Difficulty: Introductory-Intermediate | Prerequisites: Basic vector notation, component form of vectors
Vectors are the language of multivariable calculus. Before you can describe curves, surfaces, or motion in 3D, you need to know how to combine vectors: measuring angles (dot product), finding perpendicular directions (cross product), computing work done by a force, and projecting one vector onto another. This material underpins almost everything that follows in Calculus III, from parametric curves to surface integrals.
The dot product gives you a scalar and tells you angles and projections. The cross product gives you a vector perpendicular to both inputs and tells you areas. Magnitude gives you length. Most exam questions boil down to choosing the right product for the job.
Dot product (a · b)
The sum of the products of corresponding components: (a₁b₁) + (a₂b₂) + (a₃b₃). The result is always a scalar (a single number), not a vector. In simple terms, it measures how much two vectors point in the same direction.
Cross product (a × b)
A binary operation on two 3D vectors that produces a new vector perpendicular to both inputs. The result is a vector, not a scalar. Think of it as: the cross product finds the direction that sticks straight out of the plane the two vectors lie in, and its length equals the area of the parallelogram they span.
Magnitude (|a|)
The length of a vector, calculated as the square root of the sum of its squared components: √(a₁² + a₂² + a₃²). In simple terms, this is the distance from the origin to the point the vector points to.
Scalar projection (compₐ b)
The signed length of the shadow that vector b casts onto vector a. Computed as (a · b) / |a|. Think of it as: how far along a does b reach?
Orthogonal / perpendicular
Two vectors are orthogonal when the angle between them is 90°. The test: their dot product equals zero. In simple terms, orthogonal means the vectors have nothing in common direction-wise.
Work (W = F · D)
The energy transferred by a force F acting over a displacement D. It equals the dot product of the force vector and the displacement vector. The result is a scalar (energy), not a vector.
Exam questions rarely say "compute the dot product." They use phrases that map to specific operations:
"Find the angle" between two vectors: use the dot product. cos θ = (a · b) / (|a| |b|)
"Find the work": use the dot product. W = F · D
"Orthogonal" or "perpendicular": use the dot product and check whether the result is 0
"Find the area": use the cross product. Triangle area = ½ |a × b|. Parallelogram area = |a × b|
"Find a normal vector": use the cross product. n = v₁ × v₂
"Projection": use the scalar projection formula. compₐ b = (a · b) / |a|
Given a = ⟨a₁, a₂, a₃⟩ and b = ⟨b₁, b₂, b₃⟩:
a · b = (a₁b₁) + (a₂b₂) + (a₃b₃)
The result is a scalar, always
Positive result: vectors point roughly the same way (angle < 90°)
Zero result: vectors are perpendicular (angle = 90°)
Negative result: vectors point roughly opposite (angle > 90°)
Given a = ⟨a₁, a₂, a₃⟩ and b = ⟨b₁, b₂, b₃⟩:
a × b = ⟨(a₂b₃ – a₃b₂), –(a₁b₃ – a₃b₁), (a₁b₂ – a₂b₁)⟩
The result is a vector, always
The resulting vector is perpendicular to both a and b
Order matters: a × b = –(b × a)
W (Work): the energy result, a scalar number
F (Force): the push/pull vector ⟨f₁, f₂, f₃⟩
D (Displacement): the path vector, found by subtracting the start point from the end point
Compute W = F · D using the dot product
\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}\mathbf{a} \times \mathbf{b} = \langle a_2 b_3 - a_3 b_2,\; -(a_1 b_3 - a_3 b_1),\; a_1 b_2 - a_2 b_1 \rangle|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\text{comp}_{\mathbf{a}}\mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}W = \mathbf{F} \cdot \mathbf{D}Students often confuse the output types. The dot product returns a scalar. The cross product returns a vector. If your answer to a cross product is a single number, something has gone wrong.
The cross product is not commutative. a × b and b × a point in opposite directions. Swapping the order flips the sign.
Projection is not the same as the dot product. The dot product gives a · b (raw scalar). The scalar projection divides by |a| to give the signed length of the shadow. Students frequently forget the division.
A dot product of zero does not mean one of the vectors is the zero vector. It means the vectors are perpendicular.
⚠️ The "keyword finder" pattern is heavily tested. Expect word problems that never say "dot product" or "cross product" directly. You need to map the phrasing to the operation.
⚠️ Work problems are a favourite. You will be given a force vector and two points. Remember: D = end point – start point.
⚠️ Area problems using the cross product: triangle = ½ |a × b|, parallelogram = |a × b|. Know which is which.
⚠️ Angle-finding with the dot product: always take cos⁻¹ of the result. Students sometimes stop at the cosine value.
Work (W = F · D) is how physicists and engineers calculate energy transfer: a force applied along a ramp, the thrust on an aircraft, the pull on a cable. The cross product is used in physics to find torque (the rotational equivalent of force) and in computer graphics to determine which way a surface faces (the surface normal).
True or false: the dot product of two vectors always produces a vector.
False. It produces a scalar.
Fill in the blank: if a · b = 0, the vectors are ______.
Orthogonal (perpendicular).
True or false: a × b = b × a.
False. a × b = –(b × a).
Fill in the blank: the area of a triangle formed by vectors a and b is ______.
½ |a × b|.
True or false: the scalar projection compₐ b equals a · b.
False. It equals (a · b) / |a|.
Q: Given a = ⟨2, –1, 3⟩ and b = ⟨4, 0, –2⟩, compute a · b.
A: (2)(4) + (–1)(0) + (3)(–2) = 8 + 0 – 6 = 2.
Q: Are the vectors a = ⟨1, 2, –1⟩ and b = ⟨2, –1, 0⟩ orthogonal?
A: a · b = (1)(2) + (2)(–1) + (–1)(0) = 2 – 2 + 0 = 0. Yes, they are orthogonal.
Q: Find a vector perpendicular to both a = ⟨1, 0, 2⟩ and b = ⟨3, 1, 0⟩.
A: Compute a × b = ⟨(0)(0) – (2)(1), –((1)(0) – (2)(3)), (1)(1) – (0)(3)⟩ = ⟨–2, 6, 1⟩.
Q: A force F = ⟨3, 1, –2⟩ moves an object from point (1, 0, 2) to point (4, 3, 1). Find the work done.
A: D = ⟨4–1, 3–0, 1–2⟩ = ⟨3, 3, –1⟩. W = F · D = (3)(3) + (1)(3) + (–2)(–1) = 9 + 3 + 2 = 14.
Q: Find the area of the parallelogram spanned by a = ⟨1, 2, 0⟩ and b = ⟨0, 1, 3⟩.
A: a × b = ⟨(2)(3) – (0)(1), –((1)(3) – (0)(0)), (1)(1) – (2)(0)⟩ = ⟨6, –3, 1⟩. Area = |a × b| = √(36 + 9 + 1) = √46.
The dot product reappears when computing directional derivatives and the work integral in line integrals. The cross product is essential for finding normal vectors to planes (next topic: Lines and Planes) and later for computing surface integrals and curl in vector calculus.
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