Lesson 3: The Cross Product

Section 9.4Cross Products and Planes

Math 21a

February 8, 2008

Announcements

I Homework for Monday 2/11:I Section 9.4. Exercises 4, 6, 8, 10, 19, 22, 30; pp. 664–665.I Section 9.5. Exercise 1*; pp. 673–675.

Outline

Torque and the Cross Product

Properties of the Cross ProductOn a basisIn GeneralBy components

Other applicationsAreaVolume

Torque

When force is applied to a lever fixed to a point, some of the forcegoes towards rotation while the rest goes towards stretching thelever.

r

F|F | sin θ θ

τ

The magnitude of the torque is also proportional to the length ofthe lever, and has a direction depending on which direction thelever pivots.

Torque

When force is applied to a lever fixed to a point, some of the forcegoes towards rotation while the rest goes towards stretching thelever.

r

F|F | sin θ θ

τ

The magnitude of the torque is also proportional to the length ofthe lever, and has a direction depending on which direction thelever pivots.

Example

A bicycle pedal is pushed by a foot with a 60 N force. The crankarm is 180 mm long. Find the magnitude of the torque about P.

qgl

a o K c) CD CD a rA o o\ z Fl o a

(h "j o rn o o v') oo C) gq TI o 0q I

CD o CD

CF a o oc (D CD o d w o\ I c) (D 0

15 o a (h {

5

F

Fff

iEW

sa--

oQ

! t

\r

= j

:l

-r F

"-

P

f 6

€

@3

"I'a'

{

-t-r

-o

\ ffi

tPU

J(n

t+

e

g-

A.

'C

D*

iI

A

| _

F

* i

*1. t

oG

id A

E

R.

.t4

-1-

5t-

-+:.

v

o

- B

=1

I ,^

1 T

>P

--

p-

;J

NJ P t- CD tl Lft

€-1.

CD r-l 0q :-

r\t

* hS

d P

dc

(p

nl

..

!4

hS

/r

r1

-r

lN

<ll

ii

5

^ll

;l :!.

,iS

$ 13

O

i* i"+

Fi

/^

f-

!^

FA

, V

t 'F

F.

Ll,

\ H

.

: re

='

xc

.(,

\Y

r<

| --

*.

l'E

I\t 13 8e tlFa

. +

r$

h'

)o

Fs

5

qr

o-i

*ao

oO

qf

,|\.

. i\.

r CD

hS

x .l{

. u)

&UJ

N)

\T

rr

t-

<f

fi

(\,v r

v-

Ff

J.

i+r

.i.l

vi

.4

\J

lI

*J ?

o.

\J/ _

n)

€*r

o

*l

Solution

|τ | = |r| |F| |sin θ| = (0.18 m)(60 N) sin(80◦) ≈ 10.6359 N m

Example

A bicycle pedal is pushed by a foot with a 60 N force. The crankarm is 180 mm long. Find the magnitude of the torque about P.

qgl

a o K c) CD CD a rA o o\ z Fl o a

(h "j o rn o o v') oo C) gq TI o 0q I

CD o CD

CF a o oc (D CD o d w o\ I c) (D 0

15 o a (h {

5

F

Fff

iEW

sa--

oQ

! t

\r

= j

:l

-r F

"-

P

f 6

€

@3

"I'a'

{

-t-r

-o

\ ffi

tPU

J(n

t+

e

g-

A.

'C

D*

iI

A

| _

F

* i

*1. t

oG

id A

E

R.

.t4

-1-

5t-

-+:.

v

o

- B

=1

I ,^

1 T

>P

--

p-

;J

NJ P t- CD tl Lft

€-1.

CD r-l 0q :-

r\t

* hS

d P

dc

(p

nl

..

!4

hS

/r

r1

-r

lN

<ll

ii

5

^ll

;l :!.

,iS

$ 13

O

i* i"+

Fi

/^

f-

!^

FA

, V

t 'F

F.

Ll,

\ H

.

: re

='

xc

.(,

\Y

r<

| --

*.

l'E

I\t 13 8e tlFa

. +

r$

h'

)o

Fs

5

qr

o-i

*ao

oO

qf

,|\.

. i\.

r CD

hS

x .l{

. u)

&UJ

N)

\T

rr

t-

<f

fi

(\,v r

v-

Ff

J.

i+r

.i.l

vi

.4

\J

lI

*J ?

o.

\J/ _

n)

€*r

o

*l

Solution

|τ | = |r| |F| |sin θ| = (0.18 m)(60 N) sin(80◦) ≈ 10.6359 N m

In General

DefinitionGiven vectors a and b in space, the cross product of a and b isthe vector

a× b = |a| |b| (sin θ) n,

where n is a vector perpendicular to a and b such that (a,b,n) isa right-handed set of three vectors.

Example

State whether the following position is meaningful. If not, explain.If so, is the expression a scalar or a vector?

1. a · (b× c)

2. a× (b · c)

3. a× (b× c)

4. (a · b)× c

5. (a · b)× (c · d)

6. (a× b) · (c× d)

Outline




Cross products of the standard basis vectors

Fill in the table:× i j ki

0 k − j

j

− k 0 i

k

j − i 0

I Is the cross product commutative?

No

i× j = k = −j× i

I Is the cross product associative?

No

i× (i× j) = i× k = −j

(i× i)× j = 0


Fill in the table:× i j ki 0 k − jj − k 0 ik j − i 0


No

i× j = k = −j× i


No

i× (i× j) = i× k = −j

(i× i)× j = 0




No

i× j = k = −j× i


No

i× (i× j) = i× k = −j

(i× i)× j = 0



I Is the cross product commutative? No

i× j = k = −j× i


No

i× (i× j) = i× k = −j

(i× i)× j = 0




i× j = k = −j× i


No

i× (i× j) = i× k = −j

(i× i)× j = 0




i× j = k = −j× i

I Is the cross product associative? No

i× (i× j) = i× k = −j

(i× i)× j = 0

Algebraic Properties of the Cross Product

If a, b, and c are vectors and c is a scalar, then

1. a× b = −b× a

2. (ca)× b = c(a× b) = a× (cb)

3. a× (b + c) = a× b + a× c

4. (a + b)× c = a× c + b× c

Cross product by components

QuestionIf

a = 〈a1, a2, a3〉= a1i + a2j + a3k

b = 〈b1, b2, b3〉= b1i + b2j + b3k

Find a× b.

Answer

a× b = (a2b3 − b2a3)i + (a3b1 − b3a1)j + (a1b2 − b1a2)k

= 〈a2b3 − b2a3, a3b1 − b3a1, a1b2 − b1a2〉

Cross product by components

QuestionIf

a = 〈a1, a2, a3〉= a1i + a2j + a3k

b = 〈b1, b2, b3〉= b1i + b2j + b3k

Find a× b.

Answer

a× b = (a2b3 − b2a3)i + (a3b1 − b3a1)j + (a1b2 − b1a2)k

= 〈a2b3 − b2a3, a3b1 − b3a1, a1b2 − b1a2〉

Determinant formula

This is only to help you remember, in case you’ve seendeterminants of 3× 3 matrices:∣∣∣∣∣∣

i j ka1 a2 a3

b1 b2 b3

∣∣∣∣∣∣ = i

∣∣∣∣a2 a3

b2 b3

∣∣∣∣− j

∣∣∣∣a1 a3

b1 b3

∣∣∣∣ + k

∣∣∣∣a1 a2

b1 b2

∣∣∣∣= (a2b3 − b2a3)i− (b3a1 − a3b1)j + (a1b2 − b1a2)k

= a× b

Procedure check

Example

Calculate a× b if

1. a = 〈1, 2, 0〉 and b = 〈0, 3, 1〉2. a = 3i + 2j + 4k and b = i− 2j− 3k

3. a = 〈t, t2, t3〉 and b = 〈1, 2t, 3t2〉

Solution

1. 〈2,−1, 3〉2. 2i + 13j− 8k

3.⟨t4,−2t3, t2

⟩

Procedure check

Example

Calculate a× b if

1. a = 〈1, 2, 0〉 and b = 〈0, 3, 1〉2. a = 3i + 2j + 4k and b = i− 2j− 3k

3. a = 〈t, t2, t3〉 and b = 〈1, 2t, 3t2〉

Solution

1. 〈2,−1, 3〉2. 2i + 13j− 8k

3.⟨t4,−2t3, t2

⟩

Outline




Area

The magnitude of the cross product a× b is the area of theparallelogram with sides a and b.

a

b |b| sin θ

Volume

To find the volume of a paralleliped with sides a, b, c:

ab

c

We getV = |a · (b× c)|

Volume

To find the volume of a paralleliped with sides a, b, c:

ab

c

We getV = |a · (b× c)|

More determinants

a · (b× c) = 〈a1, a2, a3〉 · 〈b2c3 − c2b3, b3c1 − c3b1, b1c2 − c1b2〉= a1(b2c3 − c2b3) + a2(b3c1 − c3b1) + a3(b1c2 − c1b2)

=

∣∣∣∣∣∣a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣

Example

Example

Find the volume of the parallelepiped determined by

a = 6i + 3j− k

b = j + 2k

c = 4i− 2j + 5k.

SolutionThe volume is∣∣∣∣∣∣∣∣∣∣∣∣6 3 −10 1 24 −2 5

∣∣∣∣∣∣∣∣∣∣∣∣ = 6(5 + 4)− 3(0− 8)− 1(−4) = 54 + 24 + 4 = 82

Example

Example

Find the volume of the parallelepiped determined by

a = 6i + 3j− k

b = j + 2k

c = 4i− 2j + 5k.

SolutionThe volume is∣∣∣∣∣∣∣∣∣∣∣∣6 3 −10 1 24 −2 5

∣∣∣∣∣∣∣∣∣∣∣∣ = 6(5 + 4)− 3(0− 8)− 1(−4) = 54 + 24 + 4 = 82

Cross product jokes

I What do you get when you cross a lion with a tiger?

I What do you get when you cross a lion with a mountainclimber?

I What do you get when you cross a mosquito with a fishmonger?

I What do you get when you cross an elephant with a banana?

I What do you get when you cross a mathematician with amovie star?

Cross product jokes






Cross product jokes






Cross product jokes






Cross product jokes






Technology

Lesson 3: The Cross Product