## The Cross Product of Two Vectors In Space

Written by Dr. Julia Arnold Associate Professor of Mathematics Tidewater Community College, Norfolk Campus, Norfolk, VA With Assistance from a VCCS LearningWare Grant

In this lesson you will learn •how to find the cross product of two vectors •how to find an orthogonal vector to a plane defined by two vectors •how to find the area of a parallelogram given two vectors •how to find the volume of a parallelepiped given three vectors

## Objective 1

### Finding the cross product of two vectors

The Cross Product the dot product, represents a vector.

a

b

The cross product is defined to be for

a

a

b a

1 ,

a

2 ,

a

3 

and b

b

1 ,

b

2 ,

b

3

a

2

b

3 

a

3

b

2 ,

a

3

b

1 

a

1

b

3 ,

a

1

b

2 

a

2

b

1 You are probably wondering if there is an easy way to remember this. The easy way is to use determinants of size 3 x 3.

a

a

b a

1 ,

a

2 ,

a

3 

and b

b

1 ,

b

2 ,

b

3

a

2

b

3 

a

3

b

2 ,

a

3

b

1 

a

1

b

3 ,

a

1

b

2 

a

2

b

1

i a

1 Let’s set up a 3 x 3 determinant as follows: 1. First use the unit vectors

i

,

j

,

and k

as the first row of the determinant.

2. Use row 2 for the components of a and row 3 for the components of b.

j k a

2

a

3

b

1

a

2

b

3

b

2

b

3 

a

3

b

2 

i

a

2

b

3

a

3

b

1 

a

1

b

3

a

3

b

1

a

1

b

2

a

1

b

2  

a

2

b

1

### 

k a

3

b

2

a

1

b

3

a

2

b

1  

Find the cross product for the vectors below. Do the problem before clicking again.

a

 2 , 4 , 5

and b

 1 ,  2 ,  1

2 1

 4 2

 5 1 

### i

   4    4  10  

6

 5  5 1 1  7

 8

 

 2     2 

   4   

 5     4 

 2    

### k

 4     

• • • • • Since the cross product is determined by using determinants, we can understand the algebraic properties of the Cross Product which are: • u  v   v

##  

Which would come from the fact that if you interchange two rows of a determinant you negate the determinant.

u       c   c u  v  u  u  0  0  u  0 u  u  0 u      w

## Objective 2

### find an orthogonal vector to a plane defined by two vectors

Now that you can do a cross product the next step is to see why this is useful.

Let’s look at the 3 vectors from the last problem

a

 2 , 4 , 5 ,

b

 1 ,  2 ,  1

and a

b

 6 , 7 ,  8 What is the dot product of

a

 And 2 , 4 , 5

with a

b

 6 , 7 ,  8

b

 1 ,  2 ,  1

with a

b

 6 , 7 ,  8 correct.

perpendicular.

?

Recall that whenever two non-zero vectors are perpendicular, their dot product is 0. Thus the cross product creates a vector perpendicular to the vectors a and b. Orthogonal is another name for

##    

u  1 , 2 , 3 , v   1 ,  1 , 0 , and w  5 , 0 ,  1 Solution: u  1 , 2 , 3 , v   1 ,  1 , 0 , and w  5 , 0 ,  1 The left hand side i  1 5 j  1 0 k 0  1  i  0  0  (  5 k  j  0 )  i  j  5 k 1 , 2 , 3  1 ,  1 , 5  1  2  15  14 The right hand side i 1  1 j 2  1 k 3  0   3 j  k  (  2 k  0  3 i )  3 i  3 j  k 0 3 ,  3 , 1  5 , 0 ,  1  15  0  1  14

Geometric Properties of the Cross Product u and v theta) be the angle between u and v  u u  v  v 1. u  v is orthogonal to both u and v Remember: orthogonal means perpendicular to

Geometric Properties of the Cross Product u and v theta) be the angle between u and v  2. u  v  u v sin  u u  v  v Proof: Let u  u 1 , u 2 , u 3 and v  v 1 , v 2 , v 3 u  v  u  v  u 2 v 3  u 3 v 2 , u 3 v 1  u 1 v 3 , u 1 v 2  u 2 v 1  u 2 v 3  u 3 v 2   u 3 v 1  u 1 v 3  u 1 v 2  u 2 v 1  2 This is the left hand side

Let u  u 1 , u 2 , u 3 and v  v 1 , v 2 , v 3 u v Re call sin   cos   u 1 2  u 2 2  u 3 2 u  v u v v 1 2 so cos 2   u  v 2 2 u   v 2  v 3 2 2 2 v u v sin   u v 1  u u   2 2 v 2  u u v v 1  cos 2  u 2 v 2  u   v 2  u 2 v 2  u   2 Now all we need to do is show that the stuff under the radical is the same as the square of the magnitude of the cross product.

In other words, equate the radicands.

Let u

u u

1 , 2 ,

u

3

u and v

v v

1 , 2 ,

v

3 

u v

2 3 

u v

3 2

u

2

v

2  (

u

1 2

u

2

u

3 2 )(

v

1 2

u v

3 1 

u

1

v

3

v

2

v

3 2 ) 

u v

1 2 

u

2

v

1  2 u  2 v 2  u   2 1 1  ) 2  1 2     2 2

u v

 2

u

2

v

2 1 3    2 2 1   2 2   2 3    2 2

u v

 2 3 1    3 2   

u

3 2 2

v

3 3 3

v

) 2  2 2

u v

1 1  2 2

u v

1 2  2 2

u v

1 3  2 2

u v

2 1  2 2

u v

2 2  2 2

u v

2 3  2 2

u v

3 1  2 2

u v

3 2  2 2

u v

3 3  2 2

u v

1 1 1 2 2 2   2

u v u v

1 1 2 2 1 2 2

u v

3  2 2

u v

 2

u v u v

1 1 3 3  2 2

u v

2 2  2 2

u v

 2 2

u v

  2 2

u v

2

u v u v

2 2 3 3  2  2 2

u v

3 3  2 1 

u v

1 2 2   2

u v

1

u v

1 2 2

u v

2 1  2 1

u v

1 3  1 

u v

3 1 3  2

u v u v

1 1 3 3

u v

2 3  

u v

3 2 3 1  2  2 3   2  2

u v u v

2 2 3 3  Which concludes the proof  3 2  (Sometimes proofs are not hard but some do require patience.)

3.

### if and only if u and v are multiples of each other

4.

area of the paralleogram having

||u|| 

u and v as adjacent sides

.

y ||v|| Proof: The area of a parallelogram is base times height. A = bh sin = y/||u||   

u

v

Example problem for property 1 Find 2 unit vectors perpendicular to a= 2 i - j + 3k and b = -4i + 2j - k. Solution The two given vectors define a plane. The Cross Product of the vectors is perpendicular to the plane and is proportional to one of the desired unit vectors.

To make its length equal to one, we simply divide by its magnitude: .

Find 2 unit vectors perpendicular to a= 2 i - j + 3k and b = -4i + 2j - k. Solution

i

2  4

j

 1 2

k

3  1  (

i

 12

j

 4

k

)  ( 6

i

 2

j

 4

k

)  (  5

i

 10

j

)  5

i

 10

j

 25  100  125  5 5

Divide

5 1 5   5

i the vector

 10 

j

by

5  5

its

5

i magnitude

 5 10 5

j

  1

i

5  2 5

j

  5 5

i

 2 5 5

j

For a second unit vector simply multiply the answer by -1 5 5

i

 2 5 5

j

## Objective 3

### find the area of a parallelogram given two vectors

Example for property 4

4.

area of the paralleogram having u and v as adjacent sides

.

Area of a Parallelogram via the Cross Product

Show that the following 4 points define a parallelogram and then find the area. A z A(4,4,6), B(4,14,6), C(1,11,2), D(1,1,2) B x D C y

Solution: First we find the vectors of two adjacent sides:

AB

 4  4 , 4  14 , 6  6  0 ,  10 , 0

 4  1 , 4  1 , 6  2  3 , 3 , 4 Now find the vectors associated with the opposite sides:

DC

 1  1 , 1  11 , 2  2  0 ,  10 , 0 

AB DB

 1  4 , 11  14 , 2  6   3 ,  3 ,  4  

This shows that opposite sides are associated with the same vector, hence parallel. Thus the figure is of a parallelogram.

Continued

The area is equal to the magnitude of the cross product of vectors representing two adjacent sides: Area = |AB XAD|

i

0 3

j

 10 3 4

k

0    40

i

 0

j

 0

k

  0

i

 0

j

 30

k

   40

i

 0

j

 30

k

 40

i

 30

k

(  40 ) 2  30 2  1600  900  2500  50 The area of the parallelogram is 50 square units.

Area of a Triangle via the Cross Product

Since the area of a triangle is based on the area of a parallelogram, it follows that the area would be ½ of the cross product of vectors of two adjacent sides.

Find the area of the triangle whose vertices are P(4,4,6), Q(5,16,-2) R(1,1,2) Area parallelogram) = |PQ x PR|. The area of the triangle is half of this. z P R Q x

Area of a Triangle via the Cross Product Continued P(4,4,6), Q(5,16,-2) R(1,1,2) Area parallelogram) = |PQ x PR|

But what if we choose RP and RQ? Would the result be the same?

Let’s do both and see!

z P R x

PQ

 5  4 , 16  4 ,  2  6  1 , 12 ,  8

PR

 1  4 , 1  4 , 2  6   3 ,  3 ,  4 Q

RP

 4  1 , 4  1 , 6  2  3 , 3 , 4

RQ

 5  1 , 16  1 ,  2  2  4 , 15 ,  4

PQ

 5  4 , 16  4 ,  2  6  1 , 12 ,  8

PR

 1  4 , 1  4 , 2  6   3 ,  3 ,  4

RP

 4  1 , 4  1 , 6  2  3 , 3 , 4

RQ

 5  1 , 16  1 ,  2  2  4 , 15 ,  4

i

1  3

j

12  3

k

 8  4    48

i

 24

j

 3

k

   24

i

 4

j

 36

k

   72

i

 28

j

 33

k

1 2  72

i

 28

j

 33

k

 1 2 (  72 ) 2  28 2  33 2  1 2 5184  784  1089  1 2 7075

i

3 4 3

j

15

k

4  4    12

i

 16

j

 45

k

   60

i

 12

j

 12

k

   72

i

 28

j

 33

k

Since this is the same vector, the magnitude would be the same also.

z R P x Q

The triple scalar product is defined as:

u

 

u For u

u

1

i

u

2

j

u

3

k

,  

u

1 

v

1

w

1

u v

2

w

2 2

u v

3

w

3 3

v

v

1

i

v

2

j

v

3

k

,

and w

w

1

i

w

2

j

w

3

k

## Objective 4

### how to find the volume of a parallelepiped given three vectors

The triple scalar product is defined as: AB ·(AC x AD). A parallelepiped is a 6 sided figure whose sides are parallelograms.

The volume of the parallelepiped can be found using the absolute value of the triple scalar product and 3 adjacent vectors.

Why would the triple scalar product be the volume of the figure?

C A B D Why would the triple scalar product be the volume of the figure?

Since the base is a parallelogram we could represent its area by the cross product ||AC x AD|| Recall that the area of a parallelepiped is the base times the height.

The height would be equivalent to:

proj AC

The triple scalar product is defined as: AB ·(AC x AD). C A B D base times the height ||AC x AD||

proj AC

 

AC

AC

AB

 

AC

 The absolute value insures a positive answer for the volume.

Volume of a Parallelepiped via the Scalar Triple Product:

Find the volume of the parallelepiped with adjacent edges AB, AC, and AD, where the points are A(4, -3, -2), B(2, 0, 5), C(-3, 2, 1), and D(1, 3, 2).

The volume is given by the scalar triple product: AB · (AC X AD). First we need the three vectors: AB = [2 - 4]i + [0 - (-3)]j + [5 - (-2)]k = -2i + 3j + 7k.

AC = [-3 - 4]j + [2 - (-3)]j + [1 - (-2)]k = -7i + 5j + 3k AD = [1 - 4]i + [3 - (-3)]j + [2 - (-2)]k = -3i + 6j + 4k First, find the cross product: = i[20 - 18] - j[-28 - (-9)] + k[-42 - (-15)] = 2i + 19j - 27k Now form the dot product to get the volume: Volume = |AB · (2i + 19j - 27k)| = | (-2i + 3j + 7k) · (2i + 19j - 27k)| = | 4 + 57 - 189 | = 136 cubic units

From:http://www.jtaylor1142001.net/calcjat/Solutions/VCro ssProduct/VCPVolParallelepiped.htm

Finally, for all of you potential physicists, a real world application of the cross product.

Torque is defined by Webster as “a twisting or wrenching effect or moment exerted by a force acting at a distance on a body, equal to the force multiplied by the perpendicular distance between the line of action of the force and the center of rotation at which it is exerted”.

M If a vector force F is applied at point B of the vector AB where both vectors are in the same plane, then M is the moment of the force F about the point B.

||M|| will measure the tendency of the vector AB to rotate counterclockwise according to the right hand rule about an axis directed along the vector M.

A great example of torque is tightening a bolt with a wrench.

A few observations can be made: The farther away from the bolt that the force is applied the greater the magnitude of the torque. Thus, a longer wrench produces greater torque for a given amount of force.

Recall: Second, we can see which produces the largest torque would be when  = 90 o . 

force

length vector where

is the angle

force between length

sin 

them

.

Example: Suppose you have a 12 inch wrench and you apply a 20 lb force at an angle of 30 degrees. What is the torque in foot-pounds at the bolt? What is the maximum torque that can be applied to this bolt?

Solution:

force

length vector where

is the angle

force between length

sin 

them

.

20  1 sin 6  20  1 2  10

foot

pounds

The maximum torque is applied when    2 20  1 sin 2  20    20

foot

pounds

For comments on this presentation you may email the author Dr. Julia Arnold at [email protected]

.