MCV4U1

UNIT 7: APPLICATIONS of VECTORS

Date Day Topic Homework

1 7.1 Vectors as Forces Pg. 362 # 2, 3, 5, 6, 8, 10, 11, 13, 14,16- 17

2 7.2 Velocity Pg. 369 # 1-8

3 7.3 The Dot Product of 2 Geometric Vectors

Pg. 377 # 1, 2, 5, 6-7 (odd), 8, 9, 11, 12

4 7.4 The Dot Product of Algebraic Vectors Pg. 385 #1, 2, 4, 6-8, 10, 11

5 7.5 Scalar and Vector Projections Pg. 398 # 1-3, 6-8, 12, 14

6 7.6 The Cross Product of 2 Vectors

Pg 407 #4, 5, 8, 9, 11Pg 414 #2, 3(b,c,d), 5, 7, 8, 10

7 7.7 The application of Cross and Dot Prodcut Pg 414 #2, 3(b,c,d), 5, 7, 8, 10

8 Chapter ReviewPg. 388 #2, 4, 5, 7, 8, 11, 14 Pg. 418 # 1-7, 9-18, 20, 21, 23, 26, 37

9 Chapter Test

Learning GoalsIn this unit, we will learn:

Vectors as Forces and Velocities The Dot and Cross Products Applications of Dot Product and Cross Product

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Day1: 7.1 Vectors as ForcesA force causes an object to undergo acceleration. For example, forces push you back in your seat when the car you are in accelerates.

The magnitude of force is measured in Newtons (N). At the earth’s surface, gravity causes objects to accelerate at a rate of approximately 9.8 m/s2.

Forces are vectors. The single force F

, the resultant force, that has the same effect as

all the forces acting together can be found by vector addition. 21 FFF

When an object is in a state of equilibrium (a state of rest or a state of uniform motion), the net force is zero or R=0. For example, steady speed. The equilibrant, E

, is the opposite force of the resultant force F

, it is the force that

would counterbalance the resultant force.

FEandFE

Draw an equilibrant for the following system of forces.

Given three forces on a plane, a state of equilibrium is maintained if a triangle can be formed with the three forces. This can only be done if the triangle inequality holds true that the sum of any two sides must be greater than or equal to the third side.

Ex 1: Which of the following sets of forces acting on an object could produce equilibrium?

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a) 13N, 27N, 14N b) 12N, 26N, 13N

Ex 2: Find the magnitude and direction of the resultant and equilibrant of a system of forces of 2000N and 1000N acting at an angle of 60o to each other.

Algebraic Resultant force:To resolve a vector means taking a single force and decomposing it into two components. A vector can be resolved into its corresponding horizontal and vertical components by creating a right triangle with the given vector. The magnitudes of the vertical and horizontal components can be found using primary trigonometric ratios and a given angle.

Ex 3: In order to keep a 250 kg crate from sliding down a ramp inclined at 25 , the force that acts parallel to the ramp must have a magnitude of at least how many Newtons?

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Ex 4: An object is being towed by two ropes. The direction of forces of the ropes are WN 20 and EN 30 . If the resultant force is 1000N due north, find the magnitude of the

tensions of each rope.

Ex 5: A box with a force of 100N is hanging from two chains attached to an overhead beam at angles of 50o and 30o to the horizontal. Determine the tensions in the chains.

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Day 2: 7.2 VelocityVelocity is a vector quantity, as the direction of motion as well as magnitude is important. Speed is the magnitude of velocity.

In velocity applications, the resultant (ground speed) is the speed of a plane or boat relative to a person on the ground which includes the effect of wind or current on the air/water speed.

A key step to solving a problem is to find an angle in the triangle formed by the vectors whose directions are given. It is helpful to draw small axes at the tail or head of the vectors when drawing diagrams.

Ex 1: A plane travels due north at an airspeed of 900 km/h. It encounters a wind blowing at 80 km/h from the west. What is the resultant velocity of the plane?

Ex 2: A man can swim 3.5 km/h in still water. Find at what angle to the bank he must head if he wishes to swim directly across a river flowing at a speed of 2 km/h.

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Ex 3: A plane is steering EN 45 at an air speed of 525 km/h. The wind is from WN 60 at 98 km/h. Find the ground speed and course of the plane.

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Ex 4: A boat heads 15 west of north with a water speed of 3 m/s. Determine its velocity relative to the ground when there is a 2 m/s current from 40 east of north.

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Day 3: 7.3 The Dot Product of 2 Geometric Vectors

The Dot product of two geometric vectors

The dot product of two vectors is a scalar (also called the scalar product).

Properties of the Dot Product:

Ex1:

Calculate vu if 30,10 vu

and 32 .

Ex2: Calculate the angle between vandu given 5.175,7 vuandvu

.

1.

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1. Two nonzero vectors are perpendicular .if a⊥b then a∙ b=02. Commutative property: a ∙b=b ∙a .3. Associative property with a scalar k: (k a ) ∙b=a ∙ (k b )=k (a∙ b) .4. Distributive property: a ∙ ( b+c)=a ∙b+a∙ c 5. Magnitudes property: a ∙a=¿a∨¿2¿

Ex3: Expand and simplify:

a) vuuk b) baba

c) baba 6543

Application of the Dot Product:

Work is done when a force acting on an object causes a displacement of an object from one position to another. Work is a scalar quantity measured in joules (J).

Work is defined as the dot product: cossFWorsFW where F

is the force

acting on an object (N), s

is the displacement caused by the force (m) and is the angle between sandF

.

Ex5: A 25 kg box is located 8 m up a ramp inclined at an angle of 18 to the horizontal. Determine the work done by the force of gravity as the box slides to the bottom of the ramp.

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Day 4: 7.4 The Dot Product of Algebraic Vectors

Recall: Algebraic vectors in component form 321 ,, aaaa

in R3.

Ex 1: Given 5,4,11,2,3 vandu , find vu

.

Using Dot product to find the angle between 2 vectors

Ex2: Determine the angle between 4,3a and )3,2( b

Ex3: Determine the value of k so that 4,5,2 kvandu

are perpendicular (orthogonal).

Ex4: A parallelogram has its sides determined by a=(2 ,3 )∧b=(3 ,1) . Determine the angle between the diagonals of the parallelogram formed by these vectors.

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In R2, if .

In R3, if .

if cosθ=0⇔θ=90 °

cosθ>0⇔0°<θ<90 °

cosθ<0⇔90°<θ<180°

Ex5: Given vectors 2,4,13,1,2 banda

, determine the components of a vector perpendicular to each of these vectors.

Ex6: Suppose that a force vector given by F⃑=(−2 ,−6 ,3) moves an object from point A(3, -1, 2) to point B(1, 4, 4). Calculate the work done on the object.

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Day 5: 7.5 Scalar and Vector Projections

Projections are formed by dropping a perpendicular from the head of one vector to another vector, or an extension of another vector. (Can be thought of as a shadow)

Given two vectors a∧b, think of the projection of aon bas the shadow that acasts on b

The direction of the projection of aon bdepends on the angle θ between a∧b

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Example1: Find the scalar and vector projections of )2,3,1()5,3,1( vonu .

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The angles that a vector PO

makes with each positive axis are called direction cosines.

is the angle PO

makes with the positive x-axis. is the angle PO

makes with the positive y-axis.

is the angle PO

makes with the positive z-axis.

Example2: Determine the direction angles for )5,3,1(u .

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x

y

z

Day 6: 7.6 The Cross Product of 2 Vectors

The cross product (vector product) is defined only in R3

since the cross product of two vectors is a vector

that is perpendicular to both banda

.

The cross product of the vectors banda

is the vector whose direction is perpendicular to banda

, such that

baandba

,, form a right-handed system

The vector a× b is the opposite of b× a and points in the opposite direction.

The magnitude of the cross product of two vectors banda

is:

where is the angle between banda

, 1800 .

Algebraically:

Given: 122131132332321321 ,,,,,,, babababababababbbbandaaaa

bcaddcba

wherebbaa

bbaa

bbaa

ba

21

21

13

13

32

32 ,,

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Another view of Cross Product:

Finding a vector perpendicular to two vectors:If banda

are two non-collinear vectors in R3, then every vector perpendicular to both

banda

is in the form Rkbak ,

.

Properties of the Cross Product:

Examples:1. Find a vector perpendicular to both (1, 2, 3) and (-1, 0, 4).

2. If 58 vandu and the angle between vandu

is 30 , find vu .

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Let candba , be vectors in 3R a× b=−b× a (not commutative)

a× (b+c )=a× b+a× c (distributive law)

(k a )× b=a× (k b )=k (a× b)

Day 7: 7.7 Applications of Cross and Dot Product

Area of a Parallelogram:

bhA

Ex1: Determine the area of the parallelogram determined by the vectors

p= (−1,5 ,6 )∧q=(2 ,3 ,−1)

Ex2: Find the Area of the triangle ABC whose vertices are A(1, 3, 5), B(-2, -3, 4) and C(0, 3, -1).

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Torque is a vector quantity measured in Newton-metres (N-m) (or in joules (J)). Force causes an object to turn causing an angular rather than a linear displacement. The torque is caused by a force defined as the cross product.

FrTorque

or sinFrTorque

where F

is the applied force, F

in Newtons and r

is the vector acting on the axis of rotation, r

in meters, is the angle between F

and r

.

Ex3: A force of 90N is applied to a wrench 15 cm long at 70 to the handle. Determine the torque.

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Review7.1 – Vectors as Forces

Resultant Force: single force used to represent the combined effect of all the forces;F=F1+F2

Equilibrant force: single force that opposes the resultant force −F=−¿ Resolution: decomposing a force into its horizontal and vertical components If a+b+c=0 ,then a ,b∧c are in a state of equilibrium

7.2 – Velocity Velocity stated relative to a frame of reference ex: the ground, the river bank Planes and wind with compass directions, boats crossing rivers with currents. Can resolve into horizontal and vertical components, or can use cosine or sine law. V r=V 1+V 2

7.3 – Dot Product of Two Geometric Vectors Geometric vectors: do not have an associated coordinate system a ∙b=¿a∨¿b∨cosθ if a⊥b then a∙ b=0 if 0<θ<90 ° then a ∙b>0 if 90 °<θ<180 ° then a ∙b<0 Commutative property: a ∙b=b ∙a Distributive property: a ∙ ( b+c)=a ∙b+a∙ c Magnitudes property: a ∙a=¿a∨¿2¿ Associative property with a scalar k: (k a ) ∙b=a ∙ (k b )=k (a∙ b)

7.4 – Dot Product of Algebraic Vectors

a ∙b=a1b1+a2b2+a3b3

7.5 – Scalar and Vector Projections

Scalar projection of aon b is|a|cosθ= a ∙b¿ b∨¿¿

bon a is|b|cosθ= a ∙b¿ a∨¿¿

Vector projection of aon b is ¿

Projection angles: x axis: Cos α = a / √(a2+b2+c2) y axis: Cos β = b / √(a2+b2+c2) z axis: Cos γ = c / √(a2+b2+c2∙)

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7.6 – Cross Product of Two Vectors (only in R 3 ) Finding a vector that is perpendicular to each of the two given vectors a× b=(a¿¿2b3−a3b2 , a3b1−a1b3 , a1b2−a2b1)¿ a× b=−b× a a× (b+c )=a× b+a× c (k a )× b=a× (k b )=k (a× b)

7.7 – Applications of the Dot Product and Cross Product Work: W=¿F∨¿s∨cosθ∧W=F ∙ s Torque: T=¿ r∨¿F∨sinθ∧T=¿ r xF∨¿ (include vector signs)

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