Difficulty: Intermediate | Prerequisites: Vectors and dot product (§12.2, §12.3), basic determinant evaluation (2×2 and 3×3), trigonometry.
The cross product is the second way to multiply two vectors and the one that returns a new vector rather than a scalar. Its direction is perpendicular to both input vectors (giving you a normal to a plane), and its magnitude equals the area of the parallelogram they span. This section also introduces the scalar triple product, which gives the volume of a parallelepiped, and applies vectors to physical problems involving torque and static equilibrium. If the dot product answers "how aligned are these vectors?", the cross product answers "how much do they spread out, and in which direction?".
The cross product a × b produces a vector perpendicular to both a and b, computed via a 3×3 determinant. Its magnitude gives the area of the parallelogram formed by a and b (halve it for a triangle). The scalar triple product a · (b × c) gives the volume of the parallelepiped spanned by three vectors. Torque problems use the cross product to relate force, position, and rotational effect.
Cross product (vector product)
For vectors a = ⟨a₁, a₂, a₃⟩ and b = ⟨b₁, b₂, b₃⟩, the cross product a × b is the vector ⟨a₂b₃ − a₃b₂, a₃b₁ − a₁b₃, a₁b₂ − a₂b₁⟩. Think of it as a new vector that sticks straight out of the plane containing a and b.
Normal vector
A vector perpendicular to a surface or plane. The cross product of two vectors lying in a plane gives a normal to that plane.
Area of a parallelogram
The area of the parallelogram spanned by vectors a and b equals |a × b|. In simple terms, the cross product's magnitude is the area.
Area of a triangle
Half the area of the parallelogram: Area = ½|a × b|, where a and b are two edge vectors of the triangle.
Scalar triple product
The quantity a · (b × c), which equals the determinant of the 3×3 matrix whose rows (or columns) are a, b, c. In simple terms, it gives a signed volume.
Volume of a parallelepiped
The volume of the parallelepiped with adjacent edges a, b, c is |a · (b × c)|. The absolute value is needed because the triple product can be negative depending on orientation.
Torque
The rotational effect of a force about a point: τ = r × F, where r is the position vector from the pivot to the point of application and F is the force. Its magnitude is |r||F| sin θ. Think of it as the "twisting power" of a force.
Static equilibrium
A system is in static equilibrium when the sum of all forces and the sum of all torques equal zero. For a lever (seesaw), the balance condition is that the total torque about the fulcrum is the zero vector.
a × b is evaluated as the determinant of a 3×3 matrix with i, j, k in the first row, a's components in the second, and b's in the third
Expand along the first row using cofactors:
i(a₂b₃ − a₃b₂) − j(a₁b₃ − a₃b₁) + k(a₁b₂ − a₂b₁)
Note the minus sign on the j-component; this is the most common source of arithmetic errors
Anti-commutative: a × b = −(b × a). Order matters.
Not associative: a × (b × c) ≠ (a × b) × c in general
Distributive over addition: a × (b + c) = a × b + a × c
Scalar association: (ta) × b = t(a × b)
a × a = 0 for any vector a (a vector crossed with itself is the zero vector)
Direction: a × b is perpendicular to both a and b, with direction given by the right-hand rule (curl fingers from a toward b, thumb points in the direction of a × b)
Magnitude: |a × b| = |a||b| sin θ, where θ is the angle between a and b
|a × b| equals the area of the parallelogram with sides a and b
To find a vector orthogonal to both u and v, compute u × v
To find a unit vector orthogonal to both, normalise the result: (u × v)/|u × v|
There are always two unit vectors orthogonal to a pair of non-parallel vectors: one is the negative of the other
To find vectors orthogonal to a single vector ⟨0, −π, π⟩, pick any vector not parallel to it and take the cross product
Given vertices P, Q, R, form two edge vectors: PQ and PR
Area = ½|PQ × PR|
Example: for P(3, 5, 0), Q(4, −2, −3), R(−2, 0, 0), compute PQ = ⟨1, −7, −3⟩ and PR = ⟨−5, −5, 0⟩, then take half the magnitude of their cross product
The scalar triple product a · (b × c) can be computed as the determinant of the 3×3 matrix with rows a, b, c
Volume of the parallelepiped with edges a, b, c = |a · (b × c)|
If the triple product is zero, the three vectors are coplanar (they all lie in one plane)
Torque about a point: τ = r × F, where r is the position vector from the pivot
For a lever on the x-axis with gravity F = −mgk:
The position vector from the fulcrum at x₀ to a mass at x is r = ⟨x − x₀, 0, 0⟩
τ = r × F = ⟨x − x₀, 0, 0⟩ × ⟨0, 0, −mg⟩ = ⟨0, (x − x₀)(mg), 0⟩
Balance condition: sum of all torques = 0, which gives Σ mᵢ(xᵢ − x₀) = 0
Solving: x₀ = (Σ mᵢxᵢ) / (Σ mᵢ), the centre of mass formula
Example: Snoopy (12 kg) and Woodstock (0.8 kg) on a seesaw. Woodstock sits 8.8 m from the fulcrum. Balance requires 12 × d = 0.8 × 8.8, so d = 0.587 m
When a weight is suspended by two wires, the system is in equilibrium: the vector sum of the two tension forces and the weight equals zero
Set up coordinates, find the angles of each wire using the given geometry (distances and wire lengths, often requiring the law of cosines or Pythagorean relationships)
Decompose each tension into horizontal and vertical components
Solve the resulting system of equations: horizontal components cancel, vertical components sum to the weight
\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}|\,|\mathbf{b}|\sin\theta\text{Area of parallelogram} = |\mathbf{a} \times \mathbf{b}|\text{Area of triangle} = \tfrac{1}{2}|\mathbf{a} \times \mathbf{b}|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\text{Volume of parallelepiped} = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}x_0 = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i}Torque is central to mechanical engineering: every bolt you tighten, every engine that turns, every bridge that stays up involves torque calculations using the cross product. The centre-of-mass formula derived from lever balance is used in structural engineering to find where to support a beam, and in robotics to keep a walking robot from tipping over.
Students often think the cross product is commutative. It is not. a × b = −(b × a). Swapping the order flips the direction.
The most common arithmetic mistake is getting the sign wrong on the j-component of the determinant expansion. Remember the cofactor pattern is +, −, + across the first row.
Students sometimes forget to take the absolute value when computing volume from the scalar triple product. The triple product can be negative (it indicates orientation), but volume is always positive.
Some students confuse the cross product formula with the dot product formula. The dot product returns a scalar; the cross product returns a vector. They are different operations with different formulas.
⚠️ Computing a cross product via the 3×3 determinant is almost certainly on the exam. Practise until you can do it without errors, especially the j-component sign.
⚠️ "Find the area of the triangle with vertices P, Q, R" is a standard exam question. Form two edge vectors, cross them, take half the magnitude.
⚠️ Scalar triple product for volume is quick if you know how to evaluate a 3×3 determinant. Practise both cofactor expansion and the diagonal (Sarrus) rule.
⚠️ Torque and lever problems test whether you can set up a cross product in a physical context. Pay attention to sign conventions and coordinate directions.
True or False: a × b = b × a.
Fill in the blank: The magnitude of a × b equals the area of the ___ formed by a and b.
True or False: If a · (b × c) = 0, the three vectors lie in a single plane.
Fill in the blank: Torque is computed as τ = r × ___.
True or False: The cross product of any vector with itself is the zero vector.
Answers: 1. False (a × b = −b × a). 2. Parallelogram. 3. True (they are coplanar). 4. F (the force vector). 5. True.
Q: Find the area of the triangle with vertices P(3, 5, 0), Q(4, −2, −3), and R(−2, 0, 0).
A: PQ = ⟨1, −7, −3⟩, PR = ⟨−5, −5, 0⟩. PQ × PR = ⟨(−7)(0) − (−3)(−5), (−3)(−5) − (1)(0), (1)(−5) − (−7)(−5)⟩ = ⟨0 − 15, 15 − 0, −5 − 35⟩ = ⟨−15, 15, −40⟩. |PQ × PR| = √(225 + 225 + 1600) = √2050. Area = ½√2050 = ½ × 5√82 = (5√82)/2.
Q: Find the volume of the parallelepiped with adjacent edges PQ, PR, PS, where P(1, 2, 0), Q(3, 2, 2), R(4, 1, −1), S(3, 6, 1).
A: PQ = ⟨2, 0, 2⟩, PR = ⟨3, −1, −1⟩, PS = ⟨2, 4, 1⟩. Volume = |PQ · (PR × PS)|. First compute PR × PS, then dot with PQ, and take the absolute value.
Q: Snoopy (12 kg) and Woodstock (0.8 kg) are on a seesaw. Woodstock sits 8.8 m from the fulcrum. How far from the fulcrum must Snoopy sit for balance?
A: Balance condition: 12 × d = 0.8 × 8.8. So d = 7.04/12 ≈ 0.587 m from the fulcrum, on the opposite side.
Q: Woodstock (0.8 kg, at 8.8 m) recruits friends W₁ and W₂ (each 3.2 kg) who sit at 4.8 m and 6.8 m from the fulcrum. Where does Snoopy (12 kg) sit for balance?
A: Total torque on the right: 0.8 × 8.8 + 3.2 × 4.8 + 3.2 × 6.8 = 7.04 + 15.36 + 21.76 = 44.16. Snoopy: 12 × d = 44.16, so d = 3.68 m.
Q: Find two unit vectors orthogonal to both ⟨−2, 1, 3⟩ and ⟨6, 1, −2⟩.
A: Compute the cross product: ⟨−2, 1, 3⟩ × ⟨6, 1, −2⟩ = ⟨(1)(−2) − (3)(1), (3)(6) − (−2)(−2), (−2)(1) − (1)(6)⟩ = ⟨−5, 14, −8⟩. The magnitude is √(25 + 196 + 64) = √285. The two unit vectors are ±⟨−5, 14, −8⟩/√285. No others exist (any vector orthogonal to both must be a scalar multiple of their cross product, so only two unit vectors are possible).
The cross product is used immediately in §12.5 to find normal vectors for planes (a normal to a plane through three points is the cross product of two edge vectors). It reappears in surface integrals (Chapter 16), where the cross product of partial derivatives gives the surface area element. The torque and centre-of-mass ideas extend to moments of inertia and centre of mass for continuous distributions in later chapters.
cross product, vector product, determinant formula, right-hand rule, normal vector, area of parallelogram, area of triangle using cross product, scalar triple product, volume of parallelepiped, coplanar vectors, torque, moment of force, lever balance, Archimedes law of the lever, centre of mass, seesaw problem, tension forces, static equilibrium, wire suspension, Math 215, Calculus III, Stewart §12.4