Difficulty: Introductory to Intermediate | Prerequisites: 3D coordinate systems (§12.1), basic trigonometry (cosine function), familiarity with unit conversion (mph to m/s).
Vectors are the language of multivariable calculus. Where a scalar is a single number, a vector carries both magnitude and direction, making it the right tool for describing velocities, forces, and displacements in space. The dot product is the first way to multiply two vectors you will learn; it returns a scalar and unlocks the ability to measure angles between vectors and project one vector onto another. These operations appear in nearly every subsequent topic in the course, from equations of planes to line integrals, so getting comfortable here pays off immediately.
A vector has both length (magnitude) and direction. You add vectors component-wise, and the dot product of two vectors gives a scalar that tells you the angle between them: if the dot product is zero, the vectors are perpendicular. Projections let you decompose any vector into components along and perpendicular to a given direction, which is the basis for expressing a vector in terms of an orthonormal set.
Vector
A quantity with both magnitude and direction, represented by an ordered tuple of components, e.g. ⟨a, b, c⟩ in 3D. Think of it as an arrow from one point to another: the components tell you how far to go in each coordinate direction.
Magnitude (length, norm)
The length of a vector v = ⟨a, b, c⟩, written |v| = √(a² + b² + c²). In simple terms, it is the distance from the tail to the tip of the arrow.
Unit vector
A vector with magnitude 1. Any nonzero vector v can be turned into a unit vector by dividing by its magnitude: v̂ = v / |v|. Think of it as keeping the direction but scaling the length to exactly 1.
Standard basis vectors
i = ⟨1, 0, 0⟩, j = ⟨0, 1, 0⟩, k = ⟨0, 0, 1⟩. These are the unit vectors along the coordinate axes. Every vector in 3D can be written as a combination of these three.
Dot product (scalar product, inner product)
For vectors a = ⟨a₁, a₂, a₃⟩ and b = ⟨b₁, b₂, b₃⟩, the dot product is a · b = a₁b₁ + a₂b₂ + a₃b₃. It returns a scalar (a number), not a vector. In simple terms, it measures how much two vectors point in the same direction.
Orthogonal (perpendicular)
Two vectors are orthogonal if and only if their dot product is zero. Think of it as the vectors being at right angles to each other.
Angle between vectors
The angle θ between two nonzero vectors a and b satisfies cos θ = (a · b) / (|a| |b|). This is the geometric meaning of the dot product.
Scalar projection
The scalar projection of b onto a is the signed length of the component of b in the direction of a: comp_a b = (a · b) / |a|.
Vector projection
The vector projection of b onto a is the vector component of b in the direction of a: proj_a b = [(a · b) / (a · a)] a. In simple terms, it is the "shadow" of b onto the line defined by a.
Orthonormal basis
A set of mutually orthogonal unit vectors. In 3D, three such vectors let you decompose any vector v as v = (v · v₁)v₁ + (v · v₂)v₂ + (v · v₃)v₃. Think of it as a customised coordinate system where the axes may point in any three perpendicular directions.
A vector v = ⟨a, b, c⟩ can be thought of as the displacement from the origin to the point (a, b, c), or more generally as an arrow between any two points with that displacement
Vector from point P(x₁, y₁, z₁) to Q(x₂, y₂, z₂): PQ = ⟨x₂ − x₁, y₂ − y₁, z₂ − z₁⟩
Addition is component-wise: ⟨a₁, b₁, c₁⟩ + ⟨a₂, b₂, c₂⟩ = ⟨a₁ + a₂, b₁ + b₂, c₁ + c₂⟩
Scalar multiplication scales each component: t⟨a, b, c⟩ = ⟨ta, tb, tc⟩
When an object moves on a moving platform, the velocity relative to the ground is the vector sum of the object's velocity relative to the platform and the platform's velocity relative to the ground
Example (HW Problem 2): a football player runs due south at 23 mph on an aircraft carrier moving northeast at 10 m/s. Convert units first (1 mph ≈ 0.447 m/s), then set up coordinate directions (e.g. east = positive x, north = positive y), express each velocity as a component vector, and add them
"Due south" = ⟨0, −v₁⟩. "Due northeast" = ⟨v₂ cos 45°, v₂ sin 45°⟩. The resultant speed is the magnitude of their sum.
Algebraic definition: a · b = a₁b₁ + a₂b₂ + a₃b₃
Geometric definition: a · b = |a| |b| cos θ, where θ is the angle between the vectors
Properties:
Commutative: a · b = b · a
Distributive: a · (b + c) = a · b + a · c
Scalar association: (ta) · b = t(a · b)
a · a = |a|²
Orthogonality test: a · b = 0 if and only if a ⊥ b (assuming both are nonzero)
Rearrange the geometric definition: cos θ = (a · b) / (|a| |b|), then θ = arccos[(a · b) / (|a| |b|)]
Example (HW Problem 4): for a triangle with vertices P, Q, R, the angle at vertex P is the angle between vectors PQ and PR
Given three mutually orthogonal unit vectors v₁, v₂, v₃, any vector v in 3D can be written as v = (v · v₁)v₁ + (v · v₂)v₂ + (v · v₃)v₃
Each coefficient (v · vᵢ) is the scalar projection of v onto vᵢ
The standard basis {i, j, k} is the simplest orthonormal set, giving v = ⟨v · i, v · j, v · k⟩ = ⟨v₁, v₂, v₃⟩
This generalises: any orthonormal basis works, which is why it is so powerful in physics and engineering (you can choose axes aligned with the geometry of the problem)
To verify a set is orthonormal: check that each vector has magnitude 1, and that every pair has dot product 0
|\mathbf{v}| = \sqrt{a^2 + b^2 + c^2}\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}|\,|\mathbf{b}|\cos\theta\theta = \arccos\!\left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}\right)\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} \cdot \mathbf{a}}\,\mathbf{a}\mathbf{v} = (\mathbf{v} \cdot \mathbf{v}_1)\mathbf{v}_1 + (\mathbf{v} \cdot \mathbf{v}_2)\mathbf{v}_2 + (\mathbf{v} \cdot \mathbf{v}_3)\mathbf{v}_3The dot product is used in physics to calculate work: W = F · d, where F is the force vector and d is the displacement. Only the component of force in the direction of motion does work. In computer graphics, the dot product determines how much light hits a surface (the angle between the light direction and the surface normal), which is the foundation of shading models.
Students often think the dot product produces a vector. It does not. The dot product always returns a scalar (a single number). The cross product (§12.4) is the one that returns a vector.
A common error in relative velocity problems is forgetting to convert all quantities to the same units before adding vectors. Mixing mph and m/s will produce a meaningless answer.
Students sometimes apply the angle formula and get a negative value inside the arccos. This is fine: cos θ can be negative (meaning the angle is obtuse). The formula still works; the angle just falls between π/2 and π.
When decomposing a vector into an orthonormal basis, students occasionally forget to verify that the basis vectors are both unit length and mutually orthogonal. If either condition fails, the decomposition formula v = (v · v₁)v₁ + (v · v₂)v₂ + (v · v₃)v₃ does not hold.
⚠️ Computing the angle between two vectors using the dot product formula is one of the most common exam questions. Be prepared to do it quickly and accurately, including converting from radians to degrees if asked.
⚠️ Orthogonality checks ("verify these vectors are orthogonal") appear frequently. The test is simple: compute the dot product and confirm it equals zero.
⚠️ The orthonormal decomposition v = Σ(v · vᵢ)vᵢ is a conceptually important result. Expect a question asking you to verify it for a specific vector and basis, or to find the missing third orthonormal vector given two.
⚠️ Relative velocity problems require careful setup of coordinate axes and unit conversion. Read the problem twice before computing.
True or False: The dot product of ⟨1, 2, 3⟩ and ⟨4, −2, 0⟩ is 0.
Fill in the blank: If a · b = 0 and both vectors are nonzero, the vectors are ___.
True or False: The magnitude of ⟨3, 4, 0⟩ is 25.
Fill in the blank: The unit vector in the direction of ⟨0, 0, 5⟩ is ___.
True or False: The scalar projection of b onto a can be negative.
Answers: 1. True (4 − 4 + 0 = 0). 2. Orthogonal (perpendicular). 3. False (the magnitude is 5; 25 is the magnitude squared). 4. ⟨0, 0, 1⟩. 5. True (it is negative when the angle between a and b is greater than 90°).
Q: A football player runs due south at 23 mph on an aircraft carrier moving due northeast at 10 m/s. Find the player's speed relative to the water in m/s.
A: Convert 23 mph to m/s: 23 × 0.44704 ≈ 10.28 m/s. Set east = +x, north = +y. Player's velocity relative to carrier: ⟨0, −10.28⟩. Carrier's velocity: 10⟨cos 45°, sin 45°⟩ ≈ ⟨7.07, 7.07⟩. Resultant: ⟨7.07, −10.28 + 7.07⟩ = ⟨7.07, −3.21⟩. Speed = √(7.07² + 3.21²) ≈ √(50 + 10.3) ≈ √60.3 ≈ 7.77 m/s.
Q: Compute the three angles of the triangle with vertices (−1, −2, −3), (3, 5, −1), and (2, 4, 3), to the nearest tenth of a radian.
A: Find the vectors forming each angle. At vertex A(−1, −2, −3): AB = ⟨4, 7, 2⟩, AC = ⟨3, 6, 6⟩. Use cos θ = (AB · AC)/(|AB||AC|). AB · AC = 12 + 42 + 12 = 66. |AB| = √69, |AC| = √81 = 9. cos θ_A = 66/(9√69). Repeat at each vertex. List angles from least to greatest.
Q: Let v₁ = ⟨−1/√3, 1/√3, 1/√3⟩ and v₂ = ⟨−2/√6, 1/√6, 1/√6⟩. Verify that v₁ and v₂ are orthogonal unit vectors.
A: |v₁|² = 1/3 + 1/3 + 1/3 = 1, so |v₁| = 1. |v₂|² = 4/6 + 1/6 + 1/6 = 1, so |v₂| = 1. v₁ · v₂ = (−1/√3)(−2/√6) + (1/√3)(1/√6) + (1/√3)(1/√6) = 2/√18 + 1/√18 + 1/√18 = 4/√18 − wait, let us recompute carefully: (2/√18) + (1/√18) + (1/√18) = 4/√18. That is not zero, so check signs. v₁ · v₂ = (−1)(−2)/(√3·√6) + (1)(1)/(√3·√6) + (1)(1)/(√3·√6) = 2/√18 + 1/√18 + 1/√18 = 4/√18 ≠ 0. Re-examine the problem: if v₂ = ⟨−2/√6, 1/√6, 1/√6⟩, then v₁ · v₂ = 2/(3√2) + 1/(3√2) + 1/(3√2) = 4/(3√2) ≠ 0. The original homework likely has v₂ = ⟨2/√6, −1/√6, −1/√6⟩ or a sign variant that makes the dot product vanish. Always compute both dot products and magnitudes to verify.
Q: Let v = ⟨1, 2, 3⟩ and suppose {v₁, v₂, v₃} is an orthonormal basis. Verify v = (v · v₁)v₁ + (v · v₂)v₂ + (v · v₃)v₃.
A: Compute each dot product v · vᵢ to get the three coefficients. Multiply each coefficient by its unit vector. Add the three resulting vectors component-wise. The result should equal ⟨1, 2, 3⟩. This works because orthonormal bases allow you to read off each component independently.
The dot product is essential for equations of planes (§12.5), where the normal vector and the dot product define the plane equation. Vector projections reappear when computing the work done by a force along a path (line integrals, Chapter 16). The orthonormal decomposition idea extends to Fourier analysis and quantum mechanics, where vectors are replaced by functions but the same projection logic applies.
vectors 3D, dot product, scalar product, inner product, angle between vectors, orthogonal vectors, perpendicular vectors, unit vector, normalise a vector, magnitude of a vector, vector projection, scalar projection, component of a vector, orthonormal basis, orthogonal decomposition, relative velocity, velocity addition, direction cosines, Math 215, Calculus III, Stewart §12.2, Stewart §12.3