Difficulty: Introductory to Intermediate | Prerequisites: §12.1 (3D coordinates and distances), basic trigonometry (cosine).
Vectors are the language of multivariable calculus. Where §12.1 gave you points, this section gives you arrows: objects with both magnitude and direction. The dot product then provides a way to measure how much two vectors point in the same direction, which leads to the concept of projection (decomposing one vector along another). These tools appear in nearly every subsequent topic in the course, from equations of planes to line integrals.
Vectors have magnitude and direction. You add them component-wise and scale them by multiplying each component. The dot product of two vectors gives a scalar that encodes the angle between them, and projections let you decompose a vector into a component along another vector and a component perpendicular to it.
Vector
A quantity with both magnitude (length) and direction. Represented as an ordered list of components, e.g. v = ⟨v₁, v₂, v₃⟩ in R³.
Think of it as an arrow in space. Two arrows with the same length and direction are the same vector, regardless of where they start.
Magnitude (norm, length)
The length of a vector v = ⟨v₁, v₂, v₃⟩, written |v|, equals √(v₁² + v₂² + v₃²).
In simple terms, it is how long the arrow is.
Unit vector
A vector with magnitude 1. The unit vector in the direction of v is v/|v|. The standard unit vectors are i = ⟨1,0,0⟩, j = ⟨0,1,0⟩, k = ⟨0,0,1⟩.
Dot product (scalar product)
For vectors u = ⟨u₁, u₂, u₃⟩ and v = ⟨v₁, v₂, v₃⟩, the dot product is u · v = u₁v₁ + u₂v₂ + u₃v₃. Equivalently, u · v = |u||v|cos θ, where θ is the angle between the vectors.
Think of it as a measure of how much two vectors point in the same direction. It returns a number (scalar), not a vector.
Orthogonal
Two vectors are orthogonal (perpendicular) when their dot product is zero. This is the 3D generalisation of "at right angles."
Scalar projection (component)
The scalar projection of v onto u is comp_u v = (u · v) / |u|. It tells you the signed length of v's shadow on u.
Vector projection
The vector projection of v onto u is proj_u v = [(u · v) / (u · u)] u. It is the vector component of v that lies along u.
In simple terms, it is the shadow of v cast onto the line through u, returned as a vector.
Orthogonal projection (orth)
Defined as orth_u v = v - proj_u v. It is the component of v perpendicular to u. Together, proj_u v and orth_u v decompose v into two perpendicular pieces.
A vector from point P to point Q is PQ = Q - P, computed component-wise.
Vector addition: ⟨a₁, a₂, a₃⟩ + ⟨b₁, b₂, b₃⟩ = ⟨a₁+b₁, a₂+b₂, a₃+b₃⟩. Geometrically, place the tail of b at the head of a.
Scalar multiplication: c⟨a₁, a₂, a₃⟩ = ⟨ca₁, ca₂, ca₃⟩. This scales the length by |c| and reverses direction if c < 0.
Important identity: for any three points A, B, C, the sum AB + BC + CA = 0 (the zero vector). Walking from A to B to C and back to A is a closed loop.
When objects move in different frames (e.g. a person running on a moving ship), the velocity relative to a fixed frame is the vector sum of the individual velocities.
Convert units before adding. Direction matters: set up a coordinate system and express each velocity as a component vector.
Algebraic definition: u · v = u₁v₁ + u₂v₂ + u₃v₃.
Geometric definition: u · v = |u||v|cos θ.
Key properties:
Commutative: u · v = v · u.
Distributive: u · (v + w) = u · v + u · w.
u · u = |u|².
u · v = 0 if and only if u and v are orthogonal (or one is the zero vector).
Rearrange the geometric definition: cos θ = (u · v) / (|u||v|).
To find the angles of a triangle with vertices P, Q, R: form the vectors along two edges meeting at each vertex, compute the dot product, and solve for the angle.
Scalar projection of v onto u: comp_u v = (u · v) / |u|. This is a number (positive if v has a component in the direction of u, negative if opposite).
Vector projection of v onto u: proj_u v = [(u · v) / (u · u)] u. This is a vector parallel to u.
Orthogonal component: orth_u v = v - proj_u v. This is the leftover part of v, perpendicular to u.
Together, v = proj_u v + orth_u v. This decomposition is unique.
To verify orthogonality: check that (proj_u v) · (orth_u v) = 0.
When a weight hangs from wires, the system is in static equilibrium: the vector sum of all forces is zero.
Set up a coordinate system. Express each tension as a vector using the wire's direction (from the geometry of the setup). The vertical components must balance the weight, and the horizontal components must cancel.
\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3\vec{u} \cdot \vec{v} = |\vec{u}|\,|\vec{v}|\cos\theta\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}|\,|\vec{v}|}\text{comp}_{\vec{u}}\,\vec{v} = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}|}\text{proj}_{\vec{u}}\,\vec{v} = \frac{\vec{u} \cdot \vec{v}}{\vec{u} \cdot \vec{u}}\,\vec{u}\text{orth}_{\vec{u}}\,\vec{v} = \vec{v} - \text{proj}_{\vec{u}}\,\vec{v}The dot product and projections are central to physics: work done by a force is the dot product of the force and displacement vectors. Engineers use projections to decompose forces along specific directions (for example, resolving a load into components along each cable in a suspension system). In computer graphics, the dot product determines lighting by measuring the angle between a surface normal and a light-source direction.
The dot product is a scalar, not a vector. Students sometimes try to assign it a direction.
The projection formula divides by u · u (the dot product of u with itself), not by |u|. Confusing the scalar projection with the vector projection is a common source of error.
Orthogonal does not mean "opposite." Vectors pointing in opposite directions have a dot product of -|u||v|, not zero. Orthogonal means the angle between them is exactly π/2.
When computing angles from the dot product, the result of arccos is always between 0 and π. There is no need to worry about negative angles.
⚠️ Computing the angle between two vectors using the dot product is among the most commonly tested skills. Know the formula cold and practise computing arccos.
⚠️ Projection problems (finding proj_u v and orth_u v, then verifying orthogonality) appear regularly on exams. Be ready to sketch the decomposition.
⚠️ Equilibrium problems (weights suspended by wires) combine vector addition, geometry, and unit conversions. They often appear as multi-step exam questions.
⚠️ The identity AB + BC + CA = 0 is a favourite "explain in words" exam question. Be prepared to state why it is true geometrically.
True or false: ⟨1, 2, 3⟩ · ⟨4, 5, 6⟩ = ⟨4, 10, 18⟩.
Fill in the blank: two non-zero vectors are orthogonal if and only if their dot product equals ______.
True or false: the vector projection of v onto u is always parallel to u.
Fill in the blank: the magnitude of the unit vector in any direction is ______.
True or false: if θ is the angle between u and v, then cos θ can be negative.
Answers: 1. False (the dot product is a scalar: 4 + 10 + 18 = 32). 2. Zero. 3. True. 4. One. 5. True (when the angle is obtuse, cos θ < 0).
Q: Compute the angle between u = ⟨1, 0, -1⟩ and v = ⟨0, 1, 1⟩.
A: u · v = 0 + 0 + (-1) = -1. |u| = √2, |v| = √2. cos θ = -1/2, so θ = 2π/3 (about 2.094 radians, or 120°).
Q: Find proj_u v and orth_u v for u = ⟨3, -2⟩ and v = ⟨1, 3⟩.
A: u · v = 3 - 6 = -3. u · u = 9 + 4 = 13. proj_u v = (-3/13)⟨3, -2⟩ = ⟨-9/13, 6/13⟩. orth_u v = ⟨1, 3⟩ - ⟨-9/13, 6/13⟩ = ⟨22/13, 33/13⟩. Check: ⟨-9/13, 6/13⟩ · ⟨22/13, 33/13⟩ = (-198 + 198)/169 = 0. Confirmed orthogonal.
Q: A pilot runs due north at 18 mph on a carrier moving southwest at 14 m/s. What is the pilot's speed relative to the water, in km/h?
A: Convert 18 mph to m/s: 18 × 1609.34/3600 ≈ 8.047 m/s. North = ⟨0, 8.047⟩. Southwest (π/4 south of west) = 14⟨-cos 45°, -sin 45°⟩ ≈ ⟨-9.899, -9.899⟩. Sum ≈ ⟨-9.899, -1.852⟩. Speed = √(9.899² + 1.852²) ≈ √(97.99 + 3.43) ≈ √101.42 ≈ 10.07 m/s. Convert to km/h: 10.07 × 3.6 ≈ 36.3 km/h.
Q: Show that for any three points A, B, C, the vector sum AB + BC + CA equals the zero vector.
A: AB = B - A, BC = C - B, CA = A - C. Adding: (B - A) + (C - B) + (A - C) = 0. Geometrically, the three displacements form a closed triangle, so they sum to zero.
Projections are the foundation for §12.5 (equations of lines and planes), where you project vectors onto normal directions. The orthogonal decomposition v = proj_u v + orth_u v generalises in §12.4 via the cross product, and it reappears in linear algebra as the Gram-Schmidt process. Static equilibrium problems (tension in wires) use the same vector addition that drives force analysis throughout physics.
vectors R3, dot product, scalar product, inner product, angle between vectors, orthogonal vectors, perpendicular, vector projection, scalar projection, component along, orth, orthogonal projection, unit vector, magnitude, norm, relative velocity, static equilibrium, tension forces, MATH 215, Calculus III, §12.2, §12.3