next up previous contents
Next: Triple Products of Vectors Up: Vectors and Vector Products Previous: The Law of Cosines   Contents

The Vector, or Cross Product

There is a second way to multiply two vectors. This product of two vectors produces a third vector, which is why it is often referred to as ``the'' vector product (even though there are a number of products involving vectors). It is symbolically differentiated by the multiplication symbol used, which is a large $ \times$ sign, hence it is often referred to as the cross product both for the (cross-like) shape of this sign and because of the pattern of multiplication of components. We write the cross product of two vectors as e.g. $ \vC =
\vA \times \vB$ .

The cross product anticommutes:

$\displaystyle \vA \times \vB = - \vB \times \vA $

It is distributive:

$\displaystyle \vA \times (\vB + \vC) = \vA \times \vB + \vA \times \vC $

(although the order of the product must be maintained!)

It as noted above produces a vector (really a pseudovector, explained later) from two vectors. The magnitude of the cross product of two vectors is defined by:

$\displaystyle \vert\vA \times \vB\vert = A B \sin\theta = A B_\perp = A_\perp B $

using terms similar to those used above in our discussion of dot products.

Figure 3.1: The area between two vectors in a plane is the magnitude of the cross product of those vectors.
Note well! If the vectors both have dimensions of length, the cross product is the area of the parallelogram formed by the vectors as illustrated in figure 3.1. It is sometimes called the areal product for this reason, although one would think two names is enough (and in many contexts, areal product makes no sense).

The direction of $ \vA \times \vB$ is given by the right-hand rule. The direction is always perpendicular or normal to the plane defined by the two non-colinear vectors in the cross product. That leaves two possibilities. If you let the fingers of your right hand line up with $ \vA$ (the first) so that they can curl through the small angle (the one less than $ \pi$ that will not hurt your wrist) into $ \vB$ then the thumb of your right hand will pick out the perpendicular direction of the cross product. In the figure above, it is out of the page.


$\displaystyle \vA \times \vA = - (\vA \times \vA) = 0 $

Together with the rule for rescaling vectors this proves that the cross product of any vector with itself or any vector parallel or antiparallel to itself is zero. This also follows from the expression for the magnitude $ AB\sin\theta$ with $ \theta = 0$ or $ \pi$ .

Let us form the Cartesian representation of a cross product of two vectors. We begin by noting that a right handed coordinate system is defined by the requirement that the unit vectors satisfy:

$\displaystyle \hx \times \hy = \hz $

This is illustrated here:


You can easily check that it is also true that:

$\displaystyle \hx \times \hy = \hz \quad\quad \hy \times \hz = \hz \quad\quad \hz
\times \hx = \hy $

We use the anticommution rule on these three equations:

$\displaystyle \hy \times \hx = -\hz \quad\quad \hz \times \hy = -\hz \quad\quad \hx
\times \hz = - \hy $

And note that:

$\displaystyle \hx \times \hx = \hy \times \hy = \hz \times \hz = 0 $

This forms the full multiplication table of the orthonormal unit vectors of a standard right-handed Cartesian coordinate system, and the Cartesian (and various other orthonormal) coordinate cross product now follows.

Applying the distributive rule and the scalar multiplication rule, multiply out all of the terms in $ \vA \times \vB$ :

$\displaystyle (A_x \hx + A_y \hy + A_z\hz)$ $\displaystyle \times$ $\displaystyle (B_x\hx + B_y\hy + B_z\hz) =$  
    $\displaystyle \ \ A_x B_x \hx\times \hx + A_x B_y \hx\times \hy + A_x B_z \hx\times \hz$  
    $\displaystyle + A_y B_x \hy\times \hx + A_y B_y \hy\times \hy + A_y B_z \hy\times \hz$  
    $\displaystyle + A_z B_x \hz\times \hx + A_z B_y \hz\times \hy + A_z B_z \hz\times \hz$  

The diagnonal terms vanish. The other terms can all be simplified with the unit vector rules above. The result is:

$\displaystyle \vA \times \vB = (A_yB_z - A_zB_y)\hx + (A_zB_x - A_xB_z)\hy +
(A_xB_y - A_yB_x)\hz $

This form is easy to remember if you note that each leading term is a cyclic permutation of xyz. That is, $ A_yB_z \hx$ , $ A_zB_x\hy$ and $ A_x B_y \hz$ are yzx, zxy, and xyz. The second term in each parentheses is the same as the first but in the opposite order, with the attendant minus sign, from the cyclic permutations of zyx.

next up previous contents
Next: Triple Products of Vectors Up: Vectors and Vector Products Previous: The Law of Cosines   Contents
Robert G. Brown 2017-07-11