12.4Cross Product

Geometric description of the cross product of the vectors u and v

• u x v is perpendicular to u and v • The length of u x v is• The direction is given by the right hand side rule

sin u v u v

The cross product of two vectors is a vector!

curl them in the direction

of the second vector,

Your thumb will point in the

direction of the cross product

Right hand rule

Place your 4 fingers in the

direction of the first vector,

Algebraic description of the cross product of the vectors u and v

1 2 3 1 2 3

2 3 3 2 3 1 1 3 1 2 2 1

The of , , and , , is

, ,

u u u v v v

u v u v u v u v u v u v

cross product u v

u v

2 3 3 2 3 1 1 3 1 2 2 1 1 2 3

2 3 1 3 2 1 3 1 2 1 3 2 1 2 3 2 1 3

check: ( ) , , , ,

0

similary: ( ) 0

length......

u v u v u v u v u v u v u u u

u v u u v u u v u u v u u v u u v u

u v u

u v v

An easier way to remember the formula for the cross products is in terms of determinants:

2x2 determinant: a b

ad bcc d

1 2

4 6 23 4

3x3 determinants: An example

determinant = 2 72 30 36 15 8 40 59 19

sum of

forward

diagonal

products

sum of

backward

diagonal

products

Copy 1st 2 columns

1 6 2 1 6

3 1 3 3 1

4 5 2 4 5

1 6 2

3 1 3

4 5 2

1 2 3

1 2 3

u u u

v v v

i j k

u v 1 2 3 1 2

1 2 3 1 2

u u u u u

v v v v v

i j k i j

2 3 3 1 1 2 2 1 3 2 1 3 u v u v u v u v u v u v i j k k i j

2 3 3 2 1 3 3 1 1 2 2 1u v u v u v u v u v u v u v i j k

2 3 3 2 3 1 1 3 1 2 2 1, ,u v u v u v u v u v u v u v

Let 1, 2,1 and 3,1, 2 Find . u v u v

1 2 1

3 1 2

i j k

u v1 2 1 1 2

3 1 2 3 1

i j k i j

4 1 u v i 3 2 j 1 6 k

3,5,7 u v

Geometric Properties of the cross product:

Let and be nonzero vectors and let be the angle between and .u v u v

1. is orthogonal to both and .u v u v

2. sin u v u v

3. if and only if and are scalar

multiples of each other.

u v 0 u v

4. area of the parallelogram

determined by and .

u v

u v

u

u

vv

sinh u

12

5. area of the triangle having

and as adjacent sides.

u v

u v

u

v

Problem: Compute the area of the triangle two of whose sides are given by the vectors

1,0,2 and 1,3, 2 u v

1 0 2

1 3 2

i j k

u v1 0 2 1 0

1 3 2 1 3

i j k i j

(0 6) ( 2 2) (3 0) i j k

6 3 i k

6,0,3

1| 36 9 45 area = 45

2 u v | =

Algebraic Properties of the cross product:

Let , , and be vectors and let be a scalar.cu v w

1. u v v u

2. u v w u v u w

3. c c c u v u v u v

4. 0 v v 0 0

6. u v w u v w

7. u v w u w v u v w

5. (c ) v v 0

Volume of the parallelepiped determined by the vectors , , and .a b c

Area of the base b c

Height comp b ca cos a

Volume cos b c a

Volume a b c

is called the scalar triple product a b c

this stands

for absolute value

The vectors are in the same plane if the scalar triple product is 0.coplanar

Problem: Compute the volume of the parallelepiped spanned by the 3 vectors

Solution:

u vFrom slide 9:

Quicker:

( ) w u v 1 2 3 1 2 3

1 2 3

, ,u u u w w w

v v v

i j k

1 2 3 1 2

1 2 3 1 2

u u u u u

v v v v v

i j k i j

1 2 3 , ,w w w = Triple scalar product1 2 3

1 2 3

1 2 3

w w w

u u u

v v v

=

Volume = 6

(take absolute value)

( ) w u v

0 6 6

6,0,3

6,0,3

1,3, 4

6 12 6

1,0,2 , 1,3, 2 and 1,3, 4 u v w

1 3 4

1 0 2

1 3 2

12 6 0 6

In physics, the cross product is used to measure .torque

Q

Consider a force acting on a rigid body at a point given by a position vector .F r

The torque measures the tendency of the body to rotate about the origin point P

r F

r F sin r F is the angle between the force and position vectors

r

F

12.5

Lines and Planes

Recall how to describe lines in the plane (e.g. tangent lines to a graph):

y mx b is the slopem

is the interceptb y

Point slope formula: 0

0

y ym

x x

Two point formula:0 1 0

0 1 0

y y y y

x x x x

0 0( , ) is on the linex y

0 0 1 1( , ) and ( , )

are on the line

x y x y

Equations of Lines and Planes

In order to find the equation of a line, we need :

0 0 0 0A a point on the line , ,P x y z

B a direction vector for the line , ,a b cv

x

y

z

0 0 0 0 , ,P x y z

v0r

r

, ,P x y z0P P t v

0 t r r v

, ,x y zr

0 0 0 0, ,x y zr

0 , ,P P t t a b c v

L

vector equation of line L

0

0

Here is the vector from the origin

to a point P on the linespecific

r

is the vector from the origin to

a point P ( , , ) on the linegeneral x y z

r

is a vector which is to a vector

that lies on the line

is unique: 2 , or will also do

parallel

not

v

v v v

0 0 0, , , , , ,x y z x y z t a b c 0 t r r v or

0 0 0, , x x at y y bt z z ct

equating components we get the parametric equations of the line L

0 0 0x x y y z z

a b c

symmetric equations of the line LSolving for t we get the

vector equation of the line L

0 0 0 0A a point on the line , ,P x y z

B a direction vector for the line , ,a b cv

0 1Find the parametric equations of the line containing P 5,1,3 and P 3, 2,4 .

Problem:

0choose P 5,1,3 0

(could also choose

P 3, 2,4 )

0 0 0, , , , , ,x y z x y z t a b c

0 1 1 0 3 5, 2 1,4 3 2, 3,1P P P P v

or 2,3, 1 v

5,1,3 2,3, 1t

The line is: 5 2 , 1 3 , 3x t y t z t

Two lines in 3 space can interact in 3 ways:

A) Parallel Lines -

their direction vectors are scalar

multiples of each other

B) Intersecting Lines -

there is a specific t and s, so that the

lines share the same point.

C) Skew Lines -

their direction vectors are not parallel and

there is no values of t and s that make the

lines share the same point.

1 2Determine whether the lines and are parallel, skew

or intersecting. If they intersect, find the point of intersection.

L L

3 8 2t s

5 3 6 4t s

Check to make sure that the

values are equal for this and .

z

t s

1 4 5t s

11 4 5 2

3 3 check 4,2,3

1 : 3 , 5 3 , 1 4L x t y t z t 2 : 8 2 , 6 4 , 5 L x s y s z s

Set the x coordinate equal to each other: , or 2 5s t

Set the y coordinate equal to each other: , or 4 3 11s t

We get a system of equations: 2 5 or

4 3 11

s t

s t

4 2 10

4 3 11

s t

s t

1t 2s

1Find the point of intersection using :L

3 1

5 3 1

1 4 1

x

y

z

Problem:

In order to find the equation of a plane, we need :

0 0 0 0A a point on the plane , ,P x y z

B a vector that is orthogonal to the plane , ,a b cn

Planes

this vector is called

the

to the plane

normal vector

0 0 n r r

0 n r n rvector equation of the plane

, ,a b cn

0

0 0 0

0

0a x x b y y c z z

n r r

scalar equation of the plane

linear equation of the plane

0 0 0 orax by cz ax by cz

ax by cz d

0 0 0 0 0, ,OP x y z r

, ,OP x y z r

0 0 0 0 0- , ,P P x x y y z z r r

1 2Determine the equation of the plane that contains the lines and .L L

In order to find the equation of a plane, we need :

A a point on the plane

1 3 4

2 4 1

i j k

u v 4 6 k

1 : 3 , 5 3 , 1 4L x t y t z t 2 : 8 2 , 6 4 , 5 L x s y s z s

Problem:

0can choose e.g. (3,5, 1)P

B a vector that is orthogonal to the plane , ,a b cn

1 2We have two vectors in the plane: from : 1,3, 4 and from : 2, 4,1L L u v

1 8 j 3 16 i 13 7 2 i j k is normal

13,7,2n 13( 3) 7( 5) 2( 1) 0x y z 13 7 2 72x y z or