Difficulty: Intermediate | Prerequisites: MATH 125 or equivalent; basic vector operations (dot product, vector addition, scalar multiplication)
Big picture: The cross product is one of the central tools in multivariable calculus and shows up everywhere vectors live in three dimensions. It gives you a way to find a vector perpendicular to two given vectors, which feeds directly into computing areas, finding normal vectors to planes, and setting up surface integrals later in the course. If you are comfortable with determinants and basic vector arithmetic, this topic is approachable. If not, review 3D coordinate geometry and dot products first.
The cross product of two vectors in 3D gives a new vector perpendicular to both, whose magnitude equals the area of the parallelogram they span. To find the area of a triangle defined by three points, compute the cross product of two edge vectors and take half its magnitude. This is one of the most commonly tested calculations in MATH 126 midterms.
Cross product (vector product)
Given two vectors a and b in R³, the cross product a × b is the unique vector perpendicular to both a and b, with magnitude |a||b|sin(θ), where θ is the angle between them. Direction follows the right-hand rule.
In simple terms, this means: point your right hand's fingers along the first vector, curl them toward the second, and your thumb points in the direction of the result.
Determinant method (for cross product)
The standard way to compute a × b by expanding a 3×3 determinant with i, j, k in the first row and the components of a and b in the second and third rows.
Think of it as: you set up a grid, then do cofactor expansion along the top row to get each component of the result.
Parallelogram area
The magnitude |a × b| equals the area of the parallelogram formed by vectors a and b placed tail to tail.
In simple terms, this means: the cross product's length tells you exactly how much flat area two vectors enclose when you complete the parallelogram.
Triangle area (via cross product)
Half the magnitude of the cross product of two edge vectors: Area = ½|a × b|. A triangle is half a parallelogram, so you halve the result.
Think of it as: the triangle is one diagonal slice of the parallelogram, so its area is exactly half.
Given a = ⟨a₁, a₂, a₃⟩ and b = ⟨b₁, b₂, b₃⟩, the cross product is computed via determinant expansion:
i component: a₂b₃ − a₃b₂
j component: −(a₁b₃ − a₃b₁) = a₃b₁ − a₁b₃
k component: a₁b₂ − a₂b₁
The negative sign on the j component is the most common source of sign errors. Write the full determinant every time rather than trying to do it from memory.
Given points P, Q, R, first form two edge vectors from the same vertex, e.g. PQ = Q − P and PR = R − P.
Compute PQ × PR using the determinant method.
The area of the triangle is ½|PQ × PR|.
It does not matter which vertex you start from. The magnitude of the cross product is the same regardless of which pair of edges you choose, provided they share a common vertex.
Find the area of the triangle with vertices P(3, 0, 3), Q(−2, 1, 5), R(7, 2, 7).
PQ = Q − P = ⟨−5, 1, 2⟩
PR = R − P = ⟨4, 2, 4⟩
PQ × PR:
i: (1)(4) − (2)(2) = 4 − 4 = 0
j: −((−5)(4) − (2)(4)) = −(−20 − 8) = 28
k: (−5)(2) − (1)(4) = −10 − 4 = −14
Result: ⟨0, 28, −14⟩
|PQ × PR| = √(0² + 28² + (−14)²) = √(784 + 196) = √980
Area = ½√980
The exam asked students not to simplify, so ½√980 was the final answer. If simplification were required: √980 = √(4 × 245) = 2√245 = 2√(49 × 5) = 14√5, giving area = 7√5.
\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\text{Area of triangle} = \frac{1}{2}|\vec{PQ} \times \vec{PR}|Students often forget the negative sign on the j component of the cross product. The cofactor expansion alternates signs: +, −, +. Missing this flips one component and ruins the answer.
Students sometimes form edge vectors from different vertices (e.g. PQ and QR instead of PQ and PR). The vectors must share a common tail for the cross product to give the correct parallelogram.
The cross product is not commutative: a × b = −(b × a). The order affects the direction of the result. For area calculations this does not matter (you take the magnitude), but for normal vectors and plane equations the sign matters.
Students occasionally confuse the cross product with the dot product. The dot product gives a scalar; the cross product gives a vector. If your answer to a cross product is a single number, something has gone wrong.
⚠️ Triangle area via cross product is a near-guaranteed exam question. It appeared on this Spring 2025 midterm and shows up regularly on MATH 126 exams at UW.
⚠️ The cross product feeds directly into Problem 2 (finding plane equations), where it is used to compute normal vectors. Master this computation and you cover two problem types at once.
⚠️ Read whether the exam asks you to simplify. On this midterm the instruction was "do not simplify," so leaving √980 unsimplified was correct and saved time.
True or false: the cross product of two parallel vectors is the zero vector. (True)
Fill in the blank: the j component of a × b has a ______ sign in the cofactor expansion. (negative)
True or false: |a × b| = |b × a|. (True, the magnitudes are equal even though the directions are opposite)
Fill in the blank: to find the area of a triangle from a cross product, you multiply the magnitude by ______. (1/2)
True or false: you can use the dot product to find the area of a triangle. (False, you need the cross product)
Q: Find the area of the triangle with vertices A(1, 0, 0), B(0, 2, 0), C(0, 0, 3).
A: Form AB = ⟨−1, 2, 0⟩ and AC = ⟨−1, 0, 3⟩. Cross product: AB × AC = ⟨6, 3, 2⟩. Magnitude = √(36 + 9 + 4) = √49 = 7. Area = 7/2.
Q: If a × b = ⟨3, −1, 4⟩, what is b × a?
A: b × a = −(a × b) = ⟨−3, 1, −4⟩.
Q: Two vectors u and v satisfy |u| = 3, |v| = 5, and the angle between them is π/6. What is the area of the parallelogram they form?
A: |u × v| = |u||v|sin(π/6) = 3 × 5 × 1/2 = 15/2.
Q: Why does the triangle area formula use ½ rather than the full cross-product magnitude?
A: The cross-product magnitude gives the area of the parallelogram spanned by the two vectors. A triangle formed by two edges is exactly half that parallelogram.
The cross product is the engine behind plane equations in 3D: the normal vector to a plane is a cross product of two vectors lying in the plane (see the Lines and Planes study notes). It also appears in torque calculations in physics and in computing the tangential and normal components of acceleration for vector-valued functions. Later in MATH 126, cross products return when you study surface integrals and Stokes' theorem.
cross product, vector product, determinant method, cofactor expansion, triangle area in 3D, parallelogram area, right-hand rule, normal vector, MATH 126, Calculus III, multivariable calculus, vector operations, 3×3 determinant, anti-commutativity