The cross product is a vector operation that acts on vectors in three dimensions and results in another vector in three dimensions. In contrast to dot product, which can be defined in both 2-d and 3-d space, the cross product is only defined in 3-d space. Another difference is that while the dot-product outputs a scalar quantity, the cross product outputs another vector.
The algebraic interpretation of the cross product is defined as
If and are 2 vectors in 3-d space, then the cross product can be defined in the following ways:
- as the determinant using the first row;
- as in vector notation.
Find the cross product of and
As the determinant, we have
Simply using the formula from definition (B), we have
Find the cross product of the vectors and
We can get the cross product as the determinant of the following matrix:
Some of the main properties of the cross product are as follows:
1. The cross product is not commutative. That is, . By the right-hand screw rule, and have opposite directions, so . 2. The cross products are distributive with respect to vector addition: . 3. . 4. . 5. . 6. Under reflection (i.e. taking mirror image), all components of a vector change signs. Thus, for we have . So remains the same under reflection.
All these properties can are derived from the definition of the cross product and are left as exercises for the readers to prove.
Given two vectors and , find
We can get the cross product by calculating the determinant:
We could just follow the same steps, but a shortcut would be to use Theorem 1 above:
There are also some properties that relate the cross product and the dot product:
E) (known as Lagrange's identity)
The first two properties are easy to understand if we realize that the cross product outputs a vector perpendicular to both vectors, and that the dot product of perpendicular vectors is zero. The others are like the first two as a result of the properties of the dot and cross products, and are left for the reader to prove.
Given the vectors show that they do indeed satisfy the above four properties A) through D).
We have , which will be left for the readers to show how. Then
Consider three-dimensional vectors and , and let be the angle between them. The geometric interpretation of the cross product is a vector that is perpendicular to both and (using the right-hand rule) and has norm defined as
Given two vectors and what is the magnitude of the cross product if the angle between the two vectors is
Then using the definition for magnitude of a cross product, we have
Show the equivalence of the geometric interpretation and the algebraic interpretation of the cross product.
We need to show that the geometric and algebraic definitions give vectors with the same magnitude and direction. To check direction, we will show that both vectors are perpendicular to and . This is given in the definition of the geometric interpretation. For the algebraic interpretation, taking the dot product gives:
Furthermore, the vectors point in the same direction since the determinant
is positive by expanding along the third row; and hence the vectors are determined by the right hand rule.
Next, we check that the vectors have the same length, by calculating the square of the norm. We have
For the algebraic interpretation, we have
By comparing terms, we see that the lengths of the vectors are equal.
We know that the area of a parallelogram is given by In the figure above, the area is the product of the magnitudes of and
We can see that the area of the parallelogram is equal to the magnitude of the cross product That is, the magnitude of the cross product is equal to the area of the parallelogram formed by the two vectors.
Find the area of the triangle formed by the vectors , , and
Since the cross product of and is the area of the parallelogram, the area of the triangle is equal to half the area of the parallelogram:
Show that the cross product is equal to the volume of of the parallelepiped.
From the properties of the cross product, the area of the base of the parallelepiped is
Let be the angle formed by the vector with the vertical, then the height of the parallelpiped is
Now, since is a vector normal to the plane formed by vectors and , is also the angle between vectors and . So the area of the parallelpiped is
Consider the regular octagon centered at the origin as shown at right. Eight unit vectors are drawn from the center of the octagon to each of its vertices and labeled in the figure. For each pair of distinct unit vectors with , their cross product is computed.
What is the sum of all of these cross products?