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Vectors in Two and
Three Dimensions
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2. Copyright © Cengage Learning. All rights reserved.
9.5
The Cross Product
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3. Objectives The Cross Product Properties of the Cross Product
Area of a Parallelogram
Volume of a Parallelepiped

4. The Cross Product
In this section we define an operation on vectors that allows us to find a vector which is perpendicular to two given vectors.

5. The cross product

6. The Cross Product
Given two vectors a = a1, a2, a3 and b = b1, b2, b3 we often need to find a vector c perpendicular to both a and b. If we write c = c1, c2, c3 then a  c = 0 and b  c = 0, so a1c1 + a2c2 + a3c3 = 0 b1c1 + b2c2 + b3c3 = 0 You can check that one of the solutions of this system of equations is the vector c = a2b3 – a3b2, a3b1 – a1b3, a1b2 – a2b1.

7. The Cross Product
This vector is called the cross product of a and b and is denoted by a  b The cross product a  b of two vectors a and b, unlike the dot product, is a vector (not a scalar). For this reason it is also called the vector product. Note that a  b is defined only when a and b are vectors in three dimensions.

8. The Cross Product
To help us remember the definition of the cross product, we use the notation of determinants. A determinant of order two is defined by A determinant of order three is defined in terms of second-order determinants as

9. The Cross Product
We can write the definition of the cross product using determinants as = (a2b3 – a3b2)i + (a3b1 – a1b3)j
+ (a1b2 – a2b1)k

10. Example 1 – Finding a Cross Product
If a = 0, –1, 3 and b = 2, 0, –1, find a  b.
Solution: We use the formula above to find the cross product of a and b:

11. Example 1 – Solution = (1 – 0)i – (0 – 6)j + (0 – (–2))k = i + 6j + 2k
cont’d
= (1 – 0)i – (0 – 6)j + (0 – (–2))k
= i + 6j + 2k
So the desired vector is i + 6j + 2k.

12. Properties of the Cross Product

13. Properties of the Cross Product
One of the most important properties of the cross product is the following theorem.

14. Example 2 – Finding an Orthogonal Vector
If a = –j + 3k and b = 2i – k, find a unit vector that is orthogonal to the plane containing the vectors a and b.
Solution: By the Cross Product Theorem the vector a  b is orthogonal to the plane containing the vectors a and b. (See Figure 1.)
The vector a  b is perpendicular to a and b.
Figure 1

15. Example 2 – Solution
cont’d
In Example 1 we found a  b = i + 6j + 2k. To obtain an orthogonal unit vector, we multiply a  b by the scalar
1/|a  b|:
So the desired vector is

16. Properties of the Cross Product
If a and b are represented by directed line segments with the same initial point (as in Figure 2), then the Cross Product Theorem says that the cross product a  b points in a direction perpendicular to the plane through a and b.
Right-hand rule
Figure 2

17. Properties of the Cross Product
It turns out that the direction of a  b is given by the right-hand rule: If the fingers of your right hand curl in the direction of a rotation (through an angle less than 180) from a to b, then your thumb points in the direction of a  b (as in Figure 2). You can check that the vector in Figure 1 satisfies the right-hand rule.
The vector a  b is perpendicular to a and b.
Figure 1

18. Properties of the Cross Product
Now that we know the direction of the vector a  b, the remaining thing we need is a  b the length |a  b|.

19. Properties of the Cross Product
We have now completely determined the vector a  b geometrically. The vector a  b is perpendicular to both a and b, and its orientation is determined by the right-hand rule. The length of a  b is | a || b | sin .

20. Area of a Parallelogram

21. Area of a Parallelogram
We can use the cross product to find the area of a parallelogram. If a and b are represented by directed line segments with the same initial point, then they determine a parallelogram with base |a|, altitude |b| sin , and area A = | a |(| b | sin ) = |a  b| (See Figure 3.)
Parallelogram determined by a and b
Figure 3

22. Area of a Parallelogram
Thus we have the following way of interpreting the magnitude of a cross product.

23. Example 4 – Finding the Area of a Triangle
Find the area of the triangle with vertices P(1, 4, 6), Q(–2, 5, –1), and R(1, –1, 1).
Solution: By the Cross Product Theorem, the vector PQ  PR is perpendicular to both PQ and PR, and is therefore perpendicular to the plane through P, Q, and R.
We know that
PQ = (–2 – 1)i + (5 – 4)j + (–1 – 6)k = –3i + j – 7k
PR = (1 – 1)i + ((–1) – 4)j + (1 – 6)k = –5j – 5k

24. Example 4 – Solution We compute the cross product of these vectors:
cont’d
We compute the cross product of these vectors:
PQ  PR =
= (–5 – 35)i – (15 – 0)j + (15 – 0)k
= –40i – 15j + 15k
Therefore

25. Example 4 – Solution
cont’d
The area of the parallelogram with adjacent sides PQ and PR is the length of this cross product:
The area A of the triangle PQR is half the area of this
parallelogram, that is,

26. Volume of a Parallelepiped

27. Volume of a Parallelepiped
The product a  (b  c) is called the scalar triple product of the vectors a, b, and c. You can check that the scalar triple product can be written as the following determinant:

28. Volume of a Parallelepiped
The geometric significance of the scalar triple product can be seen by considering the parallelepiped determined by the vectors a, b, and c (see Figure 4).
Parallelepiped determined by a, b, and c
Figure 4

29. Volume of a Parallelepiped

30. Example 5 – Coplanar Vectors
Use the scalar triple product to show that the vectors a = 1, 4, –7, b = 2, –1, 4, and c = 0, –9, 18 are coplanar, that is, lie in the same plane.
Solution: We compute the scalar triple product:

31. Example 5 – Solution
cont’d
= 1(18) – 4(36) – 7(–18) = 0 So the volume of the parallelepiped is 0, and hence the vectors a, b, and c are coplanar.