Introduction: In this lesson we will examine a combination of vectors known as the cross product. Vector components in 3 dimensions will be combined in such a way as to result in another vector in 3 dimensions. Applications of the cross product will be shown.

The Lesson:

Let v = (2, 5, 1) and u = (-3, 2, 4) be two 3-dimensional vectors. We could also express these vectors in i, j, k form as: v = 2i + 5j + k and u = - 3i + 2j + 4k.

The cross product is written v ´ u and is calculated using a determinant of the matrix constructed from the vectors i, j, and k and the components of v and u as follows:

The cross product v ´ u is a new vector we have called w and it is perpendicular to both v and u. To show this, we will use the dot product formula where is the angle between two vectors r and s. Since neither v and u are vectors with 0 magnitude, if we calculate the dot product and show that it equals zero then the zero product property will allow us to conclude that  .

We have the following:

• v·w = (2, 5, 1)·(18, - 11, 19) = 2(18) + 5(- 11) + 1(19) = 0, and
• u·w = (- 3, 2, 4)·(18, - 11, 19) = - 3(18) +2(- 11) + 4(19) =0.

Our proof is done. The vector w is perpendicular to both u and v.

The vectors v and u form a parallelogram with sides of length . The area of this parallelogram can be calculated as . We will now show that this expression is also equal to the magnitude of w, the cross product of v ´ u.

Using the Pythagorean Theorem, we can calculate the magnitude of v ´ u as

To begin our proof, we will first calculate the length of each base:

• 
• .

If q is the angle between v and u, we know from their dot product that

.

Using (0.2712) gives us which lets us calculate and determine the height of the parallelogram to be

Completing our calculation of the area gives us an answer of

base ´ height =

which equals our previously stated magnitude of v ´ u.

Generalizing:

1. The cross product of two vectors is:

2. The area of the parallelogram formed by v and u is

3. The direction of the cross product is perpendicular to the two vectors which form it. This direction obeys what is known as the “right hand rule.” Point fingers of the right hand in the direction of v and curl them toward the direction of u. The thumb of your right hand now points in the direction of v ´ u. A diagram of a general v and u and their cross product is shown below.

Let's Practice:

1. The cross product can be used to find the torque applied by a force. In the diagram below, a wrench is being used to tighten a bolt. The wrench is shown as vector r and the force applied is F. The cross product of r ´ F is the result of the force being applied to the wrench. The force F is at an angle q = 250º. The angle between r and F is 90º + 20º = 110º. In our formula for the cross product we can use because .

If the force is 30 pounds and the length of the wrench is 0.3 feet then we have:

.

The direction of this vector is perpendicular to both r and F and points into the plane of the paper (-z) allowing it to tighten the bolt properly into its niche. To verify this, point the fingers of your right hand in the direction of r (0º) and curl them toward the direction of F (250º). The thumb of your right hand now points into the plane of the page, the direction of r ´ F.

2. If v = (1, 2, 3) and u = (-2, 6, 4) what is v ´ u and what is the area of the parallelogram formed by v and u?

v ´ u

The area of the parallelogram formed by v and u is

.