Linear Algebra 10 | Cross Product [dark version]

TL;DR

The cross product u × v is defined only for vectors in R3 and outputs a vector in R3.

Briefing Cornell Notes

Briefing

The cross product is a concrete way to combine two 3D vectors into a third vector: given u and v in R3, the result u × v is itself a vector that is perpendicular to both inputs. That perpendicularity makes it a practical tool for finding orthogonal directions, while its magnitude links directly to geometry—specifically the area of the parallelogram spanned by u and v.

In R3, the cross product is defined component-wise. If u = (u1, u2, u3) and v = (v1, v2, v3), then u × v = (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1). The pattern is easy to remember as “start in the middle, move down, mirror, and subtract,” producing each component from products of two coordinates with a sign change. For those who prefer a more systematic expression, the same result can be written using the Levi-Civita symbol εijk, summing over indices i, j, k from 1 to 3 and multiplying by the canonical basis vector ek so the total becomes a vector.

Three key properties anchor how u × v behaves. First is orthogonality: u × v is orthogonal to u and also orthogonal to v under the standard inner product, meaning (u × v) · u = 0 and (u × v) · v = 0. Geometrically, the cross product points in the “third direction” perpendicular to the plane containing u and v.

Second is orientation, determined by the right-hand rule. Point the thumb along u, the index finger along v, and the middle finger gives the direction of u × v. This rule matters because the cross product is not commutative: swapping the order reverses the direction.

Third is magnitude. The length (Euclidean norm) of u × v equals the area of the parallelogram formed by u and v. That area interpretation connects algebra to geometry: the parallelogram’s area can be computed as |u||v|sin(θ), where θ is the angle between the vectors. As a consequence, triangle areas follow easily too, since a triangle is half a parallelogram.

An example makes the mechanics and the properties concrete. With u = (2, 1, 0) and v = (0, 1, 0), both vectors lie in the xy-plane. Applying the component formula yields u × v = (0, 0, 2). The result is orthogonal to both inputs because it points purely in the z-direction. The right-hand rule confirms the positive z direction, and the magnitude matches the parallelogram area: the base and height in the xy-plane are 1 and 2, giving area 2. The example also reinforces non-commutativity—reversing u and v would flip the sign of the z-component.

Cornell Notes

The cross product takes two vectors in R3 and returns a third vector perpendicular to both. For u = (u1,u2,u3) and v = (v1,v2,v3), u × v = (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1). This vector is orthogonal to u and v under the standard inner product, and its direction follows the right-hand rule (thumb along u, index along v, middle gives u × v). The magnitude |u × v| equals the area of the parallelogram spanned by u and v, linking the algebra to geometry. A worked example with u = (2,1,0) and v = (0,1,0) produces u × v = (0,0,2), matching both orthogonality and area.

How is the cross product computed from components in R3?

If u = (u1,u2,u3) and v = (v1,v2,v3), then u × v is the vector (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1). Each component is built from products of two coordinates, with a subtraction that creates the correct sign pattern.

What does orthogonality mean for u × v, and how is it expressed mathematically?

Orthogonality means u × v is perpendicular to both u and v. Using the standard inner product, this becomes (u × v) · u = 0 and (u × v) · v = 0. In geometric terms, u × v points in the direction perpendicular to the plane spanned by u and v.

How does the right-hand rule determine the direction of u × v?

Place the right hand so the thumb points along u and the index finger points along v. Lift the middle finger to a right angle; it indicates the direction of u × v. Because the rule depends on the order of u and v, swapping them reverses the direction.

Why is the magnitude |u × v| tied to area?

The length of u × v equals the area of the parallelogram formed by u and v. If θ is the angle between them, the area is |u||v|sin(θ). This also implies triangle areas are half the parallelogram area.

What does the example u = (2,1,0), v = (0,1,0) show?

Both vectors lie in the xy-plane. Using the component formula gives u × v = (0,0,2), so the result points purely in the z-direction. That confirms orthogonality to both inputs. The magnitude is 2, matching the parallelogram area in the xy-plane (base 1 times height 2).

Review Questions

Given u = (1,2,3) and v = (4,5,6), compute u × v using the component formula.
Explain why u × v must be orthogonal to both u and v, and write the corresponding dot-product equalities.
If u × v has magnitude 10 and the angle between u and v is 30°, what does that imply about |u||v|? (Use the area relationship.)

Key Points

1
The cross product u × v is defined only for vectors in R3 and outputs a vector in R3.
2
For u = (u1,u2,u3) and v = (v1,v2,v3), u × v = (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1).
3
u × v is orthogonal to both u and v under the standard inner product: (u × v) · u = 0 and (u × v) · v = 0.
4
The direction of u × v follows the right-hand rule, so swapping u and v reverses the result.
5
The magnitude |u × v| equals the area of the parallelogram spanned by u and v, linking cross products to geometry.
6
A triangle’s area is half the parallelogram area, so cross products can compute triangle areas in 3D when the vectors define the base and height direction.

Highlights

u × v is perpendicular to both u and v, which is captured by dot products equaling zero.

The right-hand rule determines the sign and direction of u × v, making the operation non-commutative.

|u × v| equals the parallelogram area spanned by u and v, turning an algebraic operation into a geometric measurement.

For u = (2,1,0) and v = (0,1,0), the cross product is (0,0,2), matching both direction (z-axis) and area (2).