Vectors and curves in space · Definition

Cross product

multivariable calculusvectors

It is the tool for building a vector perpendicular to two given directions, and the first step toward describing planes and surfaces in space.

Definition

The cross product of two vectors \(\mathbf u\) and \(\mathbf v\) in \(\mathbb R^3\) is the vector perpendicular to both whose length is the area of the parallelogram they span, oriented by the right-hand rule.

\[\mathbf u\times\mathbf v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix}\]

Unlike the dot product, which returns a number, the cross product returns a vector and exists only in \(\mathbb R^3\). Its length is \(|\mathbf u||\mathbf v|\sin\theta\), where \(\theta\) is the angle between the vectors; hence it is zero exactly when \(\mathbf u\) and \(\mathbf v\) are parallel and largest when they are perpendicular. The area of a triangle with sides \(\mathbf u\) and \(\mathbf v\) is half that length.

The determinant in the formula is symbolic: it is expanded along the first row, with the vectors \(\mathbf i,\mathbf j,\mathbf k\) as entries. For example, \(\langle1,2,3\rangle\times\langle4,5,6\rangle=\langle-3,6,-3\rangle\). The minus sign of the middle term is the most frequent mistake when computing it.

Order matters: \(\mathbf v\times\mathbf u=-(\mathbf u\times\mathbf v)\), so the product is anticommutative and, in particular, \(\mathbf u\times\mathbf u=\mathbf 0\). It distributes over addition and pulls out scalars, but it is not associative: \(\mathbf i\times(\mathbf i\times\mathbf j)=-\mathbf j\), whereas \((\mathbf i\times\mathbf i)\times\mathbf j=\mathbf 0\).

Its main use will be to give the normal of a plane from two directions lying in it, and later the normal vector of a parametric surface, which is the piece needed to integrate over surfaces.

Think about it

Why does it make no sense to treat the cross product of two vectors in \(\mathbb R^2\) as a vector in \(\mathbb R^2\)?

Because the result must be perpendicular to both factors and, in the plane, that direction leaves the plane (it is vertical). That is why we work in \(\mathbb R^3\); for planar vectors only the \(\mathbf k\) component remains.

Exercises 10

References

  • James Stewart, Daniel K. Clegg, Saleem Watson, Calculus: Early Transcendentals, 9th edition, Cengage, 2021. §12.4
  • Jerrold E. Marsden, Anthony Tromba, Vector Calculus, 6th edition, W. H. Freeman, 2012. §1.3
  • Joel Hass, Christopher Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr., Thomas' Calculus, 15th edition, Pearson, 2023. §12.4