Difficulty: Intermediate | Prerequisites: 12.1 – 12.3 (3D coordinates, vectors, dot products).
The cross product gives you a vector perpendicular to two given vectors, and its magnitude equals the area of the parallelogram they span. Lines in 3D are described by parametric equations (a point plus a direction), while planes are described by a point and a normal vector. These tools combine constantly: you use cross products to find normals for planes, and you use dot products and algebra to test whether lines and planes intersect.
This material feeds directly into vector curves (Chapter 13) and surfaces (Chapter 14). If you are coming in cold, review dot products (12.3) first, since the cross product's verification step relies on them.
Cross product (a × b)
A vector operation on two 3D vectors that produces a third vector perpendicular to both inputs. Computed using a 3×3 determinant with i, j, k in the top row. Unlike the dot product, the result is a vector, not a scalar.
Right-hand rule
A physical way to determine the direction of a × b: curl the fingers of your right hand from a toward b, and your thumb points in the direction of the cross product.
Normal vector
Any vector perpendicular to a plane. The cross product of two vectors lying in the plane gives a normal to that plane.
Parametric equations of a line
x = x₀ + at, y = y₀ + bt, z = z₀ + ct, where (x₀, y₀, z₀) is a point on the line and ⟨a, b, c⟩ is a direction vector.
Equation of a plane
a(x − x₀) + b(y − y₀) + c(z − z₀) = 0, where (x₀, y₀, z₀) is a point on the plane and ⟨a, b, c⟩ is a normal vector.
Scalar triple product
a · (b × c), a scalar whose absolute value gives the volume of the parallelepiped formed by the three vectors. Not typically tested in Math 126 exams but good to know.
a × b = (a₂b₃ − a₃b₂) i − (a₁b₃ − a₃b₁) j + (a₁b₂ − a₂b₁) k
Remember the sign pattern: positive, negative, positive for i, j, k. Each 2×2 determinant covers the entries from the rows of a and b, omitting the column of that basis vector.
Verification step: after computing a × b, always check that a · (a × b) = 0 and b · (a × b) = 0. If either is not zero, you made an arithmetic error.
Anti-commutative: a × b = −(b × a). Swapping the order flips the sign.
Distributive: a × (b + c) = a × b + a × c.
Scalar factor: c(a × b) = (ca) × b = a × (cb).
Not commutative and not associative.
|a × b| = |a| |b| sin(θ), where θ is the angle between the vectors.
|a × b| equals the area of the parallelogram formed by a and b.
Area of the triangle formed by a and b = (1/2) |a × b|.
If a and b are parallel (θ = 0 or π), then a × b = ⟨0, 0, 0⟩.
The direction of a × b is given by the right-hand rule.
A line needs a point (x₀, y₀, z₀) and a direction vector ⟨a, b, c⟩.
Parametric form: x = x₀ + at, y = y₀ + bt, z = z₀ + ct.
Strategy to find a line equation:
If you have a point and a direction, you are done.
If you have two points A and B, the direction vector is AB = B − A, and use either point.
A plane needs a point (x₀, y₀, z₀) and a normal vector ⟨a, b, c⟩.
Equation: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0.
Strategy to find a plane equation:
If you have a point and a normal, plug straight in.
If you have three points A, B, C: compute AB and AC (two vectors in the plane), then take n = AB × AC as the normal.
Use different parameter names (t and u). Set the x, y, z expressions equal and solve the resulting system. If you find t and u satisfying the first two equations, check the third. If it holds, the lines intersect at that point. If not, they do not intersect (they are skew).
Two non-parallel planes meet in a line. To parameterise it:
Combine the two plane equations (add or subtract) to eliminate one variable.
Choose a free variable, set it equal to t, and solve for the other two.
Alternatively, find two specific points on both planes (by choosing convenient values for one coordinate and solving), then use those two points to write the parametric equations.
\mathbf{a} \times \mathbf{b} = (a_2 b_3 - a_3 b_2)\,\mathbf{i} - (a_1 b_3 - a_3 b_1)\,\mathbf{j} + (a_1 b_2 - a_2 b_1)\,\mathbf{k}|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta = \text{area of parallelogram}\text{Area of triangle} = \tfrac{1}{2}|\mathbf{a} \times \mathbf{b}|Line: x = x₀ + at, y = y₀ + bt, z = z₀ + ct.
Plane: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0.
The cross product is how physics defines torque: τ = r × F, where r is the lever arm and F is the applied force. The right-hand rule for cross products matches the "righty-tighty, lefty-loosy" convention for bolts and screws. Plane equations are used in computer graphics to define surfaces and in robotics to describe workspace boundaries.
Students often assume the cross product is commutative. It is not: a × b = −(b × a). Swapping the order reverses the direction.
A frequent mistake is using the same parameter variable for both lines when checking for intersection. You must use different letters (t and u), because the two objects can reach the same point at different parameter values.
Some students confuse direction vectors (for lines) with normal vectors (for planes). A line's equation uses a vector parallel to the line; a plane's equation uses a vector perpendicular to the plane.
When finding a plane through three points, students sometimes skip the cross product and try to guess the normal. This almost always leads to errors.
⚠️ "Find the equation of the plane through three points" is a standard exam question. The method is always: compute two in-plane vectors, take their cross product for the normal, then write the plane equation.
⚠️ Always verify your cross product by checking a · (a × b) = 0 and b · (a × b) = 0. This catches arithmetic mistakes before they cost you marks.
⚠️ Line intersection problems require checking the third equation after solving the first two. Many students lose marks by stopping early.
⚠️ Know the difference between lines that do not intersect because they are parallel and lines that do not intersect because they are skew (non-parallel, non-intersecting).
True or false: a × b and b × a point in the same direction. (False, they point in opposite directions.)
Fill in the blank: If a and b are parallel, then |a × b| = ___. (0)
True or false: A plane's equation uses a direction vector parallel to the plane. (False, it uses a normal vector perpendicular to the plane.)
Fill in the blank: A line in R³ is defined by a point and a ___ vector. (direction)
True or false: Two lines in 3D that do not intersect must be parallel. (False, they could be skew.)
Q: Compute ⟨1, 2, 3⟩ × ⟨4, 5, 6⟩.
A: i(2·6 − 3·5) − j(1·6 − 3·4) + k(1·5 − 2·4) = i(12 − 15) − j(6 − 12) + k(5 − 8) = ⟨−3, 6, −3⟩.
Q: Find the area of the triangle with vertices A(1, 0, 0), B(0, 1, 0), C(0, 0, 1).
A: AB = ⟨−1, 1, 0⟩, AC = ⟨−1, 0, 1⟩. AB × AC = ⟨1, 1, 1⟩. |AB × AC| = √3. Area = √3 / 2.
Q: Write the parametric equations of the line through (1, −2, 3) in the direction ⟨2, 0, −1⟩.
A: x = 1 + 2t, y = −2, z = 3 − t.
Q: Find the equation of the plane through (1, 2, 3) with normal ⟨4, −1, 2⟩.
A: 4(x − 1) − 1(y − 2) + 2(z − 3) = 0, which simplifies to 4x − y + 2z = 8.
Q: Do the lines x = 1 + 2t, y = 5 − t, z = 3t and x = 4 − u, y = −2 + u, z = 6 − 2u intersect?
A: From x: 1 + 2t = 4 − u, so u = 3 − 2t. From y: 5 − t = −2 + (3 − 2t), giving t = 0, u = 3. Check z: 3(0) = 0 and 6 − 2(3) = 0. Both equal, so yes. The intersection point is (1, 5, 0).
Cross products reappear in 13.3 when computing the binormal vector B = T × N and when using the curvature formula κ = |r'(t) × r''(t)| / |r'(t)|³. Plane equations return in Chapter 14 as tangent planes to surfaces. The scalar triple product connects to change-of-variable determinants in triple integrals.
cross product, vector product, determinant, anti-commutative, right-hand rule, parallelogram area, triangle area, torque, scalar triple product, parallelepiped volume, parametric equations, direction vector, normal vector, plane equation, skew lines, line intersection, plane intersection, Math 126, Calculus III, UW Calculus