Difficulty: Introductory to Intermediate | Prerequisites: 12.1 (3D coordinates and distance formula), basic trigonometry.
Vectors give you a way to describe quantities that have both magnitude and direction. This set of notes covers how to add, subtract and scale vectors, how to compute their length, and how the dot product lets you find angles between vectors, test for perpendicularity, and compute projections and physical work. Mastering these tools is essential because every later topic in the course (cross products, curves, surfaces, gradients) builds on them.
If you are behind, make sure you are comfortable with 3D coordinates and the distance formula from 12.1 before reading these notes. The dot product section also relies on basic trigonometry (cosine, in particular).
Vector
A quantity with both magnitude (length) and direction, written as ⟨v₁, v₂, v₃⟩ or v₁i + v₂j + v₃k. Think of it as an arrow: same length and same direction means the same vector, regardless of where you draw it.
Magnitude (length) of a vector, |v|
The length of the vector, computed as √(v₁² + v₂² + v₃²). In simple terms, it is the distance formula applied from the origin to the tip of the vector.
Unit vector
A vector with magnitude 1. You obtain it by dividing any nonzero vector by its own magnitude: (1/|v|) v. Think of it as "pure direction with no scaling."
Standard basis vectors (i, j, k)
i = ⟨1, 0, 0⟩, j = ⟨0, 1, 0⟩, k = ⟨0, 0, 1⟩. These are unit vectors along the x, y and z axes. Any vector ⟨v₁, v₂, v₃⟩ can be written as v₁i + v₂j + v₃k.
Dot product (a · b)
The sum a₁b₁ + a₂b₂ + a₃b₃. It is a scalar (a number), not a vector. In simple terms, it measures how much two vectors point in the same direction.
Orthogonal (perpendicular)
Two nonzero vectors are orthogonal when the angle between them is π/2 (90°). The test: a · b = 0.
Scalar projection (component projection)
The signed length of the shadow one vector casts onto another: comp_a(b) = (a · b) / |a|. Positive if b leans toward a, negative if away.
Vector projection
The actual vector that is the shadow of b onto a: proj_a(b) = ((a · b) / |a|²) a.
Work (physics application)
Work done by a force F over a displacement D equals F · D. It is the dot product's most common real-world appearance.
A vector v = ⟨v₁, v₂⟩ drawn from the origin ends at the point (v₁, v₂). The same vector can be drawn anywhere; only its length and direction matter.
To get the vector from point A(a₁, a₂, a₃) to point B(b₁, b₂, b₃): AB = ⟨b₁ − a₁, b₂ − a₂, b₃ − a₃⟩.
Addition: add component by component. Geometrically, place b at the head of a; the sum a + b goes from the tail of a to the head of b.
Scalar multiplication: c v = ⟨cv₁, cv₂, cv₃⟩. Positive c stretches, negative c reverses direction.
Magnitude: |v| = √(v₁² + v₂² + v₃²).
Unit vector in the direction of v: (1/|v|) v.
If a 2D vector v = ⟨x, y⟩ has magnitude r and makes angle θ with the positive x-axis:
x = r cos(θ), y = r sin(θ)
r² = x² + y², tan(θ) = y/x
This is how you convert between "length and angle" descriptions and component form.
A vector parallel to a line with slope m is simply ⟨1, m⟩ (or any scalar multiple of it). For example, ⟨1, 3⟩ is parallel to y = 3x.
Definition: a · b = a₁b₁ + a₂b₂ + a₃b₃.
Properties:
a · a = |a|²
a · b = b · a (commutative)
a · (b + c) = a · b + a · c (distributive)
c(a · b) = (ca) · b = a · (cb)
The single most important fact: for nonzero vectors a and b with angle θ between them (drawn tail to tail, 0 ≤ θ ≤ π):
a · b = |a| |b| cos(θ)
This lets you solve for the angle: cos(θ) = (a · b) / (|a| |b|).
Key consequences:
If a and b are perpendicular, θ = π/2, so a · b = 0.
If a and b are parallel and point the same way, θ = 0, so a · b = |a| |b|.
If a and b are parallel and point opposite ways, θ = π, so a · b = −|a| |b|.
Scalar (component) projection of b onto a:
comp_a(b) = (a · b) / |a|
This is a number: the signed length of the "shadow" of b on the line through a.
Vector projection of b onto a:
proj_a(b) = ((a · b) / |a|²) a
This is a vector: the actual shadow vector along a.
When a is already a unit vector T (|T| = 1), the scalar projection simplifies to T · b. This is why unit vectors are so useful in applications.
If a constant force F acts on an object that moves along displacement D:
Work = F · D
This comes directly from projecting F onto the direction of motion and multiplying by the distance.
|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\hat{\mathbf{v}} = \frac{1}{|\mathbf{v}|} \mathbf{v} \quad \text{(unit vector)}\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta\text{comp}_{\mathbf{a}}(\mathbf{b}) = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}\text{proj}_{\mathbf{a}}(\mathbf{b}) = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2} \mathbf{a}W = \mathbf{F} \cdot \mathbf{D} \quad \text{(work)}The dot product is how physics computes work done by a force (pulling a wagon at an angle, for instance). Projections are the basis of decomposing forces along and perpendicular to a surface, which is central to engineering statics and structural analysis.
Students often think the dot product produces a vector. It does not. The dot product is always a scalar (a single number).
A common error is forgetting to normalise when computing projections. The scalar projection divides by |a|, the vector projection divides by |a|². Mixing these up gives the wrong answer.
Some students assume that if a · b is positive, the vectors point in the same direction. That is too strong a claim: a positive dot product only means the angle between them is less than 90°.
When computing the angle between two vectors, students sometimes forget the absolute-value bars are not needed because the formula already handles direction through the sign of cos(θ).
⚠️ "Find the angle between two vectors" is one of the most common exam problems. The method is always the same: compute a · b, compute |a| and |b|, then use cos(θ) = (a · b) / (|a| |b|).
⚠️ Testing for orthogonality (a · b = 0) appears in many contexts: checking whether two directions are perpendicular, finding vectors normal to a plane, verifying cross product results.
⚠️ Projection problems often ask for both the scalar projection and the vector projection. Know the difference and be able to compute both quickly.
True or false: ⟨2, 4, −6⟩ and ⟨−1, −2, 3⟩ are the same vector. (False, but they are parallel, one is −2 times the other.)
Fill in the blank: The unit vector in the direction of ⟨3, 4⟩ is ___. (⟨3/5, 4/5⟩)
True or false: If a · b = 0, the vectors are parallel. (False, they are perpendicular.)
Fill in the blank: The dot product ⟨1, 2, 3⟩ · ⟨4, −5, 6⟩ = ___. (4 − 10 + 18 = 12)
True or false: The vector projection of b onto a is always shorter than b. (False, it equals b when b is already in the direction of a.)
Q: Find the angle between a = ⟨1, 0, 1⟩ and b = ⟨0, 1, 1⟩.
A: a · b = 0 + 0 + 1 = 1. |a| = √2, |b| = √2. cos(θ) = 1/2, so θ = π/3 (60°).
Q: Are the vectors ⟨2, −3, 1⟩ and ⟨1, 2, 4⟩ orthogonal?
A: Dot product = 2 − 6 + 4 = 0. Yes, they are orthogonal.
Q: Find the scalar and vector projections of b = ⟨3, 4⟩ onto a = ⟨1, 0⟩.
A: comp_a(b) = (3·1 + 4·0)/1 = 3. proj_a(b) = 3 · ⟨1, 0⟩ = ⟨3, 0⟩.
Q: A force F = ⟨2, 5, −1⟩ moves an object along displacement D = ⟨3, 1, 2⟩. Find the work done.
A: Work = F · D = 6 + 5 − 2 = 9.
The dot product's orthogonality test is used immediately in 12.4 to verify cross product calculations. Projections return in 12.5 when computing distances from points to planes. The concept of work reappears in multivariable calculus as line integrals of vector fields (Chapter 16).
vector, magnitude, length, norm, unit vector, normalise, standard basis, i j k, dot product, scalar product, inner product, angle between vectors, orthogonal, perpendicular, parallel, scalar projection, component projection, vector projection, work, force times distance, Math 126, Calculus III, UW Calculus