UNIT 2 - APPLICATIONS OF VECTORS

Date Lesson § TOPIC Homework

Feb.19 2.1

(11) 7.1

Vectors as Forces Pg. 362 # 2, 5a, 6, 8, 10 – 13, 16, 17

Feb.21 2.2

(12) 7.2

Velocity as Vectors Pg. 369 # 2,3, 4, 6, 7, 9, 11, 12

Feb.22 2.3

(13)

7.3/

7.4

The Dot Product Pg. 377 # 6ace, 7bdf, 9, 11, 14

Pg. 385 # 2, 3, 4, 6bd, 7, 9b, 10, 14

Feb.25 2.4

(14) 7.6

The Cross Product Pg. 407 # 3, 4ace, 5, 7, 8a, 9b, 11

Feb.26 2.5

(15) 7.5

Scalar and Vector Projections

QUIZ (2.1 - 2.3)

Pg. 398 # 6, 7ac, 8a, 11, 13

Feb.27 2.6

(16) 7.7

Applications of The Dot and Cross Product Pg. 414 # 1, 2,3, 5, 6, 8

Feb.28 2.7

(17)

Review for Unit 2 Test Pg. 418 # 1, 3, 4, 6, 7. 9 – 12, 15, 17,

20, 23, 30, 31, 34

Mar. 4 2.8

(18)

TEST- UNIT 2

MCV 4U Lesson 2.1 Vectors as Forces

A force can be described as a push or a pull on an object. It is measured in Newtons (N).

To describe a force it is necessary to state:

1. its direction

2. the point at which it is applied

3. its magnitude

The diagram shows 40 N and 30 N forces acting in opposite directions at point P. The combined effect of

these two forces as a 10 N force to the left. This single force that has the same combined effect as two

forces combined is called the resultant. The equilibrant is the opposite force that would exactly

counterbalance the resultant force. In this case the equilibrant force would be a 10 N force to the right.

.P40 N 30 N

Ex. 1 Two horses are pulling a load. The chains between them are at an angle of 45° to each other. One

horse pulls with a force of 260 N, the other with a force of 320 N.

a) What is the magnitude of the resultant force on the load? In what direction is this force?

b) What is the magnitude of the equilibrant force on the load? In what direction is this force?

Ex. 2 A sign is suspended in the lobby of GHS by two wires making angles with the ceiling of 50° and 30°.

If the sign weighs 2 kg, find the tensions (forces) acting in the wires. Provide an accurate vector

diagram.

N.B. The magnitude of a force is measured in Newtons (N). At the earth's surface,

acceleration due to gravity is approximately 9.8 m/s2.

Force = Mass acceleration due to gravity (where mass is measured in kg)

When calculating the force we use the constant 9.8.

Ex. 3 A lawn mower is pushed with a force of 90 N directed along the handle, which makes an angle of 36°

with the ground.

a) Determine the horizontal and vertical components of the force on the mower.

b) Describe the physical consequence of each component of the pushing force.

Ex. Find the resultant of the vectors below.

Pg. 362 # 2, 5a, 6, 8, 10 – 13, 15 - 17

MCV 4U Lesson 2.2 Velocity as a Vector

Since velocity has direction it can be represented by a vector.

When a plane flies, its velocity relative to the earth (ground velocity) is the resultant of (1) the plane's

velocity through the air (air velocity) and (2) the velocity of the wind.

Relative Velocity

The velocity of an object is its velocity relative to the frame of reference of some observer

in a given situation.

80 km/h

ie: Bus

100 km/h

WEST Car EAST

90 km/h

Truck

To someone in the bus the car is passing at 20 km/h. the velocity of the car relative to the bus

is 20 km/h east. The velocity of the truck relative to the car is 190 km/h west.

Thus we have the principle: Velocity of A relative to B = Velocity of A – velocity of B

Ex. 1 A plane is steering at N60°E at an airspeed of 450 km/h. The wind

is from the N50°W at 75km/h.

Determine the planes ground velocity.

g a w

Ex. 2 A canoeist who can paddle at 4 km/h in still water wants to cross a 500 m wide river that has a current

of 1 km/h.

a) If he steers the canoe in a direction perpendicular to the current,

determine theresultant velocity and the point on the opposite bank where he lands.

b) If he had wished to travel straight across the river, determine the direction in which he must head and

the time it takes to cross the river.

Pg. 369 # 2,3, 4, 6, 7, 9, 11, 12

MCV 4U Lesson 2.3 The Dot Product

Ex. 1 Find the angle between vectors u (3, 1) and v (2, 5).

(2, 5)

w

v

(3, 1) u

We can define the dot product of two algebraic vectors u and v as u v u v cos , where is the angle

between u and v and u v u1v1 u2v2 when in component form. ie: u (u1, u2 ) and v (v1,v2 ) .

Similarly, if u (u1, u2, u3) and v (v1,v2,v3) then u v u1v1 u2v2 u3v3

The dot product of two vectors is a scalar quantity, so it is sometimes called the scalar product.

Ex. 2 Find u v for each of the following.

a) u (1,3), v (2,5) b) u (4,2,3), v (2,3,1)

Pg. 377 # 6ace, 7bdf, 9, 11, 14

Pg. 385 # 2, 3, 4, 6bd, 7, 9b, 10, 14

If we want to find the angle between two vectors, we can use

cos u v

u v .

Ex. 3 Find between the following pair of vectors.

a) u (2,5), v (3,6) b) u (1,1,2) v (1,2,1)

Ex. 4 Find u v in each of the following cases.

a) u 6, v 8, 60 b)

u 3, v 12,

6 c)

u 2, v 9, 20

OTHER THINGS TO CONSIDER:

Properties of the Dot Product:

1. Associative: a(u v ) (au )v u (av )

2. Distributive: wuvuwvu

)(

3. u u u

2, since u u u u cos0 u u (1) u

2

Also, consider the dot product of unit vectors.

i i 1 j j 1 k k 1

i j j i 0 i k k i 0 j k k j 0

Examples:

3. If u (1,2,3) 2. If a (1,2) b (3,4) c (5,6)

2

uuu

Distributive Property:

MCV 4U Lesson 2.4 The Cross Product

The cross product is a vector quantity unlike the dot product, which is a scalar quantity. As a result, the cross

product is also known as the vector product.

1. Place vectors tail to tail.

2. Place right hand along first vector u with fingers

extended in the direction of the vector.

3. Curl the fingers in the direction of the

angle that v makes with u

4. Thumb gives direction of u v .

The magnitude of the cross product of two vectors u and v is

u v u v sin .

For non-zero vectors u and v , u and v are collinear

iff 0vu

The properties of the cross product are similar to those of the dot

product.

1. u (v w ) u v u w

2. (u v ) w u w v w

3. ku v k(u v ) u (kv )

4. a a 0

The cross product in component form, where u = ),,( 321 uuu and v = ),,( 321 vvv is given by:

u v (u2v3 u3v2 ,u3v1 u1v3,u1v2 u2v1)

For the unit vectors i , j , and k , we have the following

results.

jkijikkk

ijkikjjj

kijkjiii

0

0

0

NOTE: u v v u

For vectors u and v , we define the cross product as:

u v u v sin n where us the angle between u and v , 0 180 .

n is a unit vector perpendicular to both u and v such that u , v and n form a right handed

system as indicated below.

If u and v are two non-collinear vectors in three-dimensional space, then every vector perpendicular

to both u and v is in the form k(u v ) , k R.

u v

Ex. 1 For the following pairs of vectors, state whether u v is directed into or out of the page.

u

v u

v v

u

Ex. 2 Find a vector which is perpendicular to both u = (1, 2, 3) and v = (3, 4, 5).

Ex. 3 u = 5 and

v = 11 and the angle between them is 85 . Determine

u v .

Ex. 4 Find the cross product of kjia

342 and b 7i 3 j k .

Pg. 407 # 3, 4ace, 5, 7, 8a, 9b, 11

MCV 4U Lesson 2.5 Scalar and Vector Projections

I. Scalar Projections:

For vectors u and v , v 0 , proj v u u cos

and 0 180 .

u

scalar proj(u onto v ) v

ie:

cos adj

hyp We know that

u v u v cos .

cos Projection

u We can rewrite this formula as

u v u cos v

Projection = u cos Solving for

u cos gives

u cos =

u v

v .

So, the Scalar Projection of u onto v =

u v

v

and the Scalar Projection of v onto u =

u v

u

In general,

u v

v ≠

u v

u

II. Vector Projections:

A projection is formed by dropping a perpendicular from each point in an object onto a line or plane. A

shadow is a projection if the light rays forming the shadow meet a line or plane at right angles. A

projection is a vector. Used in physics, computer programming and astronomy to determine the lengths of

"shadows" projected onto a surface. The vector we obtain by projecting u perpendicularly onto the line

through v is called the vector projection of u onto v and is denoted proj(u onto v ) or proj v u .

For vectors u and v , v 0 , proj v u u cos(v )

where v is a unit vector in the direction of v and

0 180

u

Vector proj(u onto v ) v

If the vectors are not tail to tail, drop perpendiculars from the tail and the tip of u to meet v at O and P

and OP is the projection of u onto v .

proj v u u cos or we have:

Vector )( vontouproj

= )]ˆ)](([ vvectorvontouprojScalar

= vv

vvu

)(

Magnitude

vector proj(u onto v ) u v v

v 2

u v

v =

(u v )v

v 2

=

u v

v v

v

III. Direction Cosines

Any vector u

in space can be written as an ordered triple where u OP (a,b,c) and where

a, b, and c are its x-, y-, and z-components.

The vector u in space can also be written using the vectors kandji

,ˆ,ˆ , where kandji

,ˆ,ˆ are unit vectors along

the x-, y-, and z-axes, respectively.

i = (1, 0, 0) j = (0, 1, 0) k

= (0, 0, 1)

u OP = a i + b j + ck

Its magnitude is given by u 222 cba

To plot a vector in space, you move a units along the x-axis, then b units parallel to the y-axis, and then c units

parallel to the z-axis. Drawing a rectangular prism (box) is often helpful when doing this.

The direction angles of a vector (a, b, c) are the angles , , and that the vectors make with the positive

x-, y-, and z-axes, respectively, where 0 ,, 180

The direction cosines of a vector u are the cosines of the direction angles , , and where

cos =

a

u , cos =

b

u , cos =

c

u ., where

u 222 cba .

Thus the direction cosines are the components of a unit vector.

cos2 + cos2

+ cos2 = 1

O P

Ex. 1 Find the vector proj(b onto a ) given:

a) a (4,2) b (3,1) b) a (1,3,6) b (2,1,3)

Ex. 2 Find the scalar projections of a onto b and b onto a if a = (2, 1, –9) and b = (–3, 0, 4).

Ex 3. Find the direction cosines and the direction angles of the vector u

- (2, 3, -4).

Pg. 398 # 6, 7ac, 8a, 11, 13

MCV 4U Lesson 2.6 Applications of the Dot Product and Cross Product

I. Forces

A: Work

B: Torque (Moment of force) - is a measure of how much a force acting on an object causes that

object to rotate.

The turning effect of a force is called torque and is defined by the cross product.

T r F

r F sin(n )

where F is the applied force, r is the vector determined by the lever arm acting from the axis of rotation,

is the angle between the force and the lever arm and n is a unit vector perpendicular to both r and F . Torque is a vector quantity and is measured in Newton-metres (N–m).

While it is dimensionally correct to express joules as Newton-metres or N·m, such use is discouraged by the SI

authority to avoid confusion with torque. Torque and energy are fundamentally different physical quantities. For

example, adding 1 N·m of torque to 1 N·m of energy gives a dimensionally consistent result of 2 N·m, but this

quantity is physically meaningless.

One joule in everyday life is approximately:

the energy required to lift a small apple one meter straight up.

the energy released when that same apple falls one meter to the ground.

the energy released as heat by a person at rest, every hundredth of a second.

one hundredth of the energy a person can receive by drinking a drop of beer.

the kinetic energy of an adult human moving at a speed of about a handspan every second.

the kinetic energy of a tennis ball moving at 23 km/h (14 mph).

III. Area and Volume of a parallelogram or triangle formed by vectors a and b .

Area of a Parallelogram Area of a Triangle =

1

2(Area of a Parallelogram)

a sinah

a

b b

Area = base x height

b a sin

b a =

1

2b a

Volume of a Parallelepiped

b c

a

A parallelepiped is a six faced solid where the opposite faces are congruent parallelograms.

Let a , b , and c be the three sides as shown.

heightba

heightbaseofareaVolume

Now the height of the figure is the distance between the upper and lower faces.

ba

will give us a vector perpendicular to the plane containing a and b .

The height will be the magnitude of the projection of c onto ( ba

).

)(

)(

)]([

bac

ba

bacba

baontocprojba

heightbaVolume

This is a triple scalar product. If the triple scalar

product of three vectors is zero the vectors are

coplanar. Therefore the parallelepiped is flat and

has a zero volume.

Ex. 1 A crate on a ramp is hauled 10 m up the ramp under a constant force of 35 N applied at an angle

of 25° to the ramp. Find the work done.

Ex. 2 A 75 N force is applied to the end of a 0.3 m long wrench and makes an angle of 70° with the handle.

What is the torque on the bolt at the other end of the wrench?

Pg. 414 # 1, 2,3, 5, 6, 8