the normal vector to the plane 2x - 3y + z = 5 is ⟨2, -3, 1⟩. (True)
Fill in the blank: two lines are skew if they are ______ and ______. (not parallel; do not intersect)
True or false: a × b = b × a. (False, a × b = -b × a)
Fill in the blank: the area of a triangle with sides a and b from the same vertex is ______. (½|a × b|)
True or false: if a plane is perpendicular to the xy-plane, its normal vector must be ⟨0, 0, 1⟩. (False, its normal has z-component 0)
Q: Given L: x = t, y = 1 + 2t, z = 2 + 3t and plane P: 3x + y + 2z = 6, find the equation of the plane containing L and perpendicular to the xy-plane.
A: The direction vector of L is d = ⟨1, 2, 3⟩. A plane perpendicular to the xy-plane has normal n = ⟨a, b, 0⟩. Since L lies in the desired plane, n · d = 0, so a + 2b = 0, giving a = -2b. Choose b = 1: n = ⟨-2, 1, 0⟩. Using the point (0, 1, 2) on L: -2(x - 0) + 1(y - 1) = 0, so -2x + y = 1, or equivalently 2x - y = -1.
Q: Find parametric equations of a line that intersects L and is perpendicular to P: 3x + y + 2z = 6.
A: A line perpendicular to P has direction d = ⟨3, 1, 2⟩ (the normal of P). It must pass through a point on L. Pick t = 0 to get the point (0, 1, 2). The line is x = 3s, y = 1 + s, z = 2 + 2s.
Q: Give an example of a vector v such that proj_v k = v/2 and v is not parallel to k.
A: Need (v · k) / (v · v) = 1/2. With v = ⟨a, b, c⟩: c / (a² + b² + c²) = 1/2, so a² + b² = c(2 - c). Try c = 1: a² + b² = 1. Take v = ⟨1, 0, 1⟩. Check: proj_v k = (1/2)⟨1, 0, 1⟩ = v/2.
Q: Find a vector function r(t) for the intersection of the elliptic cylinder 4y² + z² = 16 and the plane x + 3y + z = 1.
A: Parametrise the cylinder: y = 2 cos t, z = 4 sin t (since 4(2 cos t)² + (4 sin t)² = 16 cos²t · 4 + 16 sin²t = 16). Then from the plane: x = 1 - 3(2 cos t) - 4 sin t = 1 - 6 cos t - 4 sin t. So r(t) = ⟨1 - 6 cos t - 4 sin t, 2 cos t, 4 sin t⟩.
The parametric-line and plane equations here are the foundation for describing vector functions and space curves (the next topic in this exam). Cross products reappear when computing curvature, and the dot product drives the speed-up/slow-down analysis in particle motion problems. Later in MATH 126, plane equations become the starting point for setting up double and triple integrals over bounded regions.
Parametric equations, symmetric equations, direction vector, normal vector, dot product, scalar product, cross product, vector product, projection, scalar projection, vector projection, plane equation, skew lines, parallel lines, perpendicular planes, elliptic cylinder, intersection of surfaces, curve of intersection, area of triangle cross product, MATH 126, Calculus III, UW, University of Washington, multivariable calculus, 3D geometry, vectors in space