In this explainer, we will learn how to find the cross product of two vectors in the coordinate plane.

There are two ways to multiply vectors together. You may already be familiar with the dot product, also called the scalar product. This product leads to a *scalar quantity* that is given by the product of
the magnitudes of both vectors multiplied by the cosine of the angle between
the two vectors. As for the cross product, it is a multiplication of vectors
that leads to a *vector*.

### Definition: Cross Product

The cross product of two vectors and is a vector perpendicular to the plane that contains and and whose magnitude is given by where is the angle between and .

From the definition of the cross product, we find that the cross product of two parallel (or collinear) vectors is zero as the sine of the angle between them (0 or ) is zero. Note that no plane can be defined by two collinear vectors, so it is consistent that if and are collinear.

From the definition above, it follows that the cross product of any two noncollinear vectors in the coordinate plane with and as units vectors is parallel to , where is a unit vector perpendicular to the plane containing and , as shown in the diagram.

Let us consider the two vectors and in the coordinate plane.

Vector makes an angle with and vector makes an angle . The angle between and is therefore , so we have . Using the subtraction trigonometric identity , we find that

Let us make a matrix with the components in terms of and of in the first row and those of in the second row and calculate its determinant. Recall that the determinant of a matrix is given by

Hence, we find that since .

If we combine this with the fact that is a vector parallel to , we can write the following definition of the cross product of two vectors in the coordinate plane with and as unit vectors.

### Definition: Cross Product of Two Vectors in the Coordinate Plane

For two vectors and in the coordinate plane with and as unit vectors, the cross product of and is where is the angle between and , and the vectors , , and are the fundamental unit vectors along the -, -, and -axes, respectively, as shown in the diagram.

Let us illustrate this definition of the cross product with a first example.

### Example 1: Finding a Missing Component Given the Cross Product of Two 2D Vectors

If , , and , find the value of .

### Answer

As , we have and ; and as , we have and .

We furthermore know that

Hence

Let us further practice calculating the cross product of two vectors with a further question involving vector addition.

### Example 2: Calculating the Cross Product of Two 2D Vectors

Given that , , and , determine .

### Answer

Let us first work out :

From this, we deduce that the - and -components of are and . Also, from , we know that and .

We can now calculate as

With the previous example, we can wonder whether the cross product is distributive, that is, do we have ?

It can be found out easily when looking more closely at the determinant we used to calculate :

The rule for vector addition is and ; therefore,

It follows that ; the cross product is therefore *distributive*.

### Property: Distributivity of the Cross Product

The cross product is distributive:

We will now use our knowledge of how cross products are calculated to find an unknown vector given the results of its cross product with two known vectors.

### Example 3: Finding a Vector Given its Cross Product with Two Known Vectors

If , , , and , find .

### Answer

Let us extract the components of and from their expressions in terms of and : and , and and .

We can now write their cross products with as

Hence, we have

and

We have now two linear equations with two unknowns ( and ).

From equation (1), we find that . Substituting this expression for into equation (2) gives

Substituting into , we find that

Therefore, we have

Let us now find the cross product of two vectors whose components are not explicitly given but that are defined with specific points in a rectangle.

### Example 4: Finding the Cross Product of Two Vectors in a Rectangle

is a rectangle where is a unit vector perpendicular to its plane. Find .

### Answer

To find , we have two possibilities. Either we find the components of both vectors, for instance in the coordinate plane with origin , , and , or we apply , where is the angle between and and is a unit vector perpendicular to the plane of the rectangle. This second method implies that we have to find the magnitude of both vectors and .

**First Method**

In the coordinate plane , we have , , and . From this, we find that and .

The cross product of and is as is a unit vector perpendicular to the plane of the rectangle.

**Second Method**

The magnitude of is simply the length , so it is 44 cm.

Point is the middle of the rectangleβs diagonals, so , and (applying the Pythagorean theorem in the right triangle ). Hence, .

The sine of the angle between and is given by . The negative sign comes from the fact that the angle from to goes clockwise (negative angle), and , or, for people working only with positive angles (counterclockwise turn), it is then , and .

We can now write

With the second method used to solve the last example, we see the importance of the order of the vectors in the cross product, because commuting them means that the sign of the angle changes, or that it changes from to when one uses only positive angles. The effect is that the sine changes sign. This means that

We say that the cross product is *anticommutative*.

### Property: Anticommutativity of the Cross Product

The cross product is anticommutative:

Let us now look at the geometric meaning of the magnitude of the cross product. For two vectors and , the magnitude of their cross product is .

Since we are interested here in the absolute value of , we do not need to worry if the angle is from to or from to .

If we mentally place the diagram in the unit circle, we immediately see what is.

We see that .

In a geometric context, is the height of the parallelogram spanned by and . Hence, is the area of the parallelogram .

The area of triangle is half that of . It follows that the area of a triangle is equal to half the magnitude of the cross product of two of the three vectors making its sides, that is,

As we are interested here in the magnitude of the cross product, the order of the vectors and their directions ( or ) do not actually matter. We have, however, chosen here to write the vectors starting at one of the triangle vertices to help visualize the parallelogram spanned by these two vectors and the triangle as half of the parallelogram.

Let us use the meaning of the cross product in a geometric context with the last example.

### Example 5: Finding the Area of a Triangle Given Its Three Vertices

Find the area of a triangle , where , , and .

### Answer

The magnitude of the cross product of two vectors is equal to the area of the parallelogram spanned by them. The area of the triangle is equal to half the area of the parallelogram spanned by two vectors defined by its vertices:

Hence, we just need here to choose one vertex, for instance , and find the components of the two vectors from this point, and :

Therefore,

As no specific length unit of the coordinate plane was given, we write βsquare unitsβ to show that we have calculated an area.

### Key Points

- For two vectors and in the coordinate plane , the cross product of and is where is the angle between and , and the vectors , , and are perpendicular unit vectors.
- The cross product is
*distributive*: . - The cross product is
*anticommutative*: . - The cross product of two collinear vectors is zero, and so .
- The area of the parallelogram spanned by and is given by . It follows that the area of the triangle with and defining two of its sides is given by .